UNIT 1
REAL NUMBERS AND FUNCTIONS
Structure
\
Introduction Objectives
Basic Properties of R Absolute Value Intervals on the Real Line Functions 1.5.1 Definition and Examples 1.5.2 Inverse Functions 1.5.3 Graphs o f Inverse Functions
New Functions from Old 1.6.1 Operations on Functions 1.6.2 Composite of Functions
Types of Functions 1.7.1 Even and Odd Functions 1.7.2 Monotone Functions 1.7.3 Periodic Functions
summary Solutions and Answers
1.1
INTRODUCTION
This is the first unit of the course on Calculus. We thought it would be a good idea to acquaint you with some basic results about the real number system and functions, before you actually start your study of Calculus. Perhaps, you are already familiar with these results. But, a quick look through the pages will help you in refieshing your memory, and you will be ready to tackle the course.

In the next three sections of this unit, we shall present some results about the real number system. You will find a number of examples of various types of functions in Sections 1.5 to 1.7. You should also study the graphs of these functions carefully, in order to be able to vistlalise given functions. In fact, try to draw a graph whenever you encounter a new function. We shall , systematically study the tracing of c w e s in Block 2 Unit 4.
Objectives After reading this unit you should be able to : "3
0
a
recall the basic properties of real numbers derive other properties with the help of the basic ones ichtify various types of bounded and unbounded intervals define a function and examine whether a given function is oneonelonto investigate whether a given function has an inverse or not deiine the scalar multiple, absolute value, sum, difference, product, quotien't of the given functions and determine wl$&er a given function is even odd, monotonic or periodic.
1.2
BASIC PROPERTIES OF R
In the next three sections, we are going to tell you about the set R of real numbers, which is allpervading in mathematics. The real number system is the foundation on which a large part of mathematics, including calculus, rests. Thus,before we actually start leaming calculus, it' is necessary to understand the structure of the real number system. You are already familiar with the operations of addition, subtraction, multiplication and division of real numbers, and also withGkqualities.Here we shall quickly recall some of their properties. We start with the operation of addition:
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A1 R is closed under addition. If x and y are real numbers, then x + y is a unique real number. A2 Addition is associative, x+(y+z)=(x+y)+zholdsforallx,y,zinR. A3 Zero exists. There is a real number 0 such that x+OO+x=xforallxinR. A4 Negatives exist. For each real number x, there exists a real number y (ccalled a negative or an additive inverse of x, and denoted by x) such that x + y = y + x = 0. A5 Addition is commutaPlve. x+y=y+xholdsforallx,yinR Similar to these properties of addition, we can also list some properties of the operation of multiplication:
Ml R is closed under multiplication. If x and y are real numbers, then x.y is a unique real number. M2 Multiplication is associative. x.(y.z) = (x.y).z holds for all x, y, z in R M3 Unit element exists. There exists a real number 1 such that x.1= l.x=xforeveryxinR. hZ4 Inverses exist. For each real number x other than 0, there exists a real number y (called a multiplicative inverse of x and denoted by xI, or by l/x) such that x.y y.x = 1

You may have come across a "field", in the course on Llnear Algebra
MS Multiplication is commutative. x.y = y.x holds for all x, y in R. The next property involves addition as well as multiplication. D Multiplication is distributive over addition. x.(y t z) = x.y + x.z holds for all, x, y, z in R Remark 1: The fact that the above eleven properlies are satisfied is often expressed by saying that the real numbers form afield with respect to the usual addition and multiplication operations. Remark 1 (a): Usually the operator '.' is dropped in expressions, e.g., x.y may be denoted as xy.
In addition to the above mentioned properties, we can also define an order relation on R with the help of which we can compare any two real numbers. We write x > y to mean that x is greater than y. The order relation '>' has the following properties: 0 1 Law of Trichotomyholds. For any two real numbers a, b, one and only one of the following holds: a>b,a=b,b>a. 02 '>' is transitive. I f a r b a n d b > c , t h e n a > c , v a , b , c ~R 03 Addition is monotone. Ifa,b,cinRaresuchthata>b,thena+c>b+c. 04 Multiplication is monotone in the following sense. If a, b, c in Rare such that a > b and c > 0, then ac > bc. Caution: a > b and c < 0 3 ac < bc. Remark 2: Any field together with a relation > satisfying 01 to 04 is called an ordered field. Thus R with the usual > is an example of an ordered field. Notations: We write x < y (and read x is less than y) to mean y > x. We write x I y (and read x is less than or equal to y) to mean either x < y or x = y. We write x 2 y (and read x is greater than or equal to y) if either x > y or x = y. number x is said to be positive or negative according as x > 0 or x < 0. If x 2 0, we say that x is nonnegative.
Now, you know that given any number x E R, we can always find a number y E R such that y l x. (In fact, there are infinitely many such real numbers y). Let us see what happens when we take any subset of R instead of a single real number x. Do you think that, given a set S G R, it is possible to find u E R such that u l x for all x E S ? Discuss the special case when S is empty. Before we try to answer this question, let us look at a definition.
Definition 1: Let S be a subset of R. An element u in lk is said to be an upper bound of S if u 1 x holds for every x in S. We say that S is bounded above, if there is an upper bound of S. Now we can reword our earlier questions as follows : Is it possible to find an upper bound for a given set ?
Now, each x E Z is negative. Or, in other words, x < 0 for all x E Z. It is easily seen that, in this case, we are able to find an upper bound, namely zero, for our set ZOn the other hand, if we consider the set of natural numbers, N = {I, 2,3 .......I, obviously we will not be able to find an upper bound. Thus N is not bounded above. You will, of course, realise that ifu is an upper bound for a set S then u + 1, u + 2, u + 3, ....., (in fact, u + r, where r is any positive number) are all upper bounds of S. For example, we have seen that 0 is an upper bound for Z. Check that 1,2,3,4,8, ..... are all upper bounds of Z. From among all the upper bounds of a set S, which is bounded above, we can choose an upper bound u such that u is less than or equal to every upper bound of S. It is easily seen that, if such a u exists, then it is unique. We call this u the least upper bound or the supremum of S. For example, consider the set
Now 2,3,3.5,4,4 + n are all upper bounds for this set. But you will see that 2 is less than any other upper bound. Hence 2 is the supremum or the least upper bound of T. You will agree that 1 is the 1.u.b. (least upper bound) of Z. Note that for both the sets T and Z, the 1.u.b. belongs to the set. This may not be true in general. Consider the set of all negative real numbers R = {x :x < 0).The 1.u.b. of this set is 0. But 0 c R.
r
Working on similar lines we can also define a lower bound for a given set S to be a real number v such that v I x for all x E S. We shall say that a set is bounded below, if we can find a lower bound for it. Further, from among all the lower bounds of a set S, which is bounded below, we can choose a lower bound v such that v is greater than or equal to every lower bound of S. It is easily seen that, if such a v exists, then it is unique. We call this v the greatest lower bound or the infimum of S. As in the case of l.u.b., remember that the g.1.b. of a set may or may not belong to the set. We shall say that a set S c R is bounded if it has both an upper bound and a lower bound. Based on this discussion you will be able to solve the following exercise.
E E 1) Give examples to illustrate the following : I
a) A set of real numbers having a lower bound, b) A set of real numbers without any lower bound, c) A set of real numbers whose g.1.b. does not belong to it,
Real Numbers and Functions
Elements of Differentiai Calculus
d) A bounded set of real numbers. Now we are ready to state an important property of R.
C
The order is complete
Every nmempty subset S of R that is bounded above, has a supremum. (We shall use this property in Unit 10). Many more properties are either restatements or consequences of these sixteen properties. Here is a list of some of them. 1
Zero is unique, i.e., Ifx+0'=xforallxinR,then0'=0.
2
Additive inverse is unique, i.e., For each x in R, there is a unique y in R such that x + y = y + x = 0.
3
Addition is cancellative, i.e., Ifx+y=x+z,theny=z. Unity is unique, i.e.,
4
Ifx.l'=xforallxinR,then1'=1. 5
Multiplicative inverse is unique, i.e., For each nonzero real number x, there is a unique y in R such that xy = yx = 1.
6
Multiplicationis cancellative, i.e, Ifxy=xzandx#O,theny=z.
Definitian 2: If x and y are any two real numbers, the result of subtraction of y from x is denoted by x  y and is defined as x + (y). Similarly, the division x + y (also denoted by xly) is defined as xy', provided y # 0. Now we are ready to list a few more properties. You are already aware of these. But let us quickly recall them.
(x+y)=(x)+(y)forallx,yinR. If xy = 0, then either x = 0 or y = 0. (x')I = x for all x t 0 in R. If x and y are non zero numbers such that x' = yI, then x = y. Ifa<bandc>O,thenac<bc. a is positive if and only if a is negative. Ifa<bandc<d,thena+c<b+d. If a > b and c < 0, then ac < bc. a2 is nonnegative for all a in R. If a and b are positive, then i) a 2 = b 2 w a = b . (The symbol w is read as 'if and only i f ) ii) a2> b2 w a > b iii) a2 < b2 w a < b
Ifb>0,thena2<b2wb<a<b. You are also familiar with the following subsets of R: 1)
The set N of natural numbers. Note that it is the smallest subset o f R possessing the following properties: i)
1â‚ŹN
2)
3)
ii) k ~ N + k + lE N The set Z of integers. It is the smallest subset of R possessing the following properties:
Real Numbers and Functions
i) Z 3 N ii) I f x , y ~ Z , t h e n x  ~ E Z . The set Q of rational nhmbers. We observe that it is the smallest subset of R possessing the following properties:
3
Q3Z
ii) If x, y E Q and y # 0, then xy'
E Q. You must have also studied the following properties of these sets. 1) k E N if and only if k is a positive integer, that is, k E Z and k > 0. 2) The operations of addition and multiplication on N satisfy Al, A2, A5, MI, M2, M3, M5 and D. They do not, however, satisfy A3, A4 and M4. 3) The operations on Q satisfy A1 to AS, MI to M5 and 01 to 04. Therefore Q is an ordered field. But C is not satisfied, that is, Q is not ordercomplete.
We list here some more properties of these sets which you will find useful in our study of calculus: 4) Archimedean Property: Ha and b are any real numbers and ifb > 0, then there is a positive integer n such that nb > a. 5 ) If a is any real number, there is a positive integer n such that n > a (Archimedean property applied to a and 1). 6)
7)
A real number s is the supremum of a set S c R if and only if the following conditions are satisfied. 3 sLxforallxinS. ii) For each E > 0, there is a y in S such that y > s  E. For example, consider the set A = {x E R : 8 5 x c 10). 10 is the supremum of this set. Now, if we are given any E, say, E = 0.01, we should be able to find some y E A such that y > 10  0.01 = 9.99. As you can see, y = 9.999 serves our purpose. Now 10.01 is also an upper bound for A. But 10.01 is not the supremum of A. For E =0.001, we cannot findany y E A such that y > 10.01 0.001 = 10.009. Every nonempty set of real numbers that is bounded below, has an infmum.
The exercise below can now be done easily.
E E2) I
a) Show that the set of positive real numbers is bounded below. What is its infimum? b) Write the characterisation of the infimum of a subset of R, which corresponds to 6) above. Give an example.
1.3
ABSOLUTE VALUE
In this section we shall define the absolute value of a real number. You will realise the importance of this simple concept as you study the later units.
E (epsilon) is a Greek letter used to denote small real numbers.
Definition 3 :If x is a real number, its absolute value, denoted by I x I (read as modulus of x, or mod x), is defined by the following rules:
Elements o f Differential Calculus
1x1 =
x, ifx 2 0 x, ifx c 0
For example, we get 151=5,)5(=5, ( 1.7(=1.7,(2(=2,(0(=0 i It is obvious that 1 x 1 is defined for all x E R. The following theorem gives some of the ,
important properties of I x 1.
'
;
Theorem 1: If x and y be any real numbers, then a)
(xl=max(x,x)
b) c)
Ixl=lxl (x(*=x2=(xI2
d)
Ix+yl5lxl+l~l
(the triangle inequality)
e) I ~ + Y I ~ ( I ~ I  I Y I ~ Proof: a) By the law of trichotomy (01) applied to the real numbers x and 0, exactly one of the following holds: i) x>O,ii)x=O,oriii)x~O. Let us consider these one by one. i) Ifx>O,then(x(=xandx>,x,sothat max {x, x) = x and hence I x 1 =max { x, x }

If x 0, then x = 0 =x, and thexefore, mas {x,x) =O.Also~x(=O,sothat~x~=max {x,x). iii) Ifx<O,then(x(=x,andx>x,sothat max {x,x) =x.Thus,again, (x(=max{x,x).
ii)
From this it follows that x b) c)
d)
j
IxI
=max {x,x) =max {x,x) =(XI. Ifx20,then)x)=x,sothat)x)~=~~. ~fx<O,then(xI=x,~othat(x)~=(x)~=x~. Thetefore, for all x G R, I x l2 = x2. Also 1 x l2 = 1 x 12, because I  x 1 = 1 x 1 by (b). Thus, we have 1 x l2 = x2 U e shall consider two different cases according as 1x(=rnax {(x),x)
I X + Y l = x + ~ ~ l x lI. + l ~ Letx+y<O.Then(x+y)>O, thatis, (x) + (y) > 0 and we can use the result of (i) forxandy.Now ( x + yI = I(x+y) I by(b).
Thus(x+y(=((x)+(y)(S(XI+(y(,by(i). =Ixl+ly I. Therefore,wegetIx+y 1 5 ( x I + J y1.
by (b).
Thuswefindthatforallx,y E R ( x + y( < ( x ( + (1.y e) By writing x = ( x  y ) + y and applying the triangle inequality to the numbers x  y and y, we have
I~I=I(~Y)+YI~I~YI+IY~~ sothat(x((y(<(xy(. Since (I) holds for all x and y in R, therefore, by interchanging x and y in (1) we have
...(1)
l ~ l  l x I ~ l ~  x l = l  ( x  ~=Ixyl. )l Sothat(1x1)y))S)x y ) . ...(2) From (1) and (2) we find that ( x /  1 y ( and its negative  ( ( x 1  I y 1) are both less than or at the mostequa1to)xy ).Therefore,max(1x11 y 1,(1x1I y 1)) I ( x  y l . But the left hand side of the above inequality is simply ( ( x (  1 y (
1. Therefore, we have
11x1lYll~lxYl
E
Now you should be able to prove some easy consequences of this theorem. The following exercise will also give you some practice in manipulating absolute values. This practice will come in handy when you study Unit 2. E 3) Prove the following : a) x = O e ( x l = c
i

b) c)
IxYl=ixl.lYl 1l/x~=l/\x~,
d) e)
lxYl~Ixl+lYl
f)
ifxz0
Ix+Y+z)~lxI+lYl+lzl IxYzl=lxl.IYl.lzl
e) and f ) can be ex{ended to any number of reals. Now if a E R and 6 > 0, then Ixa)<6*xa<6,and(xa)<6.
x  a < 6 , thismeans t h a t x < a + 6
!
(xa)<6,thismeansthata6<x. Thus,wegetthat/xa(<6*a6 < x < a + 6 . This means that the difference between x and a is not more than 6. In the next section, we shall see how the set { x : ( x  a / < 6) can be represented geometrically.
Real Numbers and Functions
Elements or Dilferential Calculus
1.4
INTERVALS ON THE REAL LINE
Before we define an interval let us see what is meant by a number line. The real numbers in the set R can be put into onetoone correspondence with the points on a straight line L. In other words, we shall associate a unique point on L to each real number and vice versa. Consider a straight line L [see Fig. 1 (a)]. Mark a point 0 on it. The point 0 divides the straight line into two parts. We shall use the part to the lefr of 0 for representing negative real numbers and the part to the right of 0 for representing positive real numbers. We choose a point A on L which is to the right of 0. We shall represent the number 0 by 0 and 1 by A. OA can now serve as a unit. To each positive real number x we can associate exactly one point P lying to the right of 0 on L, so that OP = 1 x 1 units ( = x units). A negative real number y will be represented by a point Q lying to the left of 0 on the straight line L, so that OQ = 1 y 1 = y y is negative). We thus find that to each real number we can associate a point on the units ( : line. Also, each point S on the line reprsents a unique real number z, such that ( z ( = OS. Further, z is positive if S is to the right of 0, and z is negative if S is to the left of 0. 
Distance is always nonnegative.
This representation of real numbers by points on a straight line is often very useful. Because of this onetoone correspondence between real numbers and thg points of a straight line, we often call a real number "a point of R . Similarly L is called a "number line". Note that the absolute value or the modulus of any number x is nothing but its distance from the point 0 on the number line. In the same way, 1 x  y 1 denotes the distance between the two numbers x and y [seeFig. I (b)].
b) Distance between x and y is I x  y (
Figure I : (a) Number line
*
Now let us consider the set of the real numbers which lie between two given real numbers a and b, where a I b. Actually, there will be four different sets satisfying this loose condition. These are : i)
]a,b[= { x : a < x < b )
ii)
[a,b]={x:a<x~b)
iii) ] a , b ] = { x : a < x < b )
iv) [ a , b [ = { x : a 5 x < b ) The representation of each of these sets is given alongside. Each of these sets is called an interval, and a and b are called the end points of the interval. The interval ]a, b[, in which the end points are not included, is called an open interval. Note that in this case we have drawn a hollow circle around a and b to indicate that they are not included in the graph. The set [a, b], contains both its end points and is called a closed interval. In the representation of this closed interval, we have put thick black dots at a and b to indicate that they are included in the set. The sets [a, b[ and ]a, b] are called halfopen (or halfclosed) intervals or semiopen (or semiclosed) intervals, as they contain only one end point. This fact is also indicated in their geometrical representation.
+
If a = b, ]a, a[ = ]a, a] = [a, a[ = and [a, a] = a. Each of these intervals is bounded above by b and bounded below by a. Can we represent the set I = {x : ( x  a ( < 6 ) on the number line? Yes, we can. We know that 1 x  a 1 can be thought of as the distance between x and a. This means I is the set of all numbers x, whose distance from a is less than 6. Thus,
is the open interval ]a  6, a + 6[. Similarly, I, = {x : 1 x  a 1 2 6) is the closed interval [a  6, a + 61. Sometimes we also come across sets like I, = {x : 0 < ( x  a I < 6). This means if x E 12, then the distance between x and a is less than 6, but is not zero. We can also say that the distance between x and a is less than 6, but x # a. Thus,
I, =]a6, a + & [ \ {a)



... ..... . .
Real Numbers and Functions
:T.
Apart from the four types of intervals listed above, there are a few more types. These are: la,
[ = {x : a < x)
(open right ray)
[a, =[ = {x : a 2 x}
(closed right ray)
] a , b[= {x : x < b)
(open left ray)
I, b] = {x : x I,= [ = R
(closed left ray)
b}
.

r.
*

+
a
a
j 4
 
b
(open interval)
4

*

As you can see easily, none of these sets are bounded. For instance, ]a, [ is bounded below, but is not bounded above, 1 , b] is bounded above, but is not bounded below. Note that and  = does not denote a real number, it merely indicates that an interval extends without
limits. We note further that if S is any interval (bounded or unbounded) and if c and d are two elements of S, then all numbers lying between c and d are also elements of S.
E E4) State whether the following are true or false. a) 06[1,81, C) 1 ~ [ 1 , 2 1
b)  1 ]=,2[ ~ d) 5e15,[
E E 5) Represent the intervals in E 4) geometrically.
1.5
FUNCTIONS
Now let us move over ro present some basic facts about functions which will help you refresh your knowledge. We shall look at various examples of functions and shall also define inverse functions. Let us start with the definition of a function.
1.5.1
I
Definition and Examples
Definition 4: If X and Y are two sets, a function f from X to Y, is a rule or a correspondence which connects every member of X to a unique member of Y. We write f: X + Y (read as "f is a function from X to Y) X is called the domain and Y is called the codomain off. We shall denote by f(x) that unique element of Y which is associated to x E X.
I
The following examples will help you in understanding this definition better.
1
Example 1: f : N + R, defined by f(x) = x. is a function since the rule f(x) = x associates a unique member (x) of R to every member x of N. The domain here is N and the codomain is R.
/
Example 2: The rule f(x) = xi2 does not define a function from N + Z as odd natural numbers like 1,3,5 ...... from N cannot be connected to any member of Z. .
Elements o f Differential Calculus
Example 3: Every natural number can be written as a product of some prime numbers. Consider the rule f ( ~=) a prime factor of x, which connects elements of N. Here since 6 = 2 x 3, f(6) has two values : f(6) = 2 and q6) = 3. This rule does not associate a unique number with 6 and hence does not give a function from N to N. Thus, you see, to describe a function completely we have to specify the following three things: a)
the domain
b)
the codomain, and
c)
the wle which associates a unique member of the codomain to each member of the domain.
The rule which defines a function need not always be in the form of a formula. But it should clearly specify (perhaps by actual listing) the correspondence between X and Y. If fi X +Y, then y = f(x) is called the image of x under for the fimage of x. The set of fimages of all members of X, i.e., {f(x) : x E X) is called the range off and is denoted by f(X). It is easy to see that f(X) c Y. Remark 3 a) Throughout this course we shall consider functions for each of which whose domain and codomain are both subsets of R. Such functions are often called real functions or realvalued functions of a real variable. We shall, however, simply use the word 'function' to mean a real function. b) The variable x used in describing a function is often called a dummy variable because it can be replaced by any other letter. Thus, for example, the rule f(x) = x, x E N can as well be written in the form f(t) = t, t E N or as f(u) = u, u E N. The variable x (or t or u) is also called an independent variable, and f(x) is dependent on this independent variable. Graph of a function: A convenient and useful method for studying a function is to study it through its graph. To draw the graph of a function f : X +Y, we choose a system of coordinate axes in the plane. For each x E X, the orderedpair (x, f(x)) determines a point in the plane (see Fig. 2). The set of all the points obtained by considering all possible values of x (remember that the domain o f f is X) is the graph of the function f. The role that the graph of a function plays in the study of the function will become clear as we proceed further. In the meantime let us consider same more examples of functions and their graphs.
Fig. 2 4
1
w
1) A constant function: The simplest example of a hnction is a constant function. A const'.t: function sends all the elements of the domain to just one element of the codomain.
'I
0;
X
For example, let f: R +R be defined by f(x) = 1. Alternatively, we may write f:x+ 1 ...V X E R
Fig. 3
The graph o f f is as shown in Fig. 3. It is the line y = 1.
In general, the graph of a constant function f : x + c is straight line which is parallel to the xaxis at a distance of 1 c ( units from it.
Real N u m b e r a n d Functior~h
2) The identity function: Another simple but important example of a funct~onis a function which sends every element of the domain to itself.
YA ,f
"
Let X be any nonempty set, and let f be the function on X defined by setting f(x) = x V x E X.
I
I
This function is known as the identity function on X and is denoted by ix.
/
/
0 /
.
,/'
The graph of i, the identity function on R, is shown in Fig. 4. It is the line y = x.
/
b
3) Absolute value Function: Another interesting function is the absolute value funct~on(or modulus function) w h ~ c hcan be defined by using the concept of the absolute value of a real number as: f(x) = I X I =
Fig. 4
x, i f x 2 O  x, if x c 0
6
f
b
The graph of this function is shown in Fig. 5. It consists of two rays, both starting at the orig~n andmaking angles 7d4 and 3x/4, respectively, with the positive direct~onof the xaxls.
E E 6) Given below are the graphs of four functions depending on the notion of absolute value. r
The functions are x +I x 1, x + ( x j + 1, x + / x + I 1, x + ( x 1 1, thoughnot necessarily in this order. (The domain in each case is R). Can you identify them?
Fig. 5
C
4) The Exponential Vunction: If a is a positive real number other than I , we call define a function f as:

f R 4 R f (x) 2
(a>O,a+l)
This function is known as the exponential function. A special case of this flmction, where a c , is often found useful. Fig. 6 shows the graph of the function f : R i R such tilai f(x)  e x . This function is also called the natural expo~lrntialfunction. Its range is the set R'of positive real numbers.


5) The Natural Logarithmic Function: This futictia~:is defined on thc set R ofpositive real numbers, with f:R'+4 R such that f(sj iil (xj. The range of this function is R . Its grnp!, i: shown in Fig. 7.
1 X

2
1
2
1
) '
3
Fig. 6
YA = 117s
6) The Greatest Integer Function: Take a real number x. Either it is an integer, say n (so that x = 11) or it is not an intzger. If if is not nil integer, we can find I by the Archimedea~iproperty of real numbers, i ~ nin~cgern. such that n < x <. n t I . Therefore. for each real number x \ve t;~!l find an integer n: such that n I x c: n + 1. Further. for a given real number x; we can f ~ n donly one such integer n. We say that n is the greatest integer not exceeding x, and denotc it by [x]. For example. [?I 3 and 13.51  3. (3.51 = 4.Let 11s consider the function defined on R by Yetrmg f ( x )  [XI.


.~ 1
it!,
(I
L
2
!
:i 4
5:ig. 7
b
3
I
Elements o f Differential Calculus
This function is called the greatest integer function. The graph of the function is as shown in Fig. 8. (It resembles the steps on an infinite staircase). Notice that the graph consists of infinitely many line segments of unit length, all parallel to the xaxis. 7)
Other Functions The following are some important classes of functions.
a)
Polynomial Functions fTx)= a,x" + a,xW'+  + a, where ao ,a,, ,a. are given real numbers (constants) and n is a positive integer.
b) Rational Functions f(x) = g (x)/k(x), where g (x) and k (x) are polynomial functions of degree n and m. This is defined for all real x, for which k (x) #O. Trigorlometric or Circular Functions flx) =sin x, f(x) = cos x, f(x) =tan x, f(x) = cos x, f(x) = sec x, f(x) = cosec x. (ex + e' ) (ex ) ,flx)=sinhx= .We shall d) Hyperbolic Funcions f(x) = coshx = 2 2 study these in detail in Unit 5. c) Fig. 8
''
1.5.2
Inverse Functions
In this subsection we shall see what is meant by the inverse of a function. But before talking about the inverse, let us look at some special categories of functions. These special types of functions will then lead us to the definition of the mverse of a function. Oneone and Onto Functions Consider the function h : x x x2, defined on the set R. Here h(2) = h (2) = 4. Thus 2 and 2 are distinct members of the domain R, but their himages are the same. (Can you find some more numbers whose himages are equal?) This may be expressed by saying that 3x, y such that ?r t y but h(x) = h(y). Now, consider txe fuilction g: x x 2x + 3 Here you will be able to see that if x, and x2 are two distinct real numbers, then g(x,) and g ( x l ) are also distinct. For,x,#x,
2xI#2x,
2x1+3#2x,+3
g(x,)#g(x,) We have considered two functions here. While one of them, namely g, sends distinct members of the domain to distinct members of the codomain, the other, namely h, does not always do so. We give a special name to functions like g above. Definition 5: A function E x x Y is said to be a oneone function (a (1  1) function or an injective function) if the images of distinct members of X are distinct members of Y. 3
3
Thus the function g above is oneone. whereas h is not oneone. Remark 4: he condition "the images of distinct members of X are distinct members of Y" in the above dt.inition can be replaced by either of the follow~ngequivalent conditions: a)
For evely pair of members x, y of X, x + y a f(x) t f(y)
b)
For every pair of members x, y of X, flx) = fly) 3 x = y.
We have observed earlier that for a func:ion EX x Y. f(X) c Y. This opens two possibilities: i)
f (X) = Y, or ii) f (X) $ Y, that is, f(X) is a proper subset of Y
The function h :x + x2 tf x E R falls in the second category. Since the square of any real number is always nonnegative, h (R) = R' u { O } , the set of nonnegative real numbers. Thus h (R),@ R. On the other hand, the function g : x + 2x + 3 belongs to the first category. Given any y E R (codomain) ifwe take x = (li2)y  312, we find that g(x) = y. This shows that every member of the codomain is a gimage of some member of the domain and thus. is in the range g(R). From this we get that g(R) = R. The following definition characterises this property of the function. Definition 6 A function f: X x Y is said to be an onto function (or a surjective function) if every member oEY is the image of some member of X. Iff is a fu~ctionfrom X onto Y. we often write: f : x 2, Y (or f :X + + Y). Thus, h is not an onto function, whereas g is an onto function. Functions which are both oneone and onto are of special importance in mathematics. Let us see what makes them special.
Consider a function f: X + Y which is both oneone and onto. Since f is an onto function, each y E Y is the image of some x E X. Also, since f is oneone, y cannot be the image of w o distinct members of X. Thus, we find that to each y E Y there corresponds a unique x E X such that f(x) = y. Consequently, f sets up a onetoone correspondence between the members of X and Y. It is this onetoone correspondence between members of X and Y which makes a oneone and onto function so special, as we shall soon see. Consider the function f : N + E defined f(x) = 2x, where E is the set of even natural numbers. We can see that f is oneone as well as onto. In fact, to each y E E there exists y/2 E N, such that f(y/2) = y. The correspondence y + yi2 defines a function, say g, from E to N such that g (Y)= ~ 1 2 . The function g so defined is called an inverse off. Since, to each y E E there corresponds, a unique x E N such that f(x) = y, only one such function g can be defined corresponding to a given function f. For this reason g is called the inverse off. As you will notice, the function g is also oneone and onto and therefore it will also have an inverse.You must have already guessed that the inverse of g is the function f. From this discussion we have the following : '"
Iff is oneone and onro function from X to Y, then there exists a unique function g : Y i X such that for each y E Y, g(y) = x w y = f(Xj. The function g so defined is called the inverse of f. Further, if g is the inverse off, then f is the inv'erse of g, and the two function f and g are said to be the inverses of each other. The inverse of a function f is usually denoted by fI . To find the inverse of a given function f, we proceed as follows: Solve the equation f(x) = y for x. The resulting expressioil for x (in terms of y) defines the Inverse function.
x"
xJ
Thus, iff (x) =  + 2, we solve  + 2 = y for x. 5 5 1
1
This gives us x = {5(y  2)) . Hence f' is the function defined by f (y) = {5(y  2))
1.5.3
Graphs of Inverse Functions
 
There is an interesting relation between the graphs of a pair of inverse functions because of which, if the graph of one of them is known, the graph of the other can be obtained easily.
,
r
Let f: X + Y be a oneone and onto function, and let g : Y +X be the inverse off. A point (p, q) lies on the graph o f f c q. = f(p) p = g (q) (q, p) lies on the graph of g. Now the points (p, q) and (q, p) are reflections of each other with respect to (w.r.t.) the line y x. Therefore, we can say that the graphs o f f and g are reflections of each other w.r.t. the line y=x. Therefore, it follow that, if the graph of one of the functions f and g is given, that of the other can be obtained by reflecting it w.r.t. the line y = x. As an illustration, the graphs of the fi~nctionsy = x3 and y = xl" are given in Fig. 9.

90you agret that these two functions are inverses of each other? If the sheet of paper on which the graphs have bezn drawn is folded along the line y = x, the two graphs will exactly coincide.
Fig. 9
Real Numbers and Functions
Elements o f Differential Calculus
E E 7) Compare the graphs of In x and ex given in Figs. 6 and 7 and verify that they are inverses of each other. If a given function is not oneone on its domain, we can choose a subset of the domain on which it is oneone, and then define its inverse function. For example, consider the function f : x + sin x. Since we know that sin (x + 2n) = sin x, obviously this hnction is not oneone on R. But if we restrict it Fo the interval [nf2, n/2], we find that it is oneone. Thus, g f ( x ) =sin x x E [n/2, n/2], then we can define
++
fI (x) = sit' (x) = y if sin y = x.
'
Sin~ilarly,we can define cos' and tan functions as inverse of cosine and tangent functions if we restrict the codomain to [0, n] and]x/2, x/2[, respectively.
E E 8) Which of the following functions are oneone? a) b) c) d)
f:R+Rdefinedbyf(x)=lx/ f : R + R defined by f (x) = 3x  1. f : R + R defined by f (x) = x2 f : R + R defined by f (x) = 1
E E9) Which of the follow~ngfunctions are onto? a) f:R+Rdefinedbyf(x)3x+7 b) f:R"+Rdefinedbyf(x)= Jj; c) f:R+Rdefinedbyf(x)=x2+1 d) f : X 4 R defined by f (x) = l/x where X stands for the set of nonzero real numbers.
E E 10) Show that the function f : X +X such that f (x)

x + l x  1
,where X is the set of all real
numbers except 1, is oneone and onto. Find its inverse.
E E 11)Give ofieexample of each of the follou~ng: a)
a oneot:c f ~ n c t i o nuhlch 1s not unto. h) onto funct~onwhich is not oneone. c) a fi~nctiontvhlch 1s ne~theroneone nor onto
Real Numbers and Functions
NEW FUNCTIONS FROM OLD
1.6
In this section we shall see how we can construct new functions from some given functions. This can be done operating upon the given f u n c t i o ~ sin a variety of ways. We give a few such ways here.
1.6.1
Operations on Functions
Scalar Multiple of a Function Consider the function f : x + 3x2 + 1
tf x E R. The functiong : x + 2 (3x2+ 1) +j x E R is such
that g (x) = 2f (x) tf x E R. We say that g = 2f, and that g is a scalar multiple o f f by 2. In the above example there is nothing special about the number 2. We could have taken any real number to construct a new function from f. Also, there is nothing special about the particular function that we have considered. We could as well have taken any other function. This suggests the following definition: Let f be a function with domain D and let k be any real number. The scalar nlultiple o f f by k is a function with domain D. It is denoted by kf and is defined by setting (kf) (x) = kf(x). Two special cases of the above definition are important. 1)
ii)
Given any function f, if k = 0, the function kf turns out to be the zero function. That is, 0.f = 0.

If k 1, the function kf is called the negative o f f and is denoted simply by f instead of the clumsy If. 
Ahsolute Value Function (or modulus function) of a given function Let f be a. fi~nctioilwith domain D. The absolute value function o f f , denoted by I f I and r e d as 111od f is defined by setting. ( fi ) (x) =: i f!X) I,for all x c D. Since fix) I = f(x); if f(x) 2 0, f and I f I have the same graph for those value of x for which f(x) 2 0. Now let us consider those values of x for which f (x) < 0. Here I f (x) / =  f (x). Therefore, the graphs o f f and ( f ( are reflections of each other w.r.t. the xaxis for those values of x for which f (x) < 0.
Fig. 10
Elements of Differential Calculus
As an example, consider the graph in Fig. 10 (a). The portion of the graph below the xaxis (that is, the portion for which f(x) < 0) has been shown by a dotted line, To draw the graph of / f I we retain the undotted portion in Fig. 10 (a) as it is, and replace the dotted portion by its reflection w.r.t. the xaxis (see Fig. lob).
Sum, difference,Product and Quotient of two functions If we are given two functions with a common domain, we can formseveral new functions by applying the four fundamental operations of addition, subtraction, multiplication and division on them. i)
Define a functions s on D by setting s(x) = f(x)+ g(x). The hnction s is called the sum of the functions f and g, and is denoted by f + g. Thus, (f+ g) (x) = f(x)+ g(x)
ii)
Define a function d on D by setting d(x) = f(x) g(x). The function d is the function obtained by subtracting g from f, and is denoted by fg. Thus, for all x E D (f  g) (x) = fix)  g(x).
iii) Define a function p on D by setting P(x) =f(x)g(x).
The function p, called the product of the functions f and g, is denoted by fg. Thus, for all XE D
iv) Define a function q on D by setting q(x) = fix)/g(x), provided g(x) # 0 for x E D. The function q is called the quotient o f f by g and is denoted by flg. ThCus, (f/g) (x) = f(x)lg(x) (g(x) # 0 for any x E Dl.
Remark 5: In case g(x) = 0 for some x E D. We can consider the set, say D, of all those values of x for which g(x) a 0, and define flg on D by setting (flg) (x) = f(x)/g(x) tf x E D. Example 4: Consider the functions f : x + x' and g : x + x3.Then the functions f + g, f  g, fg are defined as (f+gj(x)=x'+x3, (fg) (x) = x2x3. (fg) ( 4 = xS Now, g(x) = 0 e;4 x3= 0 @ x = 0. Therefore. in order to define the function flg, we shall consider only nonzero values of x. If x # 0, f(x)/g(x) = x2/x3= 1 1 ~Therefore . flg is the function. Bg : x + llx, whenever x # 0. All the operations defined on functions till now, were similar to the corresponding operations on real numbers. In the next subsection we are going to introduce an operation which has no parallel in R. Composite functions play a very important role in calculus. You will realise this as you read this course further.
1.6.2 Composite of Functions We shall now describe a method of combining two functions which is somewhat different from the ones studied so far. Uptill now we have considered functions with the same domain. We shall now consider a pair of functions such that the codomain of one is the domain of the other. Let f : X + Y and g : Y + Z be two functions. We define a function h : X +Z by setting h(x) = e(qx)). To obtain h(x), we first take the fimage, f(x), of an element x of X. This f(x) E Y. which is the domain of g. We then take the g image of f(x), that is, g(f(x)), which is an element of Z. This scheme has been shown in Fig. 11.
Real Numbers and Functions
Fig. I I
The function h, defined above, is called the composite o f f and g and is written as gof. Note the order. We first find thesfimage and then its gimage. Try to distinguish it form fog, which will be defined only when Z is a subset of X. Also, in that case, fog is a function from y to y. Example 5: Consider the functions f :x + x2 v x E Rand g :x + 8x + 1 v x E R. g f 1s ' a function from R to itself, defined by (goo(x) = g(f(x)) = g(x2)= 8x2+ 1 Q x E R. fogis a func ~ I O I I from R to itself defined by (fog)(x) = qg(x)) = q8x + 1) = (8x + I)*.Thus gofand fogare both defined, but are different from each other. The concept of composite function is used not only to combine functions, but also to look upon a give11function as made up of two simpler functions. For example, consider the function. h:x+sin(3x+7) We can think of it as the composite (gooof the functions f :x + 3x + 7 v x E Rand g : u + s i n u v u ~R. Now let us try to find the composites fog and g f of the functions: f : x + 2 x + 3 ~X E R , a n d g : x + ( l / 2 ) ~  3 / 2 v x ~R Note that f and g are inverses of each other. Now gof(x) = g(f(x)) = g(2x + 3)
Similarly,fog(x) * f(g(x)) = qx/2  312) = 2(x/2  312) + 3 = x. Thus, we see that gof(x) = x and f~g(x)= x for all x E R. Or, in other words, each of gof and fogis the identity function on R. What we have observed here is true for any two functions f and g which are inverses of each other. Thus, iff :X +Y and g :Y + X are inverses of each other, then gofand fogare identity functions. Since the domain of gof is X and that of fogis Y, we can write this as : gof = ix,fog= iy. This fact is often used to test whether two given functions are inverses of each other.
1.7
,
TYPES OF FUNCTIONS
In this section we shall talk about various types of functions, namely, even, odd, increasing, decreasing and periodic functions. In each case we shall also try to explain the concept through graphs.
1.7.1 Even and Odd Functions We shall first introduce two inlportant classes of functions: even functions and odd functions. Consider the functions f defined on R by setting
/
1
I
f(x)=x2
~ x E R .
You will noticethatqx)= (x)~= xZ= qx)
\\,
xE R
/
, ,
\.L,, ~F 0 X /'
I
,
I
This is an example ofan even functi0.n.Let7$take a look at the graph. (Fig. 12) ofthis function. We find that the graph (a parabola) is symmetrical about the yaxis. 1f we fold the paper along the yaxis, we shall see that the parts of the graph on both sides of the yaxis completely r coincide with each other. Such functions are called even functions. Thus, a function f, defined on R is even, if, for each x E R, f (x) = f(x). The graph of an even function is symmetric with respect to the yaxis. We also note that if the
~
.... . .~.
Fig. 12
~
.
' 23
Flrments o f 1)ifierential Calculus
graph of a function is symmetric with respect to the yaxis, the function must be an even f~~nction. Thus, if we are required to'draw the graph of an even function, we can use this property,to our advantage. We only need to draw that part of the graph which lies to the right of the yaxis and then just take its reflection w.r.t. the yaxis to obtain the part of the graph which lies to the left of the yaxis.
E E 12) Given below are two examples of even functions, alongwith their graphs. Try to
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convince yourself, by calculations as well as by looking at the graphs, that both the functions are, indeed, even functions. A
9
,
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 ..
0

Y
a) 
The absolute value function on R
b
X
f:x+lx/
b) (a)
The graph o f f is shown alongside.
2'
The function g defined on the set of nwzero real numbers by setting g(x) = 1/x2,x # 0. The graph of g is shown alongside.
* / 3 2  1 0
1
?\,
.
. 1 2 3
b
x
(b)
Now let us consider the function f defined by setting f(x) = x3 x E R. We observe that f(x) (x) = (x)~= x3 = f(x) v x E R. If we consider another function g glven by g(x) = sin x we shall be able to note agaln that g(x) = sin (x) = sinx = g(x).

The functions f and g above are similar in one respect: the image of x is the negative of the image of x. Such functions are called odd functions. Thus, a function f defined on R is said to be an odd function if f(x) =f(x) v x E R. If (x, Qx)) is a point on the graph of an odd function f, then (x, f(x)) is also a point on it. This can be exptessed by saying that the graph of an odd function is symmetric with respect to the origin. In other words, if you turn the graph of an odd function through 180" about the origin you will find that you get the original graph again. Conversely, if the graph of a function is symmetric with respect to the origin, the function must be an odd function. The above facts are often usefui while handling odd functions.
E E 13) We ate giving below two functions alongwith their graphs. By calculations as well as by looking at the graphs, find out for each whether it is even or odd. YA
.*
a) The identity function on R:
/b)
/
u'
The fuhction g defined on the set of nonzero real numbers by setting g(x) = Ilx, x # 0
;
YA
4
3
While many of the functions that you will come across in this course will turn out to be either even or odd, there will be many more which will be neither even nor odd. Consider, for example, the function f:x+(x+1)2 ere f(x) = (x + 1)2= x2 2x + 1. ISf(x) = f(X) v x E R?
Real Numbers and Functions
The answer is 'no'. Therefore, f is not an even function. Is f(x) = f(x) v x E R? Again, the _ answer is 'no'. Therefore f is not an odd function. The same conclusion coyld have been drawn by considering the graph o f f which is given m Fig. 13. You will observe that the graph is symmetric neither with respect to the yaxis, nor with respect to the origin. Now there should be no difficulty in solving the exercise below.
E E 14) Which of the following functions are even, which are odd, and which are neither even nor odd?
L
a) b)
x + x 2 + 1, v X E R x+~)l, V X E R
c) d)
x+cosx, x+xlx(,
1.7.2
~
X
v XE
0, ~f x is rational
RE R
1,
if x is irrational
Monotone Functions
In this subsection we shall consider two types of functions:
/ I
i) Increasing and ii) Decreasing Any function which conforms to any one of these types is called a monotone function. Does the of a company increase with production'! Does the volume of gas decrease with increase in pressure? Problems like these require the use of increasing or decreasing functions. Now let us see what we mean by an increasing function. Conslder the hnction g and h defined by x,
g (x) = x3
and h (x) =
1,
ifxlO if x > 0
Note that whenever x, > x,, we get x~~> xI3,that is, g(x,) > g(x,).
Fig.
.
14
Y4
.
In other words, as x increases. g(x) also increases. This fact can also be seen from the graph of g shown in Fig. 14. Let us find out how h(x) behaves as x increases. In this case we see that if x, > x,, then h(x,) 2 h(x,). (You can verify this by choosing any values for x, and x,). Equivalently, we can say that h (x) increases (or does not decrease) as x increases. The same can be seen from the graph of h in Fig. 15. I
I
 &a
Functions like g and h above are called increasing or nondecreasing functions. Thus, a function f defined on a domain D is said to be increasing (or nondecreasing) if, for every palr of elements x, x2E D, x2> x I a f(x2)2 f(xl). Further, we say that f is strictly increasing if x, > x, 3 f(q) > f(x,) (strict inequality). Clearly, the function g : x + x3 discussed above, is a strictly increasing function; while his not a strictly increasing function. We shall now study another concept which is, in some sense, complementary to that of an
I
t
increasing function.
Fig.
IS
Elements o f Differential Calculus
Consider the function f, defined on R by setting.
The graph off, is as shown in Fig. 16. YA 4
From the graph we can easily see that as x increases f, does not increase. Thatis,x2>x,*f, ( x 2 ) l f , (x,)orf, ( x J 2 f, (x,) Now consider the function f2 :x + x3 (x E R)
\I
The graph of f, is shown in Fig. 17. Since x, > x, 3 X: > xI33 q3< xI3 3 f2(x2)< f2(x,), we find that as x increases, f2(x) decreases. Functions like f, an; f2 are called decreasing or nonincreasing functions. The above two examples suggest the follwing definition: A fbnchon f defined on a domain D is said to be decreasing (or nonmcreasing) if for every pair of elements x,, x,, x, > x, 3 f(x2)5 f(x,).Further, f is said to be strictly decreasing if 3>XI 3 f(xz)< We have seen that, of the two decreasing functions f, and f2, f2 is strlctly decreasmg, while f, is not strictly decreasing. A function f defined on a domain D is said to be a monotone function if it is either increasing or decreasing on D.
Graph of f, Fig. 16
AY
\
4 I
i2
34
All the four functions (g, h, f,, f,) discussed above are monotone functions. The phrases 'monotorlically increasing' and 'monotonically decreasing' are often used for 'increasing' and 'decreasing', respectively.
1
While many functions are monotone, there are many others which are not monotone Consider, for example, the function. f:x+x2(xeR).
\.
.
2

 T
1 01
2
You have seen the graph o f f in Fig. 12. This function is neither increasing nor decreasing.
II
t
1
'
If we find that a given function is not monotone, we can still determine some subsets of the domain on which the function is increasing or decreasing. For example, the function f(x) = x2 is strictly decreasing in ] , 01 and is strictly increasing in [0, m [.
\
\
I
*
I
E E 15) Given below are the graphs of some funchons. Classify them as nondecreasing. strlctly decieasing, neither increasing nor decreasing:
Graph of f, Fig. 17
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k
w '\
o/

X
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+
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a
*
X
b
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1.7.3 Periodic Functions In this section we are going to tell you about yet another important class of functions, known as periodic functions. Periodic functions occur very frequently in application of mathematics to various branches of science. Many phenomena in nature such as propagation of water waves, sound waves, light waves, electromagnetic waves etc. are periodic and we need periodic functions to descrlbe them. Similarly, weather conditions and prices can also be described in terms of periodic functions. Look at the following patterns :
4
r,, ,)\,, d 0
/"
I'
"
o j  e T . e ~ . ~ . ~ l e . e . i . i .
Real Numbers and Functions l
l l
<<=<<I<z?.<:<<<< (:<< < <s<<: %
e e . ~ i e e  e i e o . . ) e . ~ . . . . e .
ee.....e.e.....*.......
>x
V~\<.Y ld # A J  ., :~;<)x><>,yld :K;
Fig. 18
You must have come across patterns similar to the ones shown in Fig. 18 on the borders of sarees, wall papers etc. In each of these patterns a design keeps on repeating itself. A similar situation occurs in the graphs of periodic functions. Look at the graphs in Fig. 19.
Fig. 19
In each of the figures shown above the graph consists of a certain pattern repeated infir!itely many times. Both these graphs represent periodic functions. To understand the situation, let us examine these graphs closely. Consider the graph in Fig. 19(a). The portion of the graph lying between x = 1 and x = 1 is the graph of the function x + I x 1 on the domain  1 Ix 5 1. This portion is being repeated both to the left as well as to the right. by translating (pushing) the graph through two units along the xaxis. That is to say, if x is any point of [1, I], then the ordinates at x, x f2, x k 4, x 6, .....are all equal. The graph therefore represents the function f defined by
+
The graph in Fig. 19(b) is fie graph of the sine function, x + sin x, y x E R. You wili notice that the portion of the graph between 0 and 2n is repeated both to the right and to hi:Icfc. You know already that sin (x + 2nj =sin x, y x E R. We now give a precise meaning to the tcrnl "a periodic function". A function f defined on a domain D is said to be a periodic function if there exists a positlvc read number p such that f(x +p) = f(xj for all x E D. The number p is said to be a pcricd off. The smallest positive ineger p with the property described above is callcd the period off. As you know, tan (x + nn) = tan x y n E N. This means that nn, n E N are all periods of the tangent function. The smallest of nn, that is n, is the period of the tangent function. See if you can do this exercise.
~
Elements of Differential Calculus
E E 16)
a) What is the period each of the functions given in Fig. 19(a) and (b)? b) Can you give one other period of each of these functions?
As another example of a periodic function, consider the function f defined on R by setting f(x)=x  [XI Let us reaall that [x] stands for the greatest integer not exceeding x. The graph of this function is as shown in Fig. 20. From the graph (as also by calculation) we can easily see that f(x + n) = f(x)
Q
x E R. and for each positive integer n.
Fig. 20
The given function is therefore periodic, the numbers 1,2,3,4 being all periods. The smallest of these, namely 1, is the period. Thus the given function is periodic and has the period 1.
Remark 6 Monotonicity and periodicity are two properties of functions which cannot coexist. A monotoile function can never be periodic, and a periodic function can never be monotone. In general, it may not be easy to decide whether a given function is periodic or not. But sometimes it can be done in a straight forward manner. Suppose we want to find whether the function f : x + x2 Q x E R is periodic or not. We start by assuming that it is periodic with period p: Then we must have p > 0 and f(x + p) = f(x) Q x
Considering x # p/2, we find that 2x + p # 0. Thus, p = 0. This is a contradiction. Therefore, there does not exist may positive number p such that f(x + p) = f(x), consequently, f is not periodic.
Q
x E R and,
E E 17) Examine whether the following functions are periodic or not. Write the periods of the periodic functions. a) x + cos x c) ,x+sin2x e) x +cos (2x + 5)
b)x+x+2 d) x +tan 3x f)x+sinx+sin2x
Elements o f Differential Calculus
I
E E 19) Is the sum of two periodic functions also periodic? Give reasons for you answer.
We end with summarising what we have discussed in this units.
1.8
SUMMARY
In this unit we have 1 briefly revised the basic properties of real numbers, 2.
defined the absolute value of a real number x as X
=
/ xif x 2 0 Ix
if x < 0
3
discussed various types of intervals in R Open: ]a,b[ = { X ER : a < x < b ) closed: [a,b] = { X ER : a l x l b ) semiopen: ]a,b] ={XER:a<x<b) or [a,b[ = { x e R : a l x c b ) , where a, b E R,
4
defined a function and discussed various types of functions along with their graphs: oneone, onto, even, odd, monotone, periodic.
5
defined composite of functions and discussed the existence of the inverse of a function.
1.9
SOLUTIONS AND ANSWERS a
E 1) a) The set { 1,2,3, .......) has a lower bound, e.g., 0. b) The set { ............., 3, 2, 1,0, 1,2, .............} does not have a lower bound. c) Theg.l.bofthesetS={l,'/,, '1 ................ '1,. ........)isO,andO@S.
,
d) {x : x E R and 1 5 x I 2) is a bounded set as it is bounded above by 2 and below by 1. E 2) a) A real number p is positive if p > 0.Hence 0 is a lower bound for the set P of positive real numbers. Thus the set P is bounded below. Its infimum is 0. b) A real number r is the infimum of a set S c R if and only if the following conditions are satisfied: i)
r l x f o r a l l x ~S.
ii)
For each e > 0 there is y E S such that y < r + E .
The set P in a) above has infimum 0, since
i)
O<pforallp~Pand ii) For each & > 0 there is d 2 E P such that d 2 < 0 + E = E
E3) a) Ix(=rnax{x,x).Hencex=O+lxl=Oand (x\=O*max. { x ,  x ) = O ~ x = O . b) There are three cases 1) x 2 0, y 2 0 2) x and y have opposite signs e)x<O,ycO. We take 2). Suppose x > 0, y < 0, then, ~ x ) = m a{x,x)=x,I r y/=max{y,y}=y.
Renl Numbers nncl Functions
xy<O*/xyI=xy=xx(y)=(xJ)yl.
1) and 3) can be proved similarly. c) Ifx>O,Ix)=xand/l / x ( = l / x = l / ( x ) Ifx<O,(xJ=xandl l/x)=l/x= 1/Ix)
d)Ixy/=lx+(Y)I~/~I+IYI=I~I+IYI e)Ix+y+zI=((x+y)+z(1(x+y)+)zJS1~I+Jy(+(~J
f)lx~zl=lxyllzl=lxlI~llzl
E4) a) False E5) a)

b) True 8
1
c) True b) 4
d) False C
2
E 8) b) is oneone E9) a) is onto
Hence f is oneone. I f y e X,putx=
Y+l . Then x E X and y = f(x). Hence, f is onto. Y1
p
Ell) a) f:R++R:f(x)= J;;
E12) a) f(x) = I x I f(x) = ( x 1 = ( x I = qx). Hence, f is even. b) g(x) = 1/x29 g(x) = 1 / (  ~ )=~g(x). Hence, g is even. Q
E13) a) f(x) = x 3 f(x) =  x =f(x). Hence, f is odd. b) g(x) =.1/X 3 g ( x) = l/x =g(x). Hence, g is odd. E14) a), c and e) are even d) is odd b) is neither even nor odd. E15) a) neither increasing, nor decreasing b) nondecreasing c) strictly decreasing E l 9 The period of the function in Fig. 19 a) is 2. Other periods are 4,6,8, ........... The period of the function in Fig. 19 b) is 2x. Other periods are 4x, 6x, ........... E17) a) Periodic with period 2x Since cos (x + 2x) = cos x for all x. b) not periodic c) Periodic with period x. d) Periodic with period x/3. e) Periodic with period x. Periodic with period 2x. El8) a) and b) are periodic, c) is not. f)
b
E19) NO.For example, x  [x] and 1 sin x I are periodic, but their sum is not.
1
UNIT 2
LIMITS AND CONTINUITY
Structure 2.1
Introduction Objectives
2.2
2.3
2.4 2.5
2.1
Lirnib Algebra of Limits Limits as x += (or  m) Onesided Limits Continuity Definitions and Examples Algebra of Continuous Functions Summary Solutions and Answers
INTRODUCTION
The last unit has helped you in recalling some fundamentals that will be needed in this course. We will now begin the study of calculus, starting with the concept of 'limit'. As you read the later units, you will realise that the seeds of calculus were sown as early as the third century B.C. But it was only in the nineteenth century that a rigorous definition of a limit was given by Weierstrass. Before him, Newton, d' Alembert and CAuchy had a clear idea abort limits, but none of them had given a formal and precise definition. They had depended, more or less, on intuition or geometry. The introduction of limits revolutionised the study of calculus. The cumbersome proofs which were used by the Greek mathematicians have given way to neat, simpler ones. You may zlrbady have an intuitive idea of limits. In Sec. 2 of this unit, we shall give you a precise defi~itionof this concept. This will lead to the study of continuous functions in Sec.3. Most of the functions that you will come across in this course will be continuous. We shall also give you some examples of discontinuous functions.
Objectives After reading this unit you should be able to : calculate the limits of functions whenever they exist, identify points of continuity and discontinuity of a function.
2.2 ,BASIC PROPERTIES OF R In this section we will introduce you to the notion of 'limit'. We start with considering a situation which a lot of us are familiar with, such as train travel. Suppose we are travelling from Delhi to Agrh by a train which will reach Agra at 10.00 a.m. As the time gets closer and closer to 10.00 a.m., the distance of the train from Agra gets closer and closer tdzero (assuming that the train is running on time!). Here, if we consider time as out independent variable, denoted by t and distance as a function of time, say f(t), then we see that f(t) approaches zero as t approaches 10. In this case we say that the limit of f(t) is zero as t tends to 10. Now consider the function f :R +R defined by f(x) = x2+ 1. Let us consider Tables l(a) and 1(b) in which we give the values of f(x) as x takes values nearer and nearer to 1. In Table l(a) we see values of x which are greater than 1. We can also express this by saying that x approaches 1 from the right. Similarly, we can say that x approaches 1 from the left in Table 1(b).
Real Numbers and
Table 1 (a)
Functions
Table 1 @)
We find that, as x gets closer and closer to 1, f(x) gets closer and closer to 2. Alternatively, we express this by saying that as x approaches 1 (or tends to I), the limit of f(x) is 2. Let us now give a precise meaning of 'limit'. Definition 1 Let f be a function defined at all points near p (except possibly at p). Let L be a real number. We say that f approaches the limit L as x approaches p ~ ffor , each real number E > 0, E can find a real number 6 > 0 such that O<Ixp(<6*(f(x)L(<&. AS youknowfromUnit 1, ( x  p I < 6 m e a n s t h a t x ~]p6,p+6[andO<I~pImeansthat x # p. That is, 0 < ( x  p 1 < 6 means that x can take any value lying betweenp  6 and p + 6 except p. The limit L is denoted by lim f(x). We also write f(x) + L as x + p.
' I ~ Idelinitton S of ~ t m was ~t first stated by Karl We~erstrass,around 1850
6 (delta) are Greek letters used to denote real numbers. (& epsilon) and
'+' denotes 'tents to'
x +P
Note that, in the above definition, we take any real number E > 0 and then choose some 6 > 0, s o t h a t L  & < f ( x ) < L + & ,whenever(xpI<6,thatis,p6<x<p+6. In unit 1 we have also mentioned that ( x  p I can be thought of as the distance between x and p. In the light of this the definition of the limit of a function can also be interpreted as:
The E  6 definition does not give us the value Of L. It lust helpts u check whether a given number L is the limit of
Given E > 0. we can choose 6 > 0 such that if we choose x whose distance from p is less than
f(x).
6, then the distance of its image from L must be less than E. The pictures in Fig. 1 may help you absorb the definition.
for each E > 0
3 a 6 > 0 s.t.
O<lxcl<S
3
Fig. 1
Remember, the number E is given first and the number 6 is to be produced. An important point to note here is that while taking the limit of f(x) as x + p, we are concerned only with the values of f(x) as x takes values closer and closer to p, but not when x = p. For x2 1 example, consider the function f(x) = .This function is not defined for x = 1, but is x1 defined for all other x E R. However, we can still about is limit as x + 1. This is because for taking the limit we will have to look at the values of f(x) as x tends to 1, but not when x = 1. Nowlet us take the following examples: Example 1 Consider the function f : R + R defined by f(x) = x3. How can we find lim =f(x)? x
9
0
I f(x)  L I < E
Elements of Differential Calculus
Fig. 2
Look at the graph off in Fig. 2. You will see that when x is small, x3is also small. Asx comes closer and closer to 0, x3 also comes closer and closer to zero. It is reasonable to expect that lirnf(x)=Oasx+ 0. Let us prove that this is what happens. Take any real number E > 0. Then, ( f(x)  0 I < E w I x3 I < E B ( x ( < &'I3. Therefore, if we choose 6 = &'I3 we get I f(x)  0 I < E whenever O<IxOI<6.Thisgivesus lim f(x)=O. x+o A useful general rule to prove lim f(x) = L is to write down f(x)  L and then express it in x+a
terms of (x  a) as much as possible.
Let us now see how to use this rule to calculate the limit in the following examples. x21 Example 2 Let us calculate lim x+l xl We know Aat division by zero is not defined. Thus, the function f(x) =
x2  1
x_1 is not defined
at x = 1. But, as we have mentioned earlier, when we calculate the limit as x approaches 1, we do not take the value of the function at x = 1. Now, to obtain lim x+l
x21=(xI)(x+l),sothat,
x21
x2 1 we first note that
x1
I
2
x  I  lim x_1 =x+lforx+l.Therefore lim  x + 1 (x+ 1). X+l x1
As x approaches 1, we can intuitively see that this limit approaches 2. To prove that the limit is 2,wefirstwriteflx)L=x+ 1 2=x1,whichisitselfintheformxa,sincea= 1 inthis case. Let us take any number E > 0. Now, ( ( x + I )  ~ ( < E w ( x (<~E Thus, if we choose 6 = E,in our defmiton of limit, we see that ) x  1 ) < 6 = e a I f ( x )  L ( = ( x  1 (<&.Thisshowsthat lim (x+.l)=2.Hence, x
9
l
x2 1 lim = 2. x+l x1
Example 3 Let us prove that lim (x2+ 4) = 13. x+3
Thatis,weshallprovethat VE>O. 36>Osuchthat)x2+4 1 3 ) < ~ w h e n v e r ( x  3 ) < 6 .
Real N u m b e r s and Functions
Nowwewrite)x2 9 / i n t r m s o f ( x  3 1 : 1x29I=lx+3)(x31 Thus, apart from 1 x  3 1, we have a factor, namely I x + 3 ( of [x2 91. To decide the limits of 1 x + 3 1, let us put a restriction on 6. Remember, we have to choose 6. So let us say we choose a 6 5 1. What does this imply? Ix3)<6 3(x3)<1331<x<3+1
= + 2 < ~ < 4 3 5 < ~ + 3 < 7 . R e c a I l S e c . 4 , U n1.i t Thus, we have / x2 9 ( < 7 1 x  3 / < E.Now when will this be true? It will be true when i x  3 1 w7. So this d 7 is the value of 6 we were looking for. But we have already chosen 6 < 1. This means that given E > 0, the 6 9 e choose should satisfy 6 5 1 and also 6 5 ~ 2 7 . In other words, 6 = min { 1, d7}, should serve our purpose. Let us verify this : ! x  3 ) < 6 * I x  3 1 < 1 a n d ~ x  3( < d 7 * I x 2  9 1 = ) x + 3 l l x  3 ) < 7 . d 7 = ~ .
Remark 1 : Iff is a constant function on R, that is, if f(x) = k
x E RI where k is some fixed real
number, then ,lim ,f(x) = k. ~ o w b l e a s try e the following exercises. E 1) Show that 1 lim  = 1 a) x,, x
b)
x3  1
lim = 3. x+l x1
Before we go further, let us ask, 'Can a function f(x) tend to two different limits as x tends to p '? The answer is NO, as you can see from the following :
Elements o f Differential calculus
*Theorem 1 If lim fix) = L and lim fix) = M, then L = M. Proof :Suppose L # M, then ( L  M I > 0. Since lim f(x) = L. If we take E = I L  M I 'J 'P 2 then 3 6, > 0 such that Similarly, since lim f(x) = M, 3
> 0 such that
P 'x
Ifwechoosie6=rnin~6,,6,1,then6>OandIxp(<6willmeanthat(xp)<6,and IXPI<~~. In this case we will have both ( f(x)  L ( < E,as well as, I f(x)  M I < e.
Sothat(LM)=)Lf(x)+f(x)M(lIf(x)LI+(f(x)MI<e+~=2&=ILM(. That is, we get I L  M 1 < ( L  M 1, which is a contradiction. Therefore, our supposition is wrong, Hence L = M. We now staite and prove a theorem whose usefulness will be clear to you in Unit 4. Theorem 2 Let f, g, and h be functions defined on an interval I containing a, except possibly at a, Suppose
i)
lirn f (x) = L = lim h (x)
x+a
x+a
Then lirn g (x) exist and is equal to L. x+ a
Proof: By the definition of limit, given E > 0 , 3 6, > 0 and 6, > 0 such that (f(x)LI<&forOI~aI<6,and Ih(x)L)<eforO(xaI<6,. 'Let 6 = min ( 4 , 6,). Then, O<(xa1<6* If(x)L(<&andIh(x)LICE 3LeIf(x)<L+&,and L&Ih(x)lL+e We also have f(x)l g(x) lh(x) t~ x E l{a) Thus,wegetO<Ix  a I < 6 3 L  & 5 f ( x ) l g ( x ) l h ( x ) l L + e Inotherwords,O<I x  a l e 6 3 (g(x)L 1 % Therefore lirn g(x) = L. X 4 8
Theorem 2 is also called the sandwich theorem (or the squeeze theorem), because g is being smdwiched between f and h. Let us see how this theorem can be used.
Example 4 Given that I f(x) 1 ( 5 3(x +
t/ x E R, can we calculate
lim f (x) ?
x+1
We know that3(x + S f(x) 1 S 3 (x + v x. This means that 3(x + + 1 5 f(x) 5 3(x + 1)2+ 1 5 t~ x. Using the sandwich theorem and the fact that lirn [3 (x+
x+1
+ 1]= 1 = lirn [3(x+ I'.?
+ I], weget lirn f(x)= 1 x'
I
In the next section we will look at the limits of the sum, product and quotient of functions.
Algebra of limits
2.2.1
Now that you are familiar with limits. Let us state some basic properties of limits. (Their proofs are beyond and scope of this course.) 'Theorem3 Let f and g be two h c t i o n s such that lim f(x) and lim g(x) exist. Then P'X
i)
P'X
lim[f(x)+g(x)]=limf(x)+limg(x) P'X
ii)
P'X
lim [f(x) g (x)] = X'P
5) lim X+P
[:yp
Sum rule
P 'x
f(x)
]
lim g(x)
1 = I' ,provided lim g (x) gdx) litn g (x) xrp
P'X
Productrule t0
Reciprocal rule
We can easily prove two more rules in addition to the three rules given in Theorem 3. These are : *' lirn k = k
iv)
Constant function rule
X+P
v)
limx=p
Identity function rule
x +P
We shall only indicate the method of proving iv) and v): iv) Here I f(x)  L I = 1 k  k I = 0 < E, whatever be the value of 8. *
Using the properties that we have just stated, we will calculate the limit in the following example. Example 5 Let us evaluate lim x+2
3x2 + 4x
2x + 1
NOW lim 2~ + 1 = lirn 2x + lim 1 by using i) x+2
x+2
lirn 2 lirn x + lirn 1 by using ii)
=
x+2
I
P
:.
x+2
x+2
= 2 x 2 + 1=5+Obyusingiv)andv) We can use (iii) of Theorem 3. Then the required limit is
lirn 3 x 2 + lirn 4x
lirn (3x2 + 4x)
x+2
lim ( 2 +~1)
!
x+2
x+2

x+2
lim 2 x + liml
x+2
x+2
b~u~ingi)
x+2
lirn 3 lim x lirn x + lim 4 lirn x
x+2

x+2
x+2
x+2
lim 2 lirn x + lim 1
x+2
x+2
x+2
by using ii)
x+2
3 x 2 x 2 + 4 x 2 = 20  4 2x2+1 5 You can similarly calculate the limits in the following exercises. 
E
E E3) t
3
E2) Show that lim  = 3 X+l
X
Calculate lirn x + l 2x +
(A)
Real Numbers and Functions
Elements o f Differential Calculus
2.2.2 Limits as x
+

(or  ) 1
Take a look at the graph of the hnction f(x) =  ,x > 0 in Fig. 3. This is a decreasing function X
of x. In fact, we see fromFig. 3 that f(x) comes closer and closer to zero as x gets larger and larger. This situation is similar to the one where we have a hnction g(x) getting closer and closer to a value L as x comes nearer and nearer to some number p, that is when lim g(x) = L. X'P
The only difference is that in the case of f(x), x is not approaching any finite value, and is just becoming larger and larger. We express this by saying that f(x) + 0 as x + w ,or lim f(x) = 0. XW
Note that, = is not a real number. We write x + w merely to indicate that x becomes larger and larger. We now formalise this discussion in the following definition. Fig. 3
Definition 2 A function f is said to tend to a limit L as x tends to w if, for each E > 0 it is possible to choose K > 0 such that I f(x)  L I < E whenever x > K. In this case, as x gets larger and larger, f(x) gets nearer and nearer to L. We now give another example of this situation. Example 6 Let f be defined by setting f(x) = 1/x2for all x E R \ (0). Here f is defined for all real values of x other than zero. Let us substitute larger and larger values of x in f(x) = 1/x2and see what happens (see Table 2). Table 2
We see that as x becomes larger and larger, f (x) comes closer and closer to ie:o: Now, let us choose any E > 0. If x > l! J;, then 1/x2< E. Therefore, by choosing K = 11 &, we find that x >K
i f(x) / < E. Thus, lim
f(x) = 0.
XSQ,
Fig. 4 gives us a graphic idea of how this function behaves as x + w.
AY
Fig. 4
Sometimes we also need to study the behaviour of a function f(x), as x takes smaller and smaller negative values. This can be examined by the following definition. Definition 3 A function f is said to tend to a limit L as x +  if, for each E > 0,it is possible to choose K > 0, such that / f(x)  L I < E whenever x <  K. The following example will help you in understanding this idea. Example 7 Consider the function f : R + R defined by
The graph o f f is as shown in Fig. 5.
LImltr and Continuity
What happens to f(x) as x takes smaller and smaller negative values? Let us make a table (Table 3) to get some idea. Table 3
We see that x takes smaller and smaller negative values, fix) comes closer and closer to zero. In fact 141 + x2)< E whenever 1 + x2> I/&,that is, whenever x2> (I/&) 1, that is, whenever either
x <  k 3 I f(x) j C E. Consequently, lim fix) = 0. X+@
In the above example we also find that lim f (x) = 0. xbm
Let us see how lim f(x) can be interpreted geometrically.

x+m
In the above example, we have the function f(x) = 141 + x2),and as x + m, or x +  m, f(x) +0. From Fig. 5 you can see that, as x + or x +  m ,the curve y = f(x) comes nearer and nearer the straight line y = 0, which is the xaxis. Similarly, if we say that lim g(x) = L, then it means that,as x ,m the curve y = g (x) comes x+m
closer and closer to the straight line y = L. x  1. Example 8Let us show that lim  X'm (1 + x2)
I
Now, 1 x2/(1+ x2) 1 1 = 141 + x2). In the previous example we have shown that 1 1/(1+ x2) ) < E for x > K, where K = 1 I/â‚Ź 1.' 1 Thus, given E > 0, we choose K = / 1 / ~1 JIJZ, so that
We show this geometrically inFig. 6. 4y
1
Fig. 6

Elements o f Differential Calculus
You must be wondering if all the properties given in Theorem 3 also hold when we take limits as x + . Yes, they do. You can use them to solve this exercise.
E
E4) Shos that
a)
lirn l/x=O xbm
b)
lim (l/x+3/x2+5)=5 Xbco
Sometimes we cannot use Theorem 2 directly, as is clear in the following example, Letus see how to overcome this problem. 3x + 1 Example 9 Suppose, we want to find xlirn , 2x + 5
'

We cannot apply Theorem 3 directly since the limits of the numerator and the denominator, as x + ,cannot be found. Instead, we rewrite the quotient by multiplying the numerator and denominator by llx. for x#O. 3x+l  3+(l/x) Then, 2 x  2 + (5 I x) , for x # 0. Now we use Theorem 3 and the fact that lirn llx = 0 (see # 4 a), to get xbm
3 + (1 / x) 3x + 1 lim = lim x+m2x+5 x  + m 2 + ( 5 / x )

lim ( 3 + l / x ) x+m
 lirn ( 2 + 5 / x ) x+m
E
3+0
 =
2+0

2
By now you knust be used to the various defmitions of limits, so can try this exercise E5) a) If for some E > 0,and for ever K, 3 x >K s.t. I f(x)  L 1 > E, what will you infer? b) If lirn f(x) ;f L, how can you express it in the E  6 form? x tP
Real Numbers and Functions
We end this section with the following important remark. Remark 2 In case we have to show that a function f does not tend to a limit L as x approaches p, we shall have to negate the definition of limit (also see E5(b)). Let us see what this means. Suppose we want to prove that lim f(x) ;t L. Then, we should find some E > 0 such that for every 6 > 0, there is some x E ]p Y8for which I (x)  L I > 6. Through our next example we shall illustrate the negation of the definition*ofthe limit of f(x) as x +00. Example 10 To show that lirn llx # I, we have to find some E 0 such that for any K (howsoever large) we can alGzys find an x > K such that ( I/x  1 I > E. 114. This clearly show that lim l/x # 1. X+(D
2.2.3
Onesided Limits
If we consider the graph of the function f(x) = [XI,shown in Fig. 7, we see that f(x) does not see to approach any fixed value as x approaches 2. But from the graph we cansay that if x approaches 2 from the left then f(x) seems to tend to 1. At the same time, if x approaches 2 from the right, then f(x) seems to tend to 2. This means that the limit o f f cxists if x approaches 2 from only onc side (left or right ) at a time. This example suggests that we in!roduce the idea of a onesided limit. Definition 4 Let f be a function defined for all x in the interval ]p, q[. f is said to approach a limit L as x approaches p from right if, given any E > 0, there exist a number 6 > 0 such that p<x<p+6qIf(x)L(<~. In symbols we denote this limit by lirn f(x) = L. x+p'
Similarly, the function f : )a, p[ + R is said to approach a limit L as x approaches p f r o p the left i f , g i v e n a n y ~ > 0 , 3 6~ O s u c h t h a t p   6 < x < p ~ I f ( x )  L I < & . This limit is denoted by lirn f (x). x+p
Note that in computing these limits the values of f(x) for x lying on only one side of p are taken ~ n t oaccount. Let uf apply this definition to the function f(x) = [ x 1, we know that for x E [l, 2[, [x] = 1. That is, [x] is a constant function on [I, 2[. Hence lim [x] = 1. Arguing similarly, we find that since [x] x+2
=:
2 for all x E [2,3[, [XIis, again, aconstant function on [2,3[, and lim [x] = 2. x+2+
Let us improve our understanding of the definition of onesided limits by lookiilg at some more examples. u
Example 11 Let f be defined on R by setting
We will show that lirn f(x) equals 1. x 4 
When x < 0 , )x 1 =  x, and therefore, f(x) = ( x)lx = 1. In order to show that lim f(x) exists x0
and equals 1, we have to start with any E > 0 and then find a 6 > 0 such that, if
6<x<O,then)f(x)(l)I<c. Since f(x) = i for all x < 0, I f(x)  ( 1) I = 0 and, hence any number 6 > 0 will work. Therefore, whateGer 6 0 we may choose, if  6 < x < 0,then 1 f(x)  (1) I = 0 < E. Hence lim f (x) x 4 
Example 12 f is a function defined on R by setting f(x)= x  [x], for all x E R. Let us examine whether lim f(x) exists. x+1
Recall (Unit 1) that this function is given by f(x) = x, if 0 5 x < 1. f(x) = x  1 if 1 5 x < 2, and, in general
= 1.
Fig. 7
Elements of Differential Calculus
Fig. 8
Since f(x) = x for values of x less than 1 but closc to 1, it is reasonabie to expect that lim f(x) = 1. Let us prove this by taking any E : O and choosing 6 = min / 1, E I. We f h d x1
16<x<l*f(x)=xandIf(x)1
I=Ixl ) < 6 < & .
Therefore. l i n ~ f(x) = 1. X+l
Proceeding exactly as above, the noting that f(x) = x  1 if 1 5 x < 2, we can similarly prove that f(x) =O.
E
5 6 ) Prove that
a)
lim x [x] 4 x3
b)
lim 1x1 =I
n+Oi
X
Before going further, !et us see how the concepts of onesided limit and limit are comected.
Real
Theorem 4 'The following statements are equivalent. lirn qx)
i)
exists
X+P
lirn f ( X )
ii)
and l i n ~ exist and are equal. x+p
Proof :'To show that i) and ii) are equivalent, we have to show that i) 3 ii) and ii) s i). We first prove that i) 3 ii). For this we assume that lirn f(x) = L. Then given *E > 0.
"36>@suchthat]f(x)L)<~forO<I~pf<6. p < x < p t6andp6<x<p.Thismeansthat
X+P
lirn f ( x ) = L z lim f(x). x+p
x+p+
We now prove the converse, that is, ii) 3 i). For this, we assume that
lirn f ( x ) : linl f(x) = L. Then, givcii E > 0,36, ,6, > 0. x+p
x+pt
such that if(x)Li<~forp6,<x<p (f(x)L(<*~for~<x<~+~,
Hence, lirn fjx) =L. X+P
Thus, we have shown that i) =, ii) and ii) 3 i), proving that they are equivalent. From Theorem 4, we can conclude that if lim qx) exists, then lirn f(x) and Aim f(x) x' P xtp* xtpalso exist and further. lim f(x)= lirn f(x)= lirn f(x) x
P
xtp+
+
x4p
Remark 3 If you apply Theorem 4 to the function f(x) = x  [x] (see Example 12), you will see that lim {x  [x]} does not exist as lirn (x  [x]) # lim {x [x]} , x1
x+1+
x+l
We shall use this concept of onesided limits to define continuous functions in the nest section.
2.3 CONTINUITY A continuous process is one that goes on smoothly without any abrupt change. Continuity of a function can also be interpreted in a similar way. Look at Fig. 9. The grapi? of the function f in Fig. 9 (a) has an abrupt cut at the point x = 3, whereas the graph of the function g in Fig. 9 (b) proceeds smoothly. We say that the function g is continuous, while f is not.
~4
y
4
(a)
(b) Fig. 9 : (a) Graph of f (b) Graph of g
Continuous functions play a very important role in calculus. As you proceed, you will be able to see that many theorems which we have stated in this course are true only for continuous functions. You will also see that continuity is a necessary condition for the derivability of a function, and that it is a sufficient condition for the integrability of a function. Bilt let us give a precise meaning to "a continuous function'' now.
umbers
and Functions
E i e n ~ e n t sof Differential Calculus
Definitions and Examples
2.3.1
In this sectLon we shall give you the definition and some examples of a continuous function. We shall also give you a short list of conditions which a function must satisfy in order to be continuous at a point.
Definition 5 Let f be a function defined on a domain D, and let r be a positive real number such that the interval ]p  r, p + r[ c D. f is said to be continuous at x = p if lim f(x) = f(p). X+P
,
By the definition of lim~tthls means that f is continuous at p is given E > 0 , 3 6 > 0 such that I fix)
 f(p) ) < F whenever 1 x  p j < 6. To clarify this concept let us look at an example
'
Example 13 L.et us check the continuity of the function

f : L R 4 W such that f(x) x at the point x = 0. Now, f(0) = 0. Thus we want to know if lirn f (x) = 0. x+O
This is true because given E > 0,,we can choose 6 = E and verify that ( x ( < 6 3 I f (x) I < E. Thus f is aontinuous at x = 0.
Remark 4 f is cont~nuousat x = p provided the following two criteria are met : i)
.
lim f(x)
x+p
exists.
Fig. 10 : (a) Graph of f
(b) G r a ~ hof g
Fig. 1U ~lioiwstwo discontinuous functions f and g. Criterion i) IS not inet by f, whereas g fails to meet criterion ii). If you read Remark 3 again, you will find that f(x) = x  [x] is not continuous as x = 1. But we have see that we can calculate onesided limits of f(x) = x  [x] at x = 1. This leads us to the following definition.
Dennition 6 A function f : lp, q[ R is said to be continuous from the right at x == p ~f lirn f(k) = f (p). We say that f is continuous from the left at q if lirn f(x) = f (q). x+pt
x+q
Thus, f(x) = x  [xl is continuous from the right but not from the left at x = 1 since lim f ( x ) x1
it f(1) and
lirn f(x) f(1) =0. x+l'
E E 7) Give E  6 definition of continuity at a point from the right as well as from the le'ft.
E
E 8) Show that function f: ]p  r, ,p + r[ + R is continuous at x = p if and only i f f is continuous from the right as well as from the left at x p (Use Theorem 4).
Real Numbers and Functions
;
Now that you know how to test the continuity of a function at a point, let us go a step further, and define continuity of a function on a set.
Definition 7 A function f defined on a domain D is said to be continuous on D, if it is continuous at every point of D. Let us see some more examples. Example 14 Let f(x) = xn for all x E R and any n E Zt. Show that f(x) is continuous at x = p for all p E R. We know that lirn x = p for any p E R. Then, by the product rule in Theorem 3, we get
z' is the set of positive integers
X+F
lirn xn = ( lirn x) ( lirn x) ..........( lim x) X+P

x+p
x+p
(n times)
X+P
pp ............. p (n times) =p". Therefore, lirn f(x) exists and equals f(p). Hence f is x+p. continuous at x = p. Since p was any arbitrary number in R, we can say that f is continuous on R. Remark 5 Using Example 14 and Theorem 3, we can also prove that polynomial a, + alx + ............ + allx", where a,, a, ..............an E R, is continuous on R, that is, l i m ( a o + a l x + ................. + a,,xn)= a, + alp + ..............+ anpnfor all p E R. 
XtP
Example 15 The greatest integer function f : R + R : f(x) = [ x ] is cliscontinuous at x = 2. To prove this we recall our discussion in Sec. 2 in which we have proved that lim f(x) = 1 x+2
and lirn f(x) = 2. Thus, since these two limits are not equal lirn f(x) does not exist. x+2+

x+2
Therefore, f is not continuous at x = 2 because the first criterion laid down in Remark 4 is not met. Example 16 Let f(x) = / x / for all x E R. This f is continuous at x = 0. Here f(x) = x, if x 2 0, and f(x) =  x if x < 0. You can show that lirn f(x)= lim x=O=f(O)and
C
x+o*
x+O'
lim f(x) = lirn (x)
x+o
f
1
x+o
= 0 = f(0). Thus
lirn f(x) exists and equals f(0). Hence f is continuous
x+o
atx=O. Note : f is also continuous at every other point of R. (Check this statement). x2  1
Example 17 Suppose we want to find whether f(x) = is continuous at x = 0. x1 In Fig. 11 you see the graph o f f . It is the line y = x + 1 except for the point (1,2) Y'
Fig. 11
We can write f(x) all x 1.'
=
x
+ 1 for
Elements of Dlfferentlal
f(o)=(02 1)/(0 I ) = 1 and limf(x)= lim (x2l)/(x1)=
Calculus
xro
1
xto
lim ( x i l ) = l
x+o
lim f(x) = f(O), so that f is continuous at x = 0.
x+O
The exponential function f(x) =ex and the logarithmic h c t i o n f(x) = In x are continuous functions. You can check this by looking at their graphs in Unit 1. Similarly,x +sinx and x + cosx are continuous, x + tanx is continuous in ]  d 2 , n;2[. This fact is quite obvious from the graphs of these functions (We have given their graphs in Unit 1). We shall not attempt a rigorous proof of their continuity here. Caution :Checking the continuity of a function from the smoothness of its graph is not a foolproof method. If you look at the graph (Fig. 12) ofthe h c t i o n x +x sin (llx), you will find that it has no breaks in the neighbourhood of x = 0. But this function is not continuous. Observe that the graph oscillates wildly near zero.
t
. Fig.
E
12
E9) Show that the function f : R + R given by f(x) = 1!(x2  9) is continuous at all points of Rexceptatx=3andx=3.
Now that we know how to check whether a function is continuous or not, let us go further, and talk about the continuity of some combinations of functions.
2.3.2 Algebra of Continuous Functions Let f and g be functions defined and continuous on a common domain D E R, and let k be any real number. In Unit 1, we defied the functions f + g, fgl Ug (provided g(x) z 0 any where in D), kif and I f I. The following theorem tells us about the continuity of these functions.
Theorem 5 Let f and g be functions defined and continuous on a common domain D, and let k be any real number. The functions f + g, kf, 1 f I and fg are all continuous on D. If g(x) # 0 anywhere in D, then the function flg is also continuous on D. We shall not prove this theorem here. In Unit 1, you have studied the important concept of composite functions. In Theorem 6, we will talk about the continuity of the composite of two continuous functions. Here again, we shall state the theorem without giving proof as it is beyond the level of this course.
Theorem 6 Let f : D, + D2and g :D2 + D, be continuous on their domains. Then gof is continuous on Dl, (Dl, D2,D3c R). Example 18 To prove that f : R +R : f (x) = (x2+ is continuous at x = 0, we consider the functions g : R +R : g(x) = x3 and h : R +R :h(x) = x2+ 1. You can check that f(x) = goh (x). Further, by Remark 5, h is continuous on R, and g is also continuous on R. Thus, goh = f is continuous on R. Let us see if the converse of the above theorems are true. For example, iff and g are defined on an interval [a, b] and iff + g is continuous on [a, b], does that mean that f and g are continuous on [a, b] ? No. Consider the functions f and g over the interval [0, 11 given by
Then neither f nor g is continuous at x = 112. (Why?) But (f + g) (x) = 1 v x E [O, 11. Therefore, f + g is continuous on [0, 11. Now, if ( f I is continuous at a point p, must f also be continuous at p? Again, the answer is No.Take, for example, the function f : R +R givenby f(x) =
1, 1,
forx1O for > 0
Then I f(x) 1 = 1 in Rand hence I f 1 is continuous. But f is not continuous at x = 0 (Why?) 1,
E10) Iff : R +R is defined by f (x) = f(x) = is f continuous at a ) x = 1
C
b) x = 3/2?
1.
i f x ~ Z ifxeZ
Limits and Continuity
,
E l e m e ~ ~ tosf Differential Calculus
One again, we won't prove this theorem here. But try to understand its statement because we shall be using it in subsequent units. Theorem 7 (Intermediate Value Theorem) Let f be continuous on the closed interval [a, b]. Suppose c ia a real number lying between f(a) and f(b). (That is, f(a) < c < f(b) or f(a) > c > (b?). Then there exists some x, E ]a, b[, such that f(xo)= c. How can we interpret this geometrically? We have already seen that the graph of a continuous function is smooth. It does not have any breaks or jumps. This theorem says that, if the points (a, f(a)) and (b, f(b)) lie on two opposite sides of a lide y = c (see Fig. 13), then the graph o f f must cross the line y = c.
Fig. 13
Note that this theorem guarantees only the existence of the number x,. It does not tell us how to find it. Another thing to note is that this x, need not be unique. That brings us to the end of this unit.
SUMMARY
2.4
Wh end this untt by summarising what we have covered in it.
.
The limit of a function f a t a point p of its domain is L is given E > 0 , 3 6 > 0, such that )f(x)L I <&whenever1xpI<6. Onesided limits
3.
p,,
1.
lim f(x) exists if and only if lim
x+p+
4.
f (x)and? ,, lim
f(x) both exist and are equal.
A fuqction f is continuous at a point x = p if lim f (x) = f (P)
,,
5.
6.
If the function f and g are continuous on D, then so are the functions f + g, fg, ( f 1, kf (where k E R) flg (where g(x) # 0 in D) The Intermediate Value Theorem : Iff is continuous on [a, b] and if f(a) < c < f(b) (or f(a) > c > f(b)), then 3x0â‚Ź ]a, b[suchthat f(xo)=c. 4
2.5
QOLUTIONS AND ANSWERS
E 1) a) Given any E > 0,if we choose 6 = min {d2,1/2) then I ~  l I < 6 < 1 / 2  x>1/2.
Real Numbers and Functions
Hence lim l/x = 1 x+l
Given E > 0, ifwe choose 6
= min
1 (217) E, 1/21,then
IxI(<1/2*~<3/2*~+2<7/2and
x3  1 Hence, lim = 3 x+l x  1 lim 3 1 E2) lirn 3 / x = + = 3 / 1 = 3 x+l hm x x+l
E3) lirn 2 x + 5 x+l
f\ I + x ) =21im x + 15+lirnl i mx2x 2 x2
X+I
E4) a) Given e > 0 if we choose K = lle, then x > K  I l/xOI=) l / x I < l / K = ~ Thus, lirn l/x=O x+m
b) Given E > 0, if we choose K = 11& ,then x > K * 1 1/x2OI=( l / x 2 ( i : 1 / K 2 = ~ . Hence, lirn 1/x2=O x+m
Now, lirn (l/x +.3/x2+ 5) x+m
=
lirn 1/x+3 lirn 1/x2+ lirn 5 = 0 + 3 x 0 + 5 = 5 x+m
x+m
E5) a)
x+m
lirn f(x)#L x+m
lirn X[XI=
x+3
b)
lirn X+O+
;;y
x2=l
I X \ / X J= ~:J+' X / X = ~ , I X I = X ~ O ~ X > O .
 x+olim (x2+2)=2 E 7) f is continuous from the right at x = p if v E > 0 there exists a 6 > 0 s.t. p<~<p+6*If(~)f(p))<~. f is continuous from the left at x = p if v E > 0 there exists a 6 > 0 s.t. p  6 < x < p * If(x)f(p))<~. E 8) f is continuous at x = p
* !!4x) = f(p)
lim f(x) = f (p) and x+plim * ,+,+
f (x) = f(p) by Theorem 4.
f is continuous from right and from left at x = p. Iff is continuous from right and left,
Elements of Differential Calculus .
lim 3
!$ fix) exists and = qp)
3f
is continuous at p,
E9) lim f(x) = x+a
lim x2  9

1
a2  9
= f (a) for all a except a = 3 f 3
X+l
Hence f is continuous at all points except at f 3, f is not defined at 23. E10) a) f is not continuous at x = 1. For E = 1 and any 8 > 0, if x is any noninteger E ] 18,1+8[,then ( f l ~ )  f ( l ) ( = I  l  lI = ~ > E . b) f is continuous at 312. Since given E > 0, if we choose 8 < 112, then
Ix(312)1< 1 1 2 3  2 < x <  1 3 x 4 Zandhence )f(x)f(312) /=O<E.
UNIT 3
DIFFERENTIATION
Structure 3.1
Introduction Objectives
3.2
The Derivative of a Function 3.2.1 Slope of a Tangent 3.2.2 Rate of Change 3.2.3 The Derivative
3.3 3.4
Derivatives of Some Simple Functions Algebra of Derivatives 3.b.l Derivative of a Scalar Multiple o f a Function 3.4.2 Derivative of the Sum of Two Functions 3.4 3 Derivative of !he P~.oductof Two Functions 3.4.4 De~'ivaliveof the Quotient of TNO Functions 3.4.5 The Chain Rule af Differentiation
3.5 3.6 3.6
3.1
Continuity versus Derivability Allmnary Solutions and Answers
INTRODUCTION
It was the seventeenth bentury. Some European mathematicians were working on two basic problems : i)
Is it possible to find the tangent to a given curve at a given point of the curve?
ii)
Is it possible to find the area under a given curve?
Two mathematical giants, Newton and Leibniz, independent of each other, solved these problems. The theory that they inventned in the process was Calculus. In this first unit on differentiation, we propose to introduce the concept of a derivative which is a basic tool of calculus, Leibniz was motivated directly by the first problem given above a problem which was of great significance for scientific applications. He recognised the derivative as the slope of the tangent to the curve at the given point. Newton, on the other hand, arrived at it by considering some physical problems such as determination of the velocity or the acceleration of a particle at a particle instant. He recognised the derivathe as a rate of change of physical quantities. We shall now show that both these considerations lead to the concept of derivative as the limit of a ratio. Of course, to understand what a derivative IS,you should have gone through Sec. 2 thoroughly.
Newton (16421727)
We shall first differentiate some standard functions using the definition of the derivative. The algebra of derivatives can then be effectively used to write down the derivatives of several functions which are algebraic combinations of the@functions. We shall also discuss the chain rule of differentiation which offers an unbelievable simplification in the process of finding derivatives. We shall also establish a relationship between differentiable functions and continuous functions which you have studied in Unit 2.
Objectives After studying this unit you should be able to : draw a tangent to a given curve at a given point determine the rate of change of a given quantity with respect to another obtain the derivatives of some simple functions such as xn, I x 1, J;; etc. from the first principles find the derivatives of functions which can be written as the sum, difference, product, quotient of functions whose derivatives you already know derive and use the chain rule of differentiation for writing down the derivatives of a composite of functions discuss the relationship between continuity and derivability of a function.
L e i b n i i (16461716)
Elements of Diffetential Calculus
3.2
THE DERIVATIVE OF A FUNCTION /
Before defining a derivetive, let us consider two problems in the next two subsections. The fmt is to find the slope of a tangent and the second is to find the rate of change of a given quantity in terms of another.
3.2.1
Slope of a Tangent
Let us consider the problem of finding a tangent to a given curve at a given point. But, what do we mean by the tangent to a curve? Euclid (300 B.C.) thought of a tangent as a line touching the curve at one point. This definition works fine in the case of a circle Fig. 1 (a), but it fails in the case of many other curves (see Fig. 1 (b)).
Fig. 1
We may define a tangent to a curve at P to be a line which best approximates the curve near P. But this definition is still too vague. Then how can we define a tangent precisely? The concept of limit which you have studied in Unit 2 comes to our aid here. Let P be a fixed point on the curve in Fig. 2 (a), and let Q be a nearby point on the curve. The line throughout P and Q is called a secant. We define the tangent line at P to be the limiting position (if it exists) of the secant PQ as Q moves towards P along the curve (Fig. 2 (b)).
(b)
Fig. 2
It may not be always possible to find the limiting position of the secant. As we shall see later, there are curves which do not have tangents at some points. In fact, there are curves hhich do not have a tangent at any point! There is another question which we can ask here. Suppose we know that a tangent to a curve exists at a point, how do we go about actually drawing the tangent? The tangent of the angle which a line makes with the positive direction of the xaxis is called the slope of the line.
We have said earlier that the tangent at P is the limiting position of the secant PQ. With reference to a system of coordinate axes OX and OY (Fig. 3), we can also say that the tangent at P is a line through P whose slope is the limiting value of the slope of PQ as Q approaches P along the cuhe. The problem of determining the tangent is, then, the problem of finding the slope of the tangent line.
Fig. 3
Suppose the curve in Fig. 3 is given by y = f(x). Let P (x, f(x)) be the point P and let Q (x + 6x, f(x + 6x)) be any other point on the curve. The prefix 6 before a variable quantity means a small change in the quantity. Thus, 6x means a small change in the variable x. (Caution: 6x is one inseparable quantity. It is not 6 x x). The coordinates(x + 6x, f(x + 6x)) indicate that Q is generally near P. If 8 is the angle which PQ makes with the xaxis, then the slope of PQ = tan 8 = QMIPM
Then limiting value of tan 8, as Q tends to P, (and hence 6x + 0) gives us the slope of the tangent at P. Thus, The slope of the tangent line at (x, f(x)) = :!?,
f(x + 6x)  f(x) 6x
This indicates that the tangent line will exist at (x, f(x)) only if the limit of
f(x + 6x)  f(x) 6x
exists as 6x ,0, Remark 1 In Fig. 3 we have taken 6x to be positive. But our discussion is valid even for negative values of 6x. Let us take an example. Example 1 Suppose we want to determine the tangent to the parabola y = xZat the point P(2,4). In Fig. 4 we give a portion of the parabola in the vicinity of P (2,4).
Fig. 4
Elements o f D i f f e r e n t i a l Calculus
Let Q (x, x2) be any other point on the parabola. The slope of y coordinate of Q  y coordinate of P PQ = x coordinate of Q  x coordinate of P
Equation o f a line passing through a point (x, y ) and having space rn is y  y , = m(x  x i )
The tangent at P (2,4) is the limiting position of PQ as x + 2. Therefore, the slope of the tangent at P is x2  4 lim x+2 x  2 =
lim
x+2
(x + 2)  (x  2) = lim (x x2 x2
+ 2) = 4
The equation of the tangent line will be (y  4) = 4 (x  2) Now, how do we draw this tangent? Just mark the point P' by moving a distance 1 unit from P, parallel to the xaxis to the right, and then, moving a distance equal to 4 units parallel to the yaxis upwacd. Join P to P' as shown inFig. 4. The coordinates of p' are (2 + 1 , 4 + 4). The resulting line will touch the parabola at P, and the slope of the tangent at P = tan 0 = 4.
E
E 1) Find the equation of the tangent to the following curves at the given points. a) y=l/xat(2,1/2)
In this subsection we have given a precise definition of a tangent to a curve. We have also seen how to draw the tangent to a given curve at a given point. Now let us consider the second problem mentioned at the beginning of this section.
3.2.2 "Rate of Change Suppose a particle is moving along a straight line, and covers a distance s in time t. The distance covered depends on the time t. That is s = f(t), a function of time. When the time changes to t + at, the distance covered changes from f(t) = s to f(t + 8t) = s + 6s. Therefore n e can say that 6s is the distance covered in the time 6t. We want to know the average velocrty of the particle during the time interval t to t + 8t (or t + 8t to t, according as t > 0 or t < 0). Now, the average velocity =
Total distance travelled Total time taken
Therefore, the average velocity in the time interval [t, t + 6t] (or [t  6t, t])
Differentiation
But this does not give us the velocity of the particle at a particular instant t, which is called the instantaneous velocity. Rather it is the average velocity over the internal 6f. How do we calculate this? To find the velocity at a particular time t, we proceed to find the average velocity in the time interval [t, t + St] (or [t + t, t]) for smaller and smaller values of 6t. If 6t is very small, then t + 6t is very near t and so the average velocity during the time interval 6t would be very near the velocity at t. It seems reasonable, therefore, to define the lim instantaneous velocity at time t to be gt+O
f(t +tit)f(t) 6t
Thus, we have where s = f(t) is the distance travelled in time t. Comparing this box with the one given at the end of the last subsection, we find that the concepts of the slope of a tangent and the instantaneous velocity are identical. Further, velocity can be considered as the rate of change of distance with respect to time. So, extending our definition of velocity to other rates of change, we can say that if a quantity y depends on x according to the rule y = f(x), then the rate of change of y with respect to x can be defined as
Example 2 Suppose we want to find the rate of change of the function f defined by f(x)=x+5, + x ~ R a t x = O . We shall first calculate the average rate of change off in an interval [O,6x]. This average rate of change off in [O,6x] is
?
Hence, the rate of change off at 0,which is the limiting value of this average rate as 6x
0,
m
Example 3 Suppose a particle is moving along a straight line and the distance s covered in time t is given by the equation F = (1/2)t2. Let us draw the curve represented by the function s = (1/2)$easuring time along xaxis and distance along yaxis. Let P and Q be points on the curve which correspond to t, = 2 and t, = 4. We shall show that the average velocity of the particle in the time interval 1241is the slope of the line PQ and the velocity at time t, = 2 is the slope of the tangent to the curve at t, = 2.
.&
The curve represented by s = (1/2)$ is a parabola, as shown in Fig. 5. P and Q correspond to the values t, = 2 and t, = 4 oft. Now s, = (1/2)tI2= 2 and S, = ( 1 1 2 ) = ~~ 8. Therefore, the coordinates of the pohts P and Q are (2,2) and (4,8), respectively.
I!
82 The slope of PQ = = 6/2 = 3. 42 Also, the distance travelled by the particle in the time (t2 t,) is s2 s, = 8  2 = 6. Therefore, the average velocity of the particle in the time (t2t,) is
i
distance Ravelled =6/2=3. time taken Hence, the slope of PQ is the same as the average velocity of the particle in the time (t2 t,).
I
Further, to calculate the slope of the tangent at P, we choose a point R
1
6t may be positive or negative
o:FS
Elements of Differential
on the curve, near P. Then the required slope is
C~ICUIUS
(slope of PR).
Fig. 5
 1
lim (2+&)'2 2  ,im (4+6t)6t  ]jm 4 + 6t  2  6t+0 26t (2 + 6t)  2  6t+0
2
Ans, what is the velocity at t,? It is
6sJ/Gt,, which is again equal to 2. Thus the velocity
at t, is the same as the slope of the tangent at P.
E
E
Remark 2 i) Example 3 is a particular case of the general result: If the path of a particle moving according to s = f(t) is shown in the tsplane and if P and Q are points on the path which correspond to t = t, and t = 5, then the average velocity of the particle in time (t,t,) is given by the slope of PQ and the velocity at time t, is given by the slope of the tangen; at P. ii) Distance is always measured in units of length (metres, centimetres, feet) and so velocity v really means v units of distance per unit of time. The slope of the tailgent is a dimensionless number, while the velocity has the dimension of lengthhime. Now you can lry some exercises on your own. E 2) A particle is thrown vertically upwards in the air. The distance it covers in time t is given by s(t) = ut (112) gt2 where u is the initial velocity and g denotes the acceleration due to gravity. Find the velocity of the particle at any time t.

E 3) Find the rate of change of the area of a circle with respect to its radius when the radius 1s 2 cm. (Hint: Eipress the area of a circle as a function of its radius first).
E E 4) Find the average rate of'change of the function f defined by f(x)
+ 1, ++x
= 2x2
E
R in the
interval [I, 1 + h] and hence evaluate the rate of change o f f at x = 1.
i t
3.2.3 The Derivative We have seen that the slope of a tangent and the rate of growth have the same basic concept behind them. Won't it be better, then, to give a separate name to this basic concept, and study it independently of its diverse applications? We give it the name "derivative".
Definition 1 Let y = f(x) be a realvalued function whose domain is a subset D on R. Let x ED. If f(x + 6x)  f(x) lim exists, then it is calledthe derivativeof f at x. 6x 6x+O Now, if we write fix + 6x) = y + Fy, then derivative of f =
6y16x. Here 6y denotes the
change in y caused by a change 6x in x. The derivative is denoted variously by f (x), dyldx or Df. The value o f f (x) at a point x, is denoted by f (x,). Thus, 1
lim If, in the expression fix,) = 6x+0
f(xo + Gx)(f(xo) we write 6x
lim x,+ 6x=x, weget 6x=xx,, and 6xo+0
lim x+x,
This is an alternative expression for the derivative o f f at the point x,.
I
Remark 3 In this definition x and y are real numbers and are two dimensionless numbers. If x and y are dimensional quantities (length, time, distance, velocity, area, volume) then the derivative will also have a dimension. For convenience, we shall always treat x and y as dimensionless real numbers. The appropriate dimensions call be added later.
Caution: 'dy' and 'dx' in the expression dyldx are not separate entities. You cannot cancel 'd' !
from dyldx to get ylx. The notation only suggests the fact that the derivative is obtained as a ratio. When f(x) exists, we say that f is differentiable (or derivable) at x. When f is differentiable at each point of its domain D, then f is said to be a differentiable function. The process of obtaining the derivative is called differentiation. The function f which associates to each point x of D, the derivative f(x) at x, is called the derived function off. Thus, the domain of the derived function is {x D: f(x) exists). The process of finding the derivative of a function by actually calculating the limit of the ratio f (x + 6x)  f (x) is called differentiating from first principles. 6x As we shall see later, it is not always necessary to find a derivative from the first principles. We
The notation dyidx is due to Leibniz and f '(x) is due to Lagrange (1 7361 81 3)
Elements of Differential Calculus
shall develop certain rules which can be used to write down the derivatives of some functions without actually finding the limit. Some such rules are contained in the next section.
3.3
DERIVATIVES OF SOME SIMPLE FUNCTIONS
In this section we shall find the derivatives of some simple functions like the constant function, the power function and the absolute value function. We shall illustrate the method of finding the derivative by the first principle method through some examples.
Example 4 Let f :R + R be a constant function, that is, f(x) = c for all x E R, c being a real number. We shall show that f is differentiable, and its derivative is zero.
Hence, a constant function is differentiable and its derivatives is equal to zero at any point of its domain. The result of the above example can be seen more easily geometrically. The constant function f(x) = c x E R represents the straightline y = c which is parallel to the x  axis (Fig. 6). If we join any two points, P and Q, on it, the line PQ is parallel to the xaxis. Hence, the angle made by PQ with the xaxis is zero. This means the slope ofPQ is tan 0 = 0. Since f (x) is the limit of this slope as Q + P, we get f (x) = 0 for all x in the domain off.
Fig. 6
Example 5 We now show that, if n is a positive integer, then D(xn)= nxw'. In order to obtain D(xn),in case it exists, we have to determine (X+ h)" lim h+O h
X"
Notice that we have used the letter h (instead of our usuai 6x) to denote the small change in the variable X. We are, in fact, free to use any notation; but 6x an$ h are the more commonly used ones. lim h+O
=
(x + h)"  xn , h
lim h+O
(xn + nhxn'+...+h n )  xn (by bionomial theorem) h
Differentiation
d
(x") = nx"' for all dx nonzero x E R even when n is a negative integer. The result also holds for all x > 0 if n is any nonzero real number. Of course, if n = 0, then x" = 1 ++ x and hence, Dx" = 0 for all x E R. This means that the result is trivially hue for n = 0. Nevertheless, right now we are in a position to prove this result for n 1/2. That is,
The result of the above example is very useful. We shall show later, that
d dx
( & ) =
(t)
x"~, and this we do in the following example.
Example 6 We shall show that the function f defined by f(x) = f(x + h) f(x) dlo h
We have,
 lim  h+O

lim h+O
=
lim h+O
& ,x > 0 is differentiable.
JT+T ,G h
(&Xi  & ) ( J x X + &) h ( i l + h + A) (x+y)x 
h ( J x + h + hi)
The result of our next example is of great significance. Recall that, in Sec. 2 we mentioned that there are functions that have no tangents at some point (or equivalently, have no derivative there). This example will illustrate this fact. Before giving the example we give some definitions. lim \I0
f ( a + h )  f (a ) , if . ~t . exists, is called the right hand derivative of f(x) at x = a and is written h
as R f (a). Likewise, lim f(a h) f(a) is called the left hand derivative of flx) at x = a and is II+Oh written as L f (a). I f f (a) exists, we must have Rf (a) = L f (a) = f ( a ) (See Unit 2, Theorem 4). +
Example 7 The function f : R + R defined by f(x) = 1 x I is not derivable at x = 0 but is derivable at every other point of its domain.
Fig. 7
Fig. 7 shows the graph of this function. To prove that the given function is not derivable at x = 0, we have to show that
&
is defined fol x < o
Elements o f Differential Calculus
lim IO+hlIOl does not exist. In fact as we shall see, RFf(0) and L f (0) both exist, but they h are not equal. h+O
.
NOW
~f (0) = lim h+O
And L f (0) = lim h+O
IO+hlIOl h
=
1 0 + h '  I OhI
=
lim l h=l h = l ( s i n c e ( h ) = h f o r h > O ) h h
h+0+
Ihl h lim= = l(smce/h/=hforh<0) h+O h h
Therefore, RF'(0) = 1 #  1 = L f (0).Hence f (0) does not exist. We shall now show that the function is derivable at every other point. First, let x > 0. Choose h so that 1 h 1 < x. This will ensure that x + h > 0 whether h > 0 or h < 0. Now, lim h+O
=

f(x + h) f(x) h
lim h+O
 lim  h+O
=
lim h+O
Ix+hl1x1 h
x+hx h
hh= 1
Thus f is derivable at x, and f (x) = 1 for all x > 0. You can now complete the solution by solving E 5).
E E 5) Show that y = ( x I is derivable and f(x)
=
1 at all points x < 0.
E E 6) Show that each of the following functions is derivable at x = 2. Find f ( 2 ) in each case. a)f:R+Rgivenbyf(x)=x b) f : R
+ R given by f(x) = ax + b where a and b are fixed real numbers.
E
E7) Find dy/dx, wherever it exists, for each of the following functions:
a)y=x3b)y=lx+l lc)y= m
,
1 x 27
So far we have obtained derivatives of certain functions by differentiating from the first
t
principles. That is, each time we have calculated lim 6x+O
f(x+Sx)f(x) 6x
But the process of taking limits is a very lengthy and complicated affair. In the section we shall see how to simplify the process of differentiation for some functions.
3.4
ALGEBRA OF DERIVATIVES
Consider the function f(x) =
t
! i ! 6
zx3 + 3 x 2 . If we try to find the derivative of this function from
x 1 the first principles, we will have to do lengthy, complicated calculations. However, a close look at thls function reveals that it is composed of several functions: constant functions like 2.3 and 1, and power functions like x'. x2and x4.We already know the derivatives of these functions. Can we use this knowledge to find the derivative of f(x)? In this section we shall state and prove some theorems which help us do just that.
3.4.1
Derivative of a Scalar Multiple of a Function
Let f R + R be a differentiable function and let c E R. Then, consider the function y,= cf(x). We call this function a scalar multiple off by c (see Unit 1). The derivative of y with respect to x is
Differentiation
Elements o f D i f f e r e n t i a l Calculus
l~nl h+O
f(x
+ h)
'h0lirn
f(x
+ h)  f(x)


f(x)
h
h
(by Theorern 3 of Unit 2)
=cf (x)
Thus, we have just proved the following theorem. 'Theorem 1 I f f is a differentiable function and c E R then cf is differentiable and (cf)'(x)

= c f (x)
Example 8 To differentiate y 7 / x I we apply the scalar multiple rule obtalned in Theoreni 1 at all polnts where the function 1 x ( is differentlable and get d d  (71x1) = 7  (1x1) dx dx t
d d . But, in view of Example 7: when x = 0,  (I x I) does not exist. When x > 0,  (1 x 1) dx dx
= 1 and
d whenxC0,  ( ) X I ) =  1 . dx Therefore,
d (71 x 1) dx
=7
d dx
 (1 x 1)
d and  (7 1 x I) does not exlst at x = 0. dx
Note: In example 8 we have used the fact that if f ( x ) does not exist at a point then (cf).(x) also does not exist at that point. Try the following exercise now.
E E8) Differentiate the following, uslng Theorem 1. b) 8
a) (5!3)x3
3.4.2
A
Derivative of the Sum of Two Functions
We know the sum of two functions f and g defined on R, denoted by f + g, is defined as (f + g) (x) = f(x) + g(x) v x R Let f and g be differentlable from R to R Let us examine whether f + g, the sum of the functions f and g, IS differentiable. Now, llrn h+O  lim  h+O
(f + g ) ( x + h )  ( f +g)(x)
h
{f(x + h ) + g(x + h)  f(x)  g(x)) h

lim
h+O
= f (x)
Differentiation
f(x + h)  f(x) h
+
lim g(x + h)  g(x) h+O h
+ g'(x)
Thus,%e have proved the following :
Theorem 2 The sum of two differentiable functions f and g is a differentiable function and each of (f +g)' (x) = f (x) + g'(x) + x E R. The above result can be easily extended to a finite sum, that is,
where f,, ..., f,, are differentiable functions. 1
Remark 4 ~ r o m Theorems 1 and 2 it fbllows that iff and g are differentiable functions, then f  g is also a differentiable function (since f  g = f + ( g)), and (f  g)' (x) = f(x)  g'(x). Let us see how Theorem 2 is useful in the following exainple.
Example 9 To differentiate 3x2+ 4 1x  9, we apply Theorem 2, and get,
d Now,  (3x2) dx
=3
cbi2 . .  (mvlew of the theorem) &
= 3 x2x=6x
d and  (9) dx
= 0 (see Example 4).
d Thus, (3x2+41x9)=6x+41 dx You are now in a position to solve this exercise.
E E9) Differentiate the following : a) 5x3+ 2
(b) an+ a,x + a2x2+ ... + anxn,where a, E R for i = 1,2, ..., n.
Elements of Differential Calculus
3.4.3
Derivative of the Product of Two Functions
Let f and g be two differentiable functions on R. We want to find out whether their product fg is also differentiable.
l'(x+h)g(x+h)f(x)p(~) 
 lim 
h+O

.
h

(We have added and subtracted f(x) g(;
7
h))
g(x + h)  g(x)
lirn h+O
 lirn .f ( x + h )  f ( x ) 
h+O
lirn h+O g(x + h) +
11
g(x + h)  g(x) lit" h h+O f(x)
iFo
(Ref. Unit 2, Theorem 3)
Thus, we get the following
Theorem 3 The product of two differentiable functions is again a differentiable function and its derivative at any point x is given by the formula.
We can extend this result to the product of three differentiable funct~ons.This gives us (fgh)'(x) = f (x)g(x)h(x)+ f(x)gf(x)h(x)+ f(x)g(x)h'(x) You see, you have to differentiate only one function at a time. This result can also be extended to the product of any finite number of differentiable functions. Thus, i f f , , ... $, are differentiable functions, then, (f,f2,...$,)'( x) = f,'(x)f,(x) ... S1(x)+ f,(x)f,'(x)f,(x) ... $,(x) + ....+ f,(x)f,(x) ...f,,'(x). Theorem 3 is very useful in simplifying calculations, as you can see in the following example.

Example 10 The differentiate f(x) x2(x i 4), nle take g(x) = x2,h(x) = x + 4. We have, f(x) = x2(x+ 4) = g(x)h(x) d d Now, gf(x) =  (x') = 2x and hf(x) =  (x + 4) dx dx

1
Thus, f (x) = g'(x)h(x) +h'(x)g(x) = 2 x ( x + 4 ) + 1 x x2 = 2x2+8x+x2=3x2+8x Remark 5 Yau could also have differentiated x2(x + 4) without using Theorem 3, as follows :
Therefore,
d (x? (X+ 4)) dx

= =
d (x3 + 4 ~ ' ) dx

d d (x3) +  (4x2) (By Theorem 2) dx dx 3x' + 4(2x)=3x2+8x 
This shows that the same function can be differentiated by using different methods. You may use any method that you find convenient. This observat~onshould also help you to check the correctness of your result. (We assume that you would not make the same mistake while uslng tu cr d~ffcrentmethods !1
E E 10)Using Theorem 3, differentiate the following functions. Also, differentiate these functions without using Theorem 3, and compare the results. a)x
3.4.4 Let @
b) ( x +~ 2~~5)2
c)(x+ l)(x+2)(x+3)
Derivative of the Quotient of Two Functions = f/g, where
 lim 

A
f and g are differentiable functions on R, and g(x) ;c 0 for any x. Then,

lim g(x) f(x+ h)  f(x) g ( x + h ) h g (x)g (x + h)
h0

 .lim
g(x>
f(x 
+ h)  f(x)  
h
f(x)
g(x
+
h)  g!x) h
(by adding and subtracting f(x)g(x) from the numerator)
lim g(x
h+O
+ h)  g(x)
lim g ( x
 h+O 
g(x) f f ( x )  f(x)g'(x) (g(xN2
Thus, we get the following.
+ h)
lim g(x) h+O
Differentiation
Elements o f Differential Calculus
Theorem 4 The quotient flg of two differentiable functions f and g such that g(x) # 0, for any x in its domain, is again a differentiable function and its derivative at any point x is given by the following formula :
This can also be written as numerator

(denominator) (derivative of numerator)  (numerator) (derivative of denominator) (denominator)
We will obtain an important corollary to Theorem 4 now.
Corollary 1 If g is a function such that g(x) # 0 for any x in its domain, then
Proof: In the result of Theorem 4, take f to be the constant function 1. Then f (x) = 0 for all x.
Therefore, 
g(x)
x
where f(x) = 1.
0  1 x g' (x)  g' (x) (g(xN2 (g(xN2
d (x") = nxnI, where n is a negative integer and x # 0. dx We have already proved this result for a positive integer n in Example 5 .
Example 11 We shall now show that

Consider the function :R \ (0)+ R given by f(x) = x  ~ where , m E N. Then f(x) = l/x" V x E R. Thus, f = l/g, where g(x) = xmfor all x E R, x # 0.g is a differentiable function and g(x) $0 as x # 0.So, except at x = 0,we find that g' (x) 2 (from Corollary 1) (g(x)) 
 mx'n1
(xmy
(g' (x) = mxn*' by using Example 5)
Denoting m by n, we get f(x) = xn,and f'(x)

= nx"'
Example 12 Let us differentiate the function f given by f(x) = ( x  ~+ 2) / (x2+ 2x) We can write f as the quotient g/h where g(x) = ( x  ~+ 2) and h(x) = x2 + 2x.
Also h'(x)
= 2x + 2.
Therefore, f (x)
=
h(x) g' (x)  g(x)hl(x) (h (x))'
\
E
E 11) Differentiate a)
E E 12)
2x + 1 x+5

b)
1
a
+ bx + cx2 + dx3 '
where a, b, c, d are fmed real numbers
Obtain the derivative of l/f(x) by differentiating from fust principles, assuming that f(x) s0 for any x.
Differentiation
Elements of Differential
Differentiate f(x) =
3.4.5
2 + 5x + 7x' by three different methods. x
Thle Chain Rule of Differentiation
The chain rule of differentiation is a rule for differentiating a composite of functions (Ref. Unit 1). It is a remarkable rule which helps us to differentiate complicated functiolls in an easy and elegant way. We establish the rule in the following theorem. Theorem 5 Let y = g(u) = g(f(x)) = (gof) (x), so that y is the composite function gof. We are given that y, regarded as a hnction of u, is differentiable and u regarded as a function of x is differentiable with respect to x. We want to prove that y, regarded as a function of x, isalso differentiablt. To do this we must show that
:To
3y16x exists where 6y is the change in the
variable y corresponding to a change 6x in the variable x. Now, 6u, be the change in the value of u corresp~ndingto a change 6x in the value of x, is given by 6u = f(x + 6x)  f(x). We have
% :/
6u
=

/:To
(E
6x)
6u lim  x lim 6x ax 6x+O
6x+0
Differentiation
This means that 6u + 0 as 6x + 0 We assume however that 6u # 0. This implies that N~~~ 6~ ' 6u
lirn 6x0
=
. 6~ i:yo
=
SY dy lim exists and is equal to 6u du
6u0
6~ 6~ , and we know that 6u 6x

6Y = dy 6u du and
6u = du lim 6x dx
"0
Hence, we get
6Y
6~ = lim lirn 6X Sx0 6U
Bx0
'
6u lim 6x
6x0
Hence dyldx exists and is equal to dyldu x duldx. You may find it more convenient to remember and use the rule in the following form: If h(x) = g(f(x)) is the c o m p ~ ~ ioft etwo differentiable functions g and f, then h is differentiable and h'(x) = g'(f(x))f (x) where g' (f (x)) denotes derivative of g(f(x))w.r.t f(x) To clarify this rule let us look at the following example
Example 13 Here we shall differentiate y = (2x +
with respect to x.
Letu=2x+ l.Theny(2x+ 1)3=u3. Now y is a differentiable function of u and u is a differentiable function of x. dyldu = 3u2and dy dy du duldx = 2. Hence we can use the chain rule to get  =  . dx du dx You might be thinking that there was really no iiecessity of using the chain rule here. We could simply expand (2x + and then write the derivative. But the situation is not always as simple as in this example. You would appreciate the power of the chain rule after using it in the next example.
Example 14 To differentiate (x' + 2x2 l)loO,

and let u = (x3+ 2x2 1). Then uto0 let = (x' + 2x2Since dyldu and duldx both exist, and dyldu = 1 0 0 and ~ ~duldx ~ = 3x2+ 4x, therefore, by chain rule, dyldx = dyldu = dy/du.du/dx. = 100 u99. (3x' =
+ 4x)
100 (x3 .i2x2
(3x21qx)
Can you really attempt to solve the above example without using the chain rule? Don't you think the rule has simplified matters a lot for you ? Instead of introducing u explicitly each time while applying the chain rule, after a little practice you would find it more convenient to do away with u and arrange the working in the above example as follows :
Elements o f Differential Calculus
Our next example illustrates that this rule can be extended to three functions. Example 15To differentiate [(5x + 2)2 + 314,we write y = ((5x +2)2 + 3j4, u = (5x + 2)2 + 3 and v=5xt2 Then y = u4 and u = v2 + 3. That is, y is a function of u, u is a function of v, and v is function of x. By extending the chain rule, we get dy  dy dy dv dx du du dx This gives, dy/dx = 4u3 . 2 v . 5 = 40 u3v =
40 [(Sx + 2)2 + 313 ( 5 +~ 2)
This example illllstrates that there m2y be situations m which we may go on using chain rule for a function of a function ..., and so on. This perhaps justifies the name 'chain' rule. Thus, if g, . . . g,, and h are functions such that h = (gIog2o. . . og,,) (x), then h'(x)
E
= g',
!g,
0
. . . ogll( 4 ) gt2(gj 0 . . .ogJ (4.. . gtlFI (g,,) (x) . g',, (xi
E 14) Find dy/dx for each of the following using the chain rule:
CONTINUITY VERSUS DERIVABILITY We end this unit with the relationship of differentiability with continuity, which we have studied in Unit 2. In Sec. 3 ofunit 2 we proved that the functon y = ( x I is continuous V x E R. We have also proved that this function is derivable at every point except at x = 0 in Example 7. This means that the function y = I x I is continuous at x = 0, but is not derivable at that point. Thus, this shows that a function can be continuous at a point without being derivable at that
point. However, we will now prove that if a function is derivable at a point, then it must be continuous at that point; or derivability * continuity. lirn f (x) = f(x,). Recall that a function f is said to be continuous at a point x, if x,xo Theorem 6 Let f be a function defined on an interval I. Iff is derivable at a point x,E I, then it is continuous at x,. Proof:
*
If x x, then we may write f(x)  f(x,)
=
f(x)  f(x,) X
 Xo
.(xxd)
Since f a derivable at x,, theiefore. lim %k existsand' equals l% (f(x,). d X
X:Xo

Thtls. taking limits as x + xo, we have,
~T:
[f(x)  KxJI
1
 lim X+XU
(x 
xO)l
=f ( s ) x 0 = 0 lim f(x)  s,so lim f(%) = 0 Therefore .,," n a t is,
lim ,,Xo
f(x)

lim
,,so
f(x,) = f(x,)
Consequently, f is continuous at x,. As we have seen, the function y = 1 x 1 is continuous but not derivable at only one point, x = 0. But there are some continuous functions which are not derivable at infinitely many points. For instance, look at Fig. 8.
Fig. 8
It shows the graph of a continuous function which is not derivable at infinitely many points. Can you mark those infinitely many points at which this function is not derivable ? You can take your hint from the graph of the functon y = 1 x I. 'The situation is, in fact, much worse. There are functions which are continuous everywhere but differentlahle nowhere. The discovery came as a surprise to the nineteeth century mathematicians who believed, till then, that if a function is so bad that it is not derivable at any point, then it can't be so good that it is continuous at every point. The first such function was put forth by Weierstrass (although he is said to have attributed the discovery to Riernann) in m
bn cos (anm),where a is an odd
18 12. He showed that the function f given by f(x) = n0
integer and b is a positive constant between 0 and 1 such that ab > 1 + 3x12, is a function which is continuous everywhere, but derivable nowhere. It will not be possible for us to prove this assertion at this stage. Sometimes we use Theorem 6 to prove that a given function is continuous at a given point. We prove tbat its derivative exists at that point. By Theorem 6 then, the community automatically follows.
E
. (2x + 3yo
E 15) Is the function f : [0, 11 +R : f(x) =
9x + 2
co~~tinuous at x = 0.1?
Differentlation
/
Elentents of Differential . Calculus
3.6 SUMMARY We conclude this unit by summarising what we have covered in it. 1.
For any function y = f(x) f(x +ax)  f(x) (if it exists) is called the derivative off at x, denoted by f(x). 6x The function f is the derived function. The derivative f(x) is the slope of the tangent to the curve y = f(x) at the point (x, y). The derivative also gives the rate of change of the function with respect to the independent variable. lirn
SX+O
2.
The defivative of a constant function is 0.
where n is any integer (and x # 0 if n 10).
1
x j is derivable at every point except at x = 0.
3.
The fuhction y
4.
(cf)' = c f , c a constant.
(fg)' = f g' (fg)'
fg' + g f
=!=
g' f  f' g
(fig)' =
5.
g
Eve@ derivable function is continuous. The converse is not true, that is, there exist functions which are continuous but not differentiable.
3.7 SOLUTIONS AND ANSWERS El) a)
zl
dY
=114.
Equation : (y  112) = (114) (x  2) or x + 4y = 4
= 3.
Equation: (y  1) = 3(x  1)
x=l
x=I
~("1
Differentiation
=4n
E3) area = A = nr2 . dr
,=,
El) average ratc of change off in [ I , 1 + h]
rate of changc. ,f f at x = 1 = 4
lim I x
=
f (1 + h)  f(l) h
f(l + h)  f(l) i%  , where h may be positive or negative. h
+ hl  1x1  (x + h)  (X)
h h h h Thus f (x) =1 if x < 0. f is derivable for x < 0. h+0
  =1
ah  =a.
 lim
h
h+O
(X + h  lim h+O
+ 1)  ( X + 1) h
= 1.
Thus dyldx exists when x >I. It does not exist when x < 1. It does not exist when x.= 1 since R f (1) = 1 and Lf (I)=1. )
dy  lim
dx
h+O
J

h
Jm + ,/G mi + J K F i
Elements o f Differential Calculus
=
1 f(x)  f(x+ h) lim h f(x) f(x + h)
h+O
f(x + h)lim f(x) . h f(x) lim f(x + h) h+O
h+O
f '(x)
 
fW2
~ i 5f)(x) =
50(2x + 3j492 (9x + 2)  9 (2x + 3)" (9x + 212
exists at x = 0.1. Heilce the function is continuous at x = 0.1
UNIT 4
DEFUVATIVES OF TRIGONOMETRIC FUNCTIONS
Structure 4.1
Introduction Objectives
4.2
Derivatives of Trigonometric Functions 4.2.1 Some Useful Lirn~ts 4.2.2 Derivatives o f Sin x and Cos x 4.2.3 Derivatives of other Tr~gonometric Funct~ons
4.3
Derivatives of Inverse Functions The Inverse Function Theorem
Derivatives of Inverse Trigonometric Functions
4.4
4.4.1 Derivatives o f sin' x and COS' x 4.4.2 Der~vativeso i ~ e c  'x and ~ o s e c  'x
4.5 4.6
Use of Transformations S m y
4.7
Solutions/Answers

4.1 INTRODUCTION In Unit 3 we have introduced the concept of derivatives. We have also talked about the algebra of derivatives and the chain rulc which help us in calculating the derivatives of some complex functions. This unit will take you a step further in your study of differential ca~culus. In this u n ~ twe shall first find the derivatives of standard trigonometric functions. We shall then go on to study the inverse function theorem and its applications in finding the derivatives of inverses of some standard functions. Finally, we shall see how the use of transformations can simplify the problem of differentiating some functions.
After study~ngthis unit you should be able to : find the derivatives of trigonometric functions @ @
@
state and prove the inverse function theorem use the inverse function theorem to find the derivatives of inverse trigo~lometric functions use su~tabletra~lsformationsto differentiate given functions.
4.2 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS In this section we shall calculate the derivatives of the six trigonometric functions; sin x, cos x, tan x, cot x, sec x and cosec x. You already know that these six functions are related to each other. For example, we have: i) sinx + cos'x = 1 ii) tan x = sin d c o s x, and many more identities which express the relationships between these functions. As you will soon see, our job of finding the derivatives of all trigonometric functions becomes a lot easier because of these identities. But let us first evaluate some important limits which will prove to be very useful later.
4.2.1
Some Useful Limits
In the next subsection we shall come across lim (+I1
sin t !
.
: > !
5
st;let's
try to
Elements of Differential Calcutus
calculate these limits. For this, we first assume that 0 < t < n/2 and consider a circle with radius 1 unit, given by x2 + y2 = 1 as shown in Fig. 1. The line OT passes through the origin and has slope = tan t. Therefore, we can write its equation as y = x tan t. This means that the ycoordinate of the point T is tan t. since its xcoordinate is 1.
Fig. 1
From the figure we can see that If the sectorial angle is 0, the area o f a sector o f a circle o f radius r is (1/2)r28
area of AOPA < area of sector OPA < area of AOTA Now, the area of AOPA
1
I
2
2
 x 1 x PB =  sin t,
=
I The area of sector OPA = 
x
2
1 The area of OTA = 2
x
1
x
1x t
=
1 2
 t,
tant t
Thus, inequality (1) can be written as: sint < t < tad t
... (2)
Since 0 < t < n12, sin t > 0, therefore, from the lefthand inequality in (2) we get 0 < sin t < t ...(3) Now, if 1~12< t < 0, then 0 < t < n12, and applying (3) to t,;0 < sin (t) < t or 0 <  sin t < t since sin (t) =  sin t. This means that if 1~12< t < 0, then t < sin t < 0 ...(4) Here we arc using various results about the order relation from Unit 1.
0
sin t
xl2
t
7~12
t
sin t
0
(a)
Fig. 2
In Fig. 2(a) and (b) you can see the representation of (3) and (4), respectively. We can prove that lim sin2 t0
t/2 = 0 by using Theorem 3 o f Unit 2 and by noting that t 4 0 o t / 2 4 0
We can combine (3) and (4) and write
 1 t 1 < sin t < 1 t 1 for n/2 < t < xI2, t # 0. You have seen in Unit 2 that lim ( t ( = 0. From this we can also say that lim 1
0
1
Now applying the sandwich theorem (Theorem 2 of Unit 2) to the functions 1 t and I t I, we get that lim sin t
=
0
t (
=
0.
tbn
1, sin t
We shall use this result to calculate lim cost. As you know, cos t t0
means lirn cos t
=
lirn (1  2 sin2 t/2)
=
1  2 lim sin2 t/2
10
=
1 2 sin2 t/2. This
t+O
=
1
t+O
Thus, we get lirn cos t
=
1
t+O
Now, let's get back to inequality (2): sin t < t < tan t. for 0 < t < 7d2. Since 0 < t < n12, sin t > 0, and therefore, after dividing by sin t, (2) becomes;
or cos t < sin t/t<l ...(5 ) , 0 < t < n/2 Now, since sin (t) =  sin t, we see that sin ( t)/(t) = sin t/t. This alongwith the result cos (t) = cos t, shows that the inequality (5) holds even when n/2 < t < 0. Thus, cos t < sint/t < 1,n/2 < t <n/2. t f O Now, let us apply the sandwich theorem to the functions cos t, sin t/t and 1, and take the limits at t + 0. This give us: lirn sin t t/t
=
1
t+O
Example 1. Suppose we want to find out sin 3 x

sin 3 x
;2 sin7x sin 3 x sin 3 x sin 3 x Let us first calculate l3 . For this we shall write  lirn t+o
X
and
X
X
replace 3x by t in the right haad side, and take the limit as x sin 3x also tends to zero, and lirn x+o x
lim x+O
t
t+O
lim sin t c+o
'3
To calculate
sin t lim 
=
x
X
x
3. If we
+ 0, we find that t

3x
3
(See Theorem 3 of Unit 2)
Is sin 5x we start by writing
sin 5 x sin 7 x
sin 5 x x sx
lirn x+O
5 11m . sin  5x 7
5x
x+O
7x 7 srn7x
lim x+O
X
3 
7
7x sin 7 x
f


7
Remark 1 in
I
since lim x+o
lim sin t
,+, = 1 or t
7x sin7x
1 = 1by Theorem 3of Unit 2 lim (sin7x1 7x) x+o
lim cost
=
t+O
1, the angle t is measured in radians. If in a
particular problem, the angles are measured in degrees, we have to first convert these into radians before using these formulas. Thus, sin to  lim sin.(nt 1180) n sin (nt I 180) n lim  1 8 0 t lim =t+o t t+O t +o ntl180 180 See if you can solve this exercise now.
E E 1) Prove that
a) lirn cos (a + x) x+o
b) lirn sin (a x+o
=
cos a
+ x) = sin a
Derivatives of Trigonometric Functions
Derivatives of Sin x and Cos x
4.2.2
We shall now find out the derivative of sin x from the first principles. If y then by definition dy dx
sin (x
 = lim Remember the formula sin A  sin B = 2 sin
h 2sin (h / 2)cns (x h
lim h4
(A+W ?
COS
2
sin (h12)
lim
=
h/2
h+o
=
1
x
f(x) = sin x,
+ h)  sinx
h+o

(A9) 2
=
cos
X
+ h/ 2)
lim cos (x + h/2) h+o
= cos
X
Thus, we get
d
(sin x)

dx
= cos x
Now, let us consider the function y
=
f(x) = cos x and find derivative. In this case,
cos (x + h)  cosx dy  li~n dx
h
h0
=
2 sin (h / 2) sin
lim
+ h / 2)
h
h0
=  lim h+O
(X
sin (h / 2) lim sin (x h 12 h+O
4
hi2)
  sin x .
Thus, we have shown that d  (cos x) = sin x dx
d Actually, having first calculated  (sin x), we could found out the derivative of cos x dx by using the formula : d
(sin
(X
+
cos x = sin (x i. ~ 1 2 )This . gives us, d2) = cos (x +
d2) can be proved by the chain rule.
using
d
(cos x) dx
= =
d
 (sin (X+ 7ti2))
dx cos (x
+ x/2) =  sin x
In the next subsection we shall find the derivatives of the other four trigonometric functions by using similar formulas. But before that it is time to do some exercises.
E
E 2) Find the derivatives of the following: a) sin2x b) cos2 x c) 5 sin7 x sin3x e) cos (sin x)
d) x3 cos 9x
Derivatives o f Trigonometric Functions
Derivatives of other Trigonometric Functions
4.2.3
We shall now find the derivatives of i) tan x ii) cot x iii) sec x iv) cosec x. i)
Suppose y
Hence,
dy dx
=

f(x)
=
tan x. We know that tan x
d (tan x) dx
= 
sin x =
=
=
d (sin x) d (cos x) cos x  sin x dx dx
 cos2 x
+ sin2 x
c0s2 X
1 cos 2 X
=,
2
= sec x
ii)
Now, suppose y
=
fix) = cot x. Since cot x

)
Now, let y ii), we get d dx
=
2
tan x
=
cosecL x
f(x) = sec x. Since we know that sec x = llcos x, proceeding as in
 (sec x) = I
lltan x, we get
tan x d(1) 1 dx  1 d (tan x) 1 dx tan2 x
=
1
=
d dx

(L)
cosx
sin x =
c0s2
=
sec x tan x
If you have followed i), ii) and iii) above, you sould not have any difficulty in finding the derivative of cosec x by using cosec x = llsin x.
d (u/v)= dx
vduldxudvldx v2
Elements o f Differential Calculus
E
d E3) Show that  (cosec x) dx
=
cosec x cot x.
Let us summarise our results. Table 1
Function
Derivative
sin x
cos x
cos x
 sin x
tan x
sec2 x
cot x

sec x
sec x tan x
C O S ~ Cx
 cosec x cot x
cosec2 x
Remark 2 Here again we note that the angle is measured in radians. Thus d dx
cos
dx
(E)= 180
x
 COS
180
x0
We shall now see how we can use these results to find the derivatives of some more complicated functions. The chain rule and the algebra of derivatives with which you must have become quite familiar by now, will come in handy again.
Example 2 Let us differentiate i) sec3 x ii) sec x tan x i)
Let y dy dx

=
3
sec x. If we write u
dy du du dx
= 3u' =
ii)
=
sec x, we get y
=
+ cot x u3. 'rhus,
sex x tan x 3
3 sec x tan x
If y = sec x tan x + cot x, then, dy  d d (sec x tan x) +  (cot x) dx dx dx d d 2 = sec x  (tan x) + tan x  (sec x)  cosec x dx _ dx . = sec x (sec2 x + tan' x)  cosecZ x
Remark 3. sin x, cos x, sec x, cosec x are periodic functions with period 2x. Their derivatives are also periodic with period 2x. tan x and cot x are periodic with period x . Their derivatives are also periodic with period x. We have been considering variables which are dimensionless. Actually, in practice, we may have to consider variables having dimensions of mass, length, time etc., and we have to be careful in interpreting their derivatives. Thus, we may be given that the distance x travelled by a particle in time t is x = a cos bt. Here, since bt has to be dimensionless 1 (being an angle), b must have the dimension  Similarly, x/a = cos bt has to be
T'
Derivatives of I r ~ g b n o m e t r i cFunctions
dimensionless. This means that a must have the same dimension as x. That is dimension of a is L. Now M d t =  ab sin bt has the ension as ab = L x 1/T = LIT, which is not unexpected, since dddt is nothing but the velocity of that particle. See if you can do these exercises now.
E
E 4) Find the derivatives of : a)
4.3
b) cot x +
cosec 2x
,/=
c) Scot 9x
DERIVATIVES OF INVERSE FUNCTIONS
We have seen in Unit 1 that the graphs of a function and its inverse are very closely related to each other. If we are given the graph of a function, we have only to take its reflection in the line y = x, to obtain the graph of its inverse. In this sectioihwe shall establish a relation between the derivatives of a function and its inverse.
The Inverse Function Theorem Let us take two functions f and g, which are inverses of each other. We have already seen in Unit 1 that in this case, gof(x) = g(f(x) = x, for all x for which f is defined, fog(y) = f(g(y)) = y for all y for which g is defined. Now, suppose that both f and g are differentiable. Then, by applying the chain rule to differentiate g(f(x)) = x, we get g'(f(x)).f(x) = 1 or g' (y).f (x) = 1, where g = f(x) where g' (f(x)) is derivative w.r.t f(x). This means that if f ' (x) # 0,we can write g'(y) = l/f '(x). So we have been able to find some relation between the derivatives of these inverse functions. Let us state our results more precisely. Theorem 1 (The Inverse Function Theorem) Let f be differentiable and strictly monotonic on an interval I. If f '(x) I, then f' is differentiable at y = f(x) and ('IT
(y) = llf )(XI.
Thus, we have the inverse function rule: 1 ,
('I)'
(Y) =
dx 
Or
1
dy dy / d x
#
0 at a certain x in
.
The strict monotonicity condition in this theorem implies that f is oneone and thus ensures the existence of
,
1,
Elements o f Differential Calculus
The derivative of the inverse function is the reciprocal of the derivative of the given function. Soon we shall see that this rule is very useful if we want to find the derivative of a function when the derivative of its inverse function is already known. This will become clear when we consider the derivative[ af the inverses of some standard hctions. But t~ ÂŁidt k&ti!&%e off = 2,where r is a rational number. In &LH L29e d unit 3 we have already proved that (xn) = nxwl when n is an integer. We shall use
dx
this fact in proving the general case. x' may not be always defined. For if
I and r = 112, x' = not defined in R.
fi
=
Theorem 2 If y = f(x)
is
defined, then
= xr, where
r is a rational number for which xr and xr' are both
d
(xr) = rxrI. dx
Proof: Let us first consider the case when r = llq, q being any nonzero integer. In thls case, y
1
= f(x) = x
. Its inverse function g will be given by x = g(y) = yq. This means
9 Thus, by the inverse function rule, we get
So far, we have seen that the theorem is true when r is of the form llq, where q is an integer. Now, having proved this, let us take the general case when r = plq, p, q E Z (q is, of course, nonzero). Here,
Now,
d
dy Thus, dx
d ( x " ~ )=~ p(~liq)pl (xliq), by chain rule dx = p(X"4)P' (l/q) x(ll*l
d = dx
(xr) = (p1q)x(plqtl = ,r.'1
This completes the proof of the theorem. The usefulness of this theorem is quite clear from the following example.
Example 3. Suppose we want to differentiate =
(X5/6+
JX
)1111
We write u = x5I6 + & . This gives us, y By chain rule, we get
=
u""
d~ . Thus, dx
gj (X
+
J; ),,11 (5x1!6
+
,<~n)
Why don't you try these exercises now ?
E
E 5) Differentiate a) 5(x3+x'I3)
4.4
b) (&s
 &9 )x2
DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
In the last section we have seen how the inverse function theorem helps us in finding the derivative of xn where n is a rational .number. We shall now use that theorem to find the derivatives of inverse trigonometric functions. We have noted in Unit 1, Section 5, that sometimes when a given function is not oneone, we can still talk about its inverse, provided we restrict its domain suitably. Now, sin x is neither a one~one,nor an onto function from R to R. But if we restrict its domain [ x12, 7~ '1, and GOdomainto [I, 11, then it becomes a oneone and onto function, and hence the existence of its inverse is assured. In a similar manner we can talk about the inverses of the remaining trigonometric functions if we place suitable restrictions on their domains and codomains. Now that we are sure of the existence of inverse trigonometric functions, let's go ahead and find their derivatives.
4.4.1 Derivatives o f Sin' x and CosI x Let us consider the function y = f(x) = sin x on the domain [d2, x/2]. Fig. 3(a) shows We can see clearly the graph of this function. Its inverse is given by g(y) = that sin x is strictly increasing on [x/2, x2].
Fig. 3
Elements of Differential Calculus
d We also know that the derivative  (sin x)
dx
=
cos x exists and is nonzero for all
x E ] x/2, x/2[. Since
sin
x
= Y,
cos x
=
This means that sin x satisfies the conditions of the inverse function theorem. We can, therefore, conclude that sin' y is differentiable on. 11, 1[, and
for
J1 y2 rd2 < x < rd2.
d  (sin' (y))
=
ff(x)
dy
cosx
Thus, we have the result
Remember, sin'x is not the same as (sin x)' = llsin x br sin x' = sin l/x. >
1 1  
1
a . u
,
 (sin' t) dt
=
Jg
Fig. 3(b) shows the graph of sin' x. We shall follow exactly the same steps to find out the derivative of the inverse cosine function. Let's start wlih the funciton y = f(x) = cos x, and restrict its domain to [0, x] and its codomain to [I, 11. Its inverse h c t i o n g(y) = cos' y exists and the graphs of cos x and cos' x are shown in Fig. 4 (a) and 4(b), respectively.
Fig. 4
As in the earlier case, we can now check that the conditions of the inverse function theorem are satisfied and conclude that cos' y is differentiable in 11, l[. Further
Since cos x sin x =
=
y,
for
This gives us the result
You can apply these two results to get the derivatives in the following exercises.
E
E 6) Differentiate a) sin' (5x) b) cos' & c) sin x cog' (x3 + 2)
Derivatives o f Trigor~omrtric Fur~ctions
E
E 7) a) By looking at the graph of tan x given alongside, indicate the interval to which the domain of tan x should be restricted so that the existence of its inverse is guaranteed. b) What will be the domain for tan' x'? d I c) Prove that  (tan x) = ll(1 + x2) in its domain. dx
In this section we have calculated the derivatives of sin' x and cos' x and if you have done E 7), you will have calculated the derivative of tan' x also. Proceeding along exactly similar lines, we shall be able to see that d (cot' x) dx

1 =
x
4.4.2 Derivatives of Sec' x and Cosec' x Let's tackle the inverses of the remaining two trigonometric functions now. To find sec:'
x, we proceed as follows : Remember, we have seen
If y = sec' x, then x = sec y or llcos y y = cos' (llx), where, I x I I1.
From this we get
=
x, which means that Ilx
=
cos y. This gives us
that cos' t is detined in the interval [  I , I ] .
Elements of Differential Calculus
1
=
d
J C ' G 2 .m (llx) 4x1
Note that although sec' x is defined for 1 x 2 1, the derivative of ses x does not exist when x = I.
/
Thus, we have
E
E 8) Following exactly similar steps, show that
Example 4 Suppose we want to find the derivative of y By chain rule, we get
= sec'
2
.
Now, you will be able to solve these exercises using the results about the derivatives of inverse trigonometric functions.
E
E 9) Differentiate, a) cot' (d2)
b,
cot' ( x + 1) tan' (x + I ) x sin 8
c) cos' (5x + 4)
h)
secI
(
]
1 x cos 8
USE OF TRANSFORMATIONS
4.5
Sometimes the process of finding derivatives is simplified to a large extent by making use of some suitable transformations. Jn this section we shall see some examples which will illustrate this fact.
Example 5 Suppose we want to find the derivative of
As you know, we can differentiate this function by using the formula for the derivative of cos' x and the chain rule. But suppose we put x = cos 8, then we get y

(4 cos3 8  3 cOs 8) cos' (cos 38) (cos 38 = 4 c0s38  3 cos 8) = 38 = 3 cos' x. Now this is a much simpler expression, and can be differentiated easily as: = COS'
2
Example 6 To differentiate y This gives us,
=
tan'
[' :

'1
, we use the transformation x
=
tan 8.
Elements o f Differential Calculus
sec 0  1
( ) 1
cos 0 sin
=
tan'
=
tan' (tan 012)
=
012
=t
=
[
n
1  ( 1  2 s i n 2 012 2 sin 012 cos 012
I
tan' x 2
 .
1 dy Now, we c q write ~ dx 2(1 + x2) Let's tackle another problem.
Example 7 Suppose we want to differentiate tad'
For this, let y =tan'
(s]
with respect to sin'
(5).
(3)(5)
. Our aim is to find dyidz. We shall
and z = sin'
use the transformation x = tan 0. This gives us
Y
= tan'
2 tan 0 [l 
z
= sin'
(I
= tan'
(tan 20) = 20, and
:z:~]
=sin'(sin20)=20.
Now if we differentiate y and z with respect to 0, we get dy/d0 = 2 and &d0 Therefore,
dy
= 2.
dy/ d0
= = 1. dz d z / d 0
Alternatively,' we have y
= z.
Hence, dy/dz = 1
So, you see, a variety of complex problems can be solved easily by using transformations. The key to a successful solution is, however, the choice of a suitable transformation. We are giving some exercises below, which will give you the necessary practice in choosing the right transformation.
E
E 10) Find the derivatives of the following functions using suitable transformation : a) sin' (3x  4x3)
b) cos' (1  2x2) d) tan'
1(
+
Derivatives o f Trigonometric Functions




Now let us summarise the points covered in this unit.
SUMMARY
4.6
In this unit we have 1. calculated the derivatives of trigonometric functions:
Function sin x cos x tan x cot X sec x C O S ~ Cx
2.
cos x  sin x
sec2 x 2 cosec x sec x tan x cosec x cot x
discussed the inverse function theorem and used the rule d dx
f
3.
Derivative
1
X )= 
f ' (Y)
d (x') = rxrI , where r is a rational number. to prove that dx
used the inverse function theorem to find the derivatives of inverse trigonometric functions:
Elements o f Differential Calculus
4.
used transformations to simplify the problems of finding the derivatives of some functions.
4.7
SOLUTIONS AND ANSWERS lim cos (a + x)
xro
b) similar
=
cos a lim cos x  sin a lim sin x
.=
cos a
xro
b) 2cos x
E2) a) 2 cos 2x
xbo
d dx
 (cos x) = 2 sin x cos x
c) 5(3 sin7x cos 3x + 7 sin6 x cos x sin 3x) = 5 sin6 x (3 sin x cos 3x + 7cos x sin 3x) d) 3x2 cos 9x  9x3 sin 9x e) sin (sin x) cos x
& (A)
sin x
d E 3)  (cosec x) dx
=

=
COS X
  CoseC
 sin2
x
0  I cos x
sin2 x
X
Cot X.
E4) a)  2 cosec 2x cot 2x
b)  cosec2 x +
j
1
( c o s s x cot x)
c)  45 cosec2 9x.
E S) a) 5(3x2 + x~"!
c) sinx
j
3x2
+ cos x cosd (x3 + 2)
E7) a) tan x restricted to 1 7~12,d 2 [ is a strictly increasing oneone b c t i o n of x. Thus, its inverse exists when restricted to 1d2, x!2[. b) The domain of tan' x is ] 4,@[. c) If y

b
f(x) = tan x,
1
Hence didx (tan' x) =
1 .
*
E 8) y = cosec' x cosec y = x = sin y y = sin' (Ilx) where ( x I 2 1.
dy
T ~ U S, dx
d dx
=  (sin . I
(llx))
=
l/x
*
Derivatives o f Trigonometric Functions.
"I
(tan' (x
E 10) a) Put x = =
b)
=
sin 8
+ y = sin..'
+I))~
(3x  4x3)
sin' (3 sin 0  4 sin3 0) sir' (sin3 8) = 30 = 3sin' x.
x
cos' (1  2x2) cos1 (1  2 cos2 012) 'y = cos' (cos 0) = cos' (cos (rc  0)) = rc  0 = rc  2 cos' x. =
cos 8!2
3
y
=
=
e 3 y = sin'
c) put x
=
tan
d) Put x
=
tan 0
e) Put x
=
tan 0
(3) =
2 tan' x
UNIT 5
DERIVATIVES OF SOME STANDARD FUNCTIONS
Structure Introduction Objectives
Exponential Functions 5.2.1 Definition o f an Exponential Function 5.2.2 Derivative of an Exponential Function
Derivatives of Logarithmic Functions 5.3.1 Differentiating the Natural Log Function 5.3.2 Differentiating the General Log Function
Hyperbolic Functionls 5.4.1 Definitions and Basic Properties 5.4.2 Derivatives of Hyperbolic Functions 5.4.3 Derivatives of Inverse Hyperbolic Functions
Methods of Differentiation 5 5.1 5.5.2 5.5.3 5.5.4
Derivative of xr Logarithmic Differentiation Derivatives o f Functions Defined in Terms o f a Parameter Derivatives of Implicit Functions
summary
5.1 INTRODUCTION Exponential functions occupy an important place in pure and applied science. Laws of growth and decay are very often expressed in terms of these functions. In this unit we shall study the derivatives of exponential functions. The inverse function theorem which was stated in Unit 4 will then help us to differentiate this inverse, the logarithmic functions. In particular, you will find that the natural exponential function in its own derivative. Further we shall introduce and differentiate hyperbolic functions and their inverses since they hold special significance for physical sciences. We shall demonstrate the method of finding derivatives by taking logarithms, and also that of drfferentiating implicit functions. With this unit we come to the end of our quest for the derivatives of some standard, frequently used functions. In the next block we shall see the geometrical significance of derivatives and shall use them for sketching graphs of functions.
Objectives After studying this unit you should be able to : find the derivatives of exponential and logarithmic functions define hyperbolic functions and discuss the existence of their inverses differentiate hyperbolic functions and inverse hyperbolic functions use the method of logarithmic differentiation for solving some problems differentiate implicit functions and also those functions which are defined with the help of a parameter.
5.2 EXPONENTIAL FUNCTIONS Our main aim, here, is to find the derivatives of exponential functions. But let us first recall the definition of an exponential function.
5.2.1
Derivatives of Some Standard Functions
Definition of an Exponential Function d
A function f defined on R by f(x) = ax, where a > 0, is known as an exponential function. Now to find the derivative of i:we shall have to take the limit: ax+b
lim
h+O
SO,

ax
h
a" may not be defined for all
ah  I h
...
= a x . lim h,O
x if a<O. For example, (2)"' i s not defined In
ah  I
: !A
 k, we get
if we put
k
h
cl
a' = k . a'. We can also Interpret k as the derivative of a" at x dx see the graphs of exponential functions for various values of a. 

Fig. 1
All these curves pass through (0, 1) as a0 = 1 for all a. Now from all these curves, we shall choose that one, whose tangent at (0, 1) has slope = 1. (We assume that such a curve exists). The value of a corresponding to this curve is then denoted by e. Thus, we have singled out the exponential function : x + ex, so that its derivative at x = 0 is 1. Thus,

e h 1 lim  1 h>O h This also means that dex  ex el'~  lim . e x .lim = e dx h>O h h+O .h That is, the derivative of thls function is the funct~onitself. Thls special exponential function IS called the natural exponential function.  
5.2.2 Derivative of an Exponential Function In Unit 1, we compared the graphs of natural exponential function ex and the natural logarithmic function Inx and found that they are reflections of each other w.r.t. the line y = x (see Fig. 2). We concluded that ex and lnx are inverses of each other. This also means that el"" x y x > 0.

big. ?
=
h a  I lim hto h
=
 (a")
0. In Fig. 1 you can
A Y

H.
d
dx
I
I
 0
Elements o f Differential CnIcuIus
F om this we can write
C
ax = elna' , or ax = exlna,where a > 0.
h a b = blna I
ex In a
d (x In a) by chain rule
dx = e x In a I n a = ax In a.
Remark 1 If we compare this result with (1) which we derived at the beginning of this section, we find that
,
ah  1 In a = lim h Thus, we have
d ex = ex, and dx
Example 1 Let us use these formula to find the derivatives of
ii)
9
Let y
,

bc.
E
E
1,
I ur
dpp
.
ill)
asin' r
. Then, by chain rule

i) W s
ex + eX ex eX
(ex e') (ex  eX) .(e' +e') (r' +eX) (e\  e' ) =
'he chain rule again to different~atc,"" " x
vou can solve these exercises now the derivatives of :
Derivatives of Some Standard Functions
E 2) How much faster is f(x) = 2x increasing at x
=
112 than at x
=
O?
E
lim In this section we have defined e as that real number for which h+o
h
h
=
1.
Alternatively, e can also be defined as a limit: e
lim (1 + lin)", or as the sum of an infinite series: e
= h+O
=
1 1 1 +  +  + .... l! 2!
But all these definitions give the same value, e = 2.718281828 ... e is an irrational number. In many situations the rate of growth (of human beings, or bacteria or radioactive particles) is proportional to the present population. That is, if x(t) is the population at dx time t, then .c x. In such situations the exponential function is of great relevance dt d since  (e') = et dt Now let us turn our attention to logarithmic functions.
1
5.3
DERIVATIVES OF LOGARITHMIC FUNCTIONS
In Unit 4, we studied the inverse function theorem, (Theorem 1, Unit 4) and used it to find the derivatives of various functions such as sin'x, cos'x, and so on. Here, we shall, yet again, apply this theorem to calculate the derivative of the natural logarithmic function.
5.3.1 Differentiating the Natural Log Function Consider the function y = lnx. This is the inverse of the natural exponential function, that is, y = lnx if and only if x = eY. From Fig. 2, you can see that the natural exponential function is a strictly increasing function. (You will be able to rigorously prove this result by the end of this course). Further, the derivative of the function x = eYis
Inx is defined on 10 [
Elements o f Differential Calculus
d) d =  (ey) = e' , 0 for all y E R. dx dx Thus, all the conditions of the inverse function theorem are satisfied. This means we can conclude that the derivative of the natural logarithmic function (which is the inverse of the natural exponential function) exists, and dy dx

d (Inx)
dx
1
1  1
=  
dxldy
eY
x
Thus, we have
Let's see how we can use this result. Example 2: Suppose we want to differentiate y
Note that x2  2x
+
=
In (x"
2x + 2).
2 = (x  I ) +~ 1 and hence, is nonzero for all x.
Therefore, ln(x2 2x + 2) is welldefined.
Example 3: If we want to differentiate y
=
In
I 1,
1 + x2 j;i I x I t 1. we will have to
consider two cases: i) I x I > 1 and ii) ( x ( < 1
since ( x I > 1 makes 1  x2 negative. So in this rase.

ii)
4
, after simplification.
whe111 x 1 < 1,
a' i s a constant function for a = I Hence, 11 does not have any lnverse The log functions are thus defined aQtra+ I
4x Ix
11
1 + x2
1 + x2 =
and so.
4x 1 x 4
dy 4x = 4 for all x such that 1 x 1 # l dx I  u Now, let us turn our attention to logarithmic funct~onswith So. wc see that

n1171traty
bases
Differentiating the General Log Function
5.3.2
Let us consider any positive number a # 1. We say logax = y if and only if x Obviously. the natural logarithm~cfi~nctionInx can be written as logex.

a'
Further, we know that logax = loc~,c Inx This rule gives a connection between the natural function, ' v c <hall usc thls relat~onshipto find the derivative of and general Iogar~thm~c loga x So if v
=
logax = logae lnx.
d dx If we put a
Derivatives of Some Standard Functions
1 x = 3 in this, we get our earlier result:
(log, x ) = log,e
.
dy d lnx dx dx Thus, we arrive at
1 x
 = log,e  = log,e 
d dx If we put a
.1
(log, x) = log, e =
x e in this, we get our earlier result:
1 d lnx =  , since logee = 1 dx x

Example 4 Let us differentiate y = log,tan3x
dy dx
I d 3  (tan x) tan3 x dx 1 = log, e tan3 x 3 tan2 x sec2 x sec2 x = 3 log, e tan x
 = log, e

If you have followed the solved examples in this section you should have no difficulty in solving these exercises.
E E 3) Find the derivatives of : a) log, 2x
b) 71oglI (5x2 + 2)
c) e"lnx
Elements o f Differentlal Calculus
5.4
HYPERBOLIC FUNCTIONS
In applications of mathematics to other sciences, we, very often, come across certain combinations of eXand e'. Because of their importance, these combinations are given special names, like the hyperbolic sine, the hyperbolic cosine etc. These names suggest that they have some similarity with the trigonometric functions. Let's look at their precise definitions and try to understand the points of similarity and dissimilarity between the hyperbolic and the trigonometric functions.
5.4.1
Defitlitions and Basic Properties
Definition 1 The hyperbolic sine function is defined by sinh x
=
ex + e' 2
for all x E R.
The range of this function is also R. Definition 2 The hyperbolic cosine function is defined by cosh x
R. The range of this function is [ I ,
 eX =
2
for all x E
1.
You will notice that +
sinh
(x)
=
cosh
e(~)

=
+ eX
=
2 sinh (x), and
 e(~) 
(x)=
2
2  eX

2
=
cosh x
In other words, the hyperbolic sine is an odd function, while the hyperbolic cosine is an even function. Fig. 3(a) and (b) show the graphs of these two functions.
Fig. 3 : Graph of (a) sinh x (b) cosh x
We also define four other hyperbolic functions as  e(~) e x  eX4 tanh x = eX , coth x = ex + e' ' +
C

sech x
E
=
ex
+
ex , cosech x
=
e x  ,x
t
E 4) Verify that
a) cosh2x  sinh2x = 1 sinh x b) tanh x = c) 1  taih2 x = sech2 x.
.
Derivatives o f Somr S t a n d a r d Functions
E E 5) Derive an identity connecting coth x and cosech x.
You must have noticed that the ider~titiesinvolving these hyperbolic functions are similar to those invo!ving trigonometric functions. It is possible to extend this analogy and get some more fc.~nulas.
* cosh x sinh x cosh (x h y) = cosh x cosh y * sinh x sinh y tanh x + tanh y sinh (X h y)
=
tanh (x
=
y)
sinh x cosh y
1 k tanh x tanh y
Since we have seen that cosh2t  sinh2 t = 1, it is obvious that a point with coordinates (cosh t, sinh t) lies on the unit hyperbola: x2  y2 = 1. (Hence the name, hyperbolic functions). We have a similar situation in the case of trigonometric functions. The point (cost, sint) lies on the unit circle: x2 + y2 = 1. That is why trigonometric fbnctions are also called circular functions.
I
There is one major point of difference between the hyperbolic and circular functions, though. While t in sint, cost, etc. is the measure of an angle, the t which appears in sinht, cosht, etc. cannot be interpreted as the measure of an angle. However, it is sometimes called the hypberbolic radian.
5.4.2 Derivatives of Hyperbolic Functions L
Since the hyperbolic functions are defined in terms of the natural exponential function, whose derivative we already know, it is very easy to calculate their derivatives, For example,
Elements of Differential Calculus
 eX sinh x
=
2
. This means,
d (ex ;e') dx
d dx
 (sinh x)

= 
Similarly, cosh x =
e"
2
= cosh
x
ex + e" gives us 2
 eX
d dx
 (cosh x) =
+
2
In the case of tanh x
=
=
sinh x
(ex  e") (ex + e') , we get
(ex + e') (ex + e')  (ex  e') (ex  e')
d dx
 (tanh x) =
(ex + eX)2
=
1  tanh2 x
=
sech2 x
We can adopt the same method for finding the derivatives of coth x, sech x and cosech x. In Table 1 we have collected all these results. Table 1
Function
Derivative
sinh x cosh x tanh x coth x sech x cosech x
cosh x sinh x sech2 x  cosech2x  sech x tanh x  cosech x coth x
Example 5: Suppose we want to find dyldx when y
=
tanh (1  x2)
See if you can solve these exercises on your own.
E
E 6) Fhd f(x) when f(x) = i ~ )
tanh
4x
+1

5
d) sech (ln x)
#
b) sinh e2"
e) ex cosh x
C)coth (llx)
Derivatives of Some Standard Functions
5.4.3 Derivatives of Inverse Hyperbolic Functions We shall try to find the derivatives of the inverse hyperbolic functions now. Let us start with the inverse hyperbolic sine functions. From Fig. 3(a) you can see that the hyperbolic sine is a strictly increasing function. This means that its inverse exists, and
We have used the formula
for finding the roots of a quadratic equation here. Note that if eY
Thus sinh'x
=
In (x + J1 + x2 ), x E ]  m,
m
[. In Fig. 4, we have drawn the graph of
 m, x 
then eY < 0, which is impossible. Therefore we ignore this root.
sinhlx. Now,
d Thus,  (sirh' x) dx
1
=i.

1 ,I= Again we ignore the root
In the case of the hyperbolic cosine function, we see from Fig. 3 (b), that its inverse will exist if we restrict its domain to [0, [. The domain of this inverse function will be [ l , w[, and its range will be [0, w[. Now y = cosh' x
o x
= cosh
y=
ey
+ ey
,
eY = x 
f i , because
then eY < I , which is impossible since y > 0.
Elements of Differential
Calculus
~ h u cash' s x

~n (x *
\ I / ), x 2 1.
Fig. 5 shows the graph of cosh'x. d Further  (cosh' x) dx
=
1 x+1
d dx ( x +

,,LEI
Note that the derivative of cosh' x does not exist at x
1.
=
Fig. 6 (a), (b) and (c) show the graphs of tanh x, coth x and cosech x. You can see that each of these functions is oneone and strictly monotonic. Thus, we can talk about the inverse in each case. Arguing as for sinh' and cosh'x, we get
(b) Fig. 6
y = cosech'x

x
=
cosechy w y = in
[i '
,x#O I
1
Since sech x =
we shall have to restrict the domain of sech x to [0, m[ before cosh x '
talking about its inverse, as we did for coshx. ~ech'xis defined for all x
E
10, I], and we
can write sech' x = In [ l + ( y ) , O < x s l Now, we can find the derivatives of each of these inverse hyperbolic functions. We proceed exactly as we did for the inverse hyperbolic sine and cosine funcitons and get
L.erivntives o f Some Standard Functions
Let us use these results to solve some problems now.
Example 6 Suppose we want to find the derivatives of (a) f(x) = sinh'(tanx), and (b) g(x) tad'(cos ex).
I

I
Let's start with f(x) = s i c ' (tan x).
f (x)
d
1
=

JG dx
 1  Iec Now if g(x)
=
sec' x
=
(tan x).
1 sec x 1
tad' (cos ex), this means that
  ex
sin e x
=
ex cosec ex
We are now listing some functions for you to differentiate.
E
E7) Differentiate the following functions on their respective domains. a) cosech' (5 & )
b) [sech' (cos2 x)]"~
C) cothl e ( ~+5x6) 2
d) t a d  ' (coth x) + coth' (2x)
e) sinh'
& + cosh'
(2x2)
Elements o f Differentlrl Calculus
5.5
METHODS OF DIFFERENTIATION
In this section, we shall study different methods of finding derivatives. We shall also see that the problem of differentiating some functions is greatly simplified by using these methods. Some of the results we derived in the earlier sections will be useful to us here.
5.5.1
Derivative of X'
In Unit 4 we have seen that AS we have mentioned in Unit 4, if x < 0, xr may not be a
real number. For example, 3%
,m
E
R
d
(xr ) = rxr' dx 
when r is a rational number. Now, we are in a
position to extend this result to the case when r is any rea1,number. So if y x > 0 and r E R, we can write this as = eln x r
xr, where
,,r Inx , since the natural exponential and logarithmic functions are inverses of
each other.

re'.'"x
1  rxr  r', X
This proves that d  (xr1 = dx
l
X
for x > 0,
r E R.
We are surd, you will be able to solve this exercise now.
E
=
E 8) Differentiate a) x~
b) xe
Derivatives o f Some Standard Functions
5.5.2 Logarithmic Differentiation Sometimes we find that the process of taking derivatives becomes simple if we take logarithms before differentiating. In this section we shall illustrate this point through some examples. But to take the logarithm of any quantity we have to be sure that it is nonnegative. To overcome this difficulty, let us first try to fmd the derivative of In ( I x I ). Now you can check easily that 1 x I =
0.
Therefore, In ( 1 x 1 ) = ln
G ,and
We get,
d Using chain rule we can now conclude that if u is any function of x, then In dx
(1
u
1)
du  1 
u ' dx Let us see how this result helps us in simplifying the differentiation of some functions.
Example 7 To differentiate
we start by taking y =
( x +~ 119 (X (X
 5)213 (
( x +~ 1)9 (X (X
 51213(
 31314 
x + ~ 2~
+ 1)113
 31314
x + ~ 2~
+ 1)113
Then taking logarithms of both sides, we get
Differentiating throughout we get, 9 3 1 dy 2x

ydx
x2+1
2 2x+2 +   +4(x  3) 3(x  5) 3(x2 + 2x + 1)
~ x a m ~8l Suppose e we want to differentiate x""', x > 0.
Let us write y = x ~ ' ' ' ~Then . y > 0 and so we can take logarithms of both sides to the base e and write Iny
=
lnx""" = sin x.ln x
Differentiating throughout, we get.
 sin x + cos x In x
Elements of Differential
heref fore
CSICUIUS
3 dx
=y
sin x' (T + cosx lnx
Example 9 Ta differentiate xCoSX + (cosx)' let f(x) = xCoSX and g(x) = cosxX.To ersure that f(x) and g(x) are well defined, let us restrict their domain to [0, ~ 1 2 1 .
y =x
cosx
+ ( c o ~ x= ) ~f(x) + g(x)
0 for x E
[o, ~ / 2 ]
,
Let us differentiate both f(x) and g(x) by taking logarithms. We have, f(x) = xW" Therefore lnf(x) = cosx lnx. Thus,
1 f (x) f(x)
_
=  sin
xcmx
1 x lnx + casx X
( sin x In x + cos x X
 X~~~  I (cos x  sin x In x)
â€˘
Similarly, g(x) I
= (cos x)'
and so Ing(x) = x ln cos x
g' (x) = In cos x
X
+  ( sin x) COS X
cos x In cos x  x sin x
 (COSX)' (cosx In cosx  x sin x) dy Hence,  = f (x) + g' (x) dx
'
 xcow  1 (cosx  xsinx lnx) + cosxX (cosx lncosx  xsinx)
If you have followed these examples you should have no difficulty in solving these exercises by the same method.
E
E 9) Differentiate.
Derivatives o f Some Standard Functions
5.5.3 Derivatives of Functions Defined in Terms of a Parameter Till now we were concerned with functions which were expressed as y = f(x). We called x an independent variable, and y, a dependent one. But sometimes the relationship between two variables x and y may be expressed in terms of another variable, say t. That is, we m i y b a p k a f m x = t$ (t), y = w (t), where the functions t$ and yj have a common domain. For example, we know that the circle x2 + y2 = a2 is also described by the pair of equations, x = a cost, y = a sint, 0 It I 2%.
In such cases the auxiliary variable t is called a parameter and the equations x = t$(t). y = JIJ (t) are called parametric equations. Now, suppose a function is defined in terms of a parameter. To obtain its derivative, we need only differentiate the relations in x and y separately. The following examples illustrates this method.
i
dy Example 10 Let us try to find  if x
dx
=
a cos 8 and y
(Here the parameter is 8) We differentiate the given equations w.r.t. 8, and get d~ do
,
= b cos 8, and
dy
dy Id0
dx = do
a sin 0
b cos 0

cot 0 Now~dx=dx/deZX a ZTry to apply this method now.
E
dy
E 10) Find ; i; if a cos 0, y = a sin 8 at2, y = 2at x = acos3 8, y = bsin3 8 C) d) x = a(8  sin 8), = a(l  cos 8) a)
,
=b
sin 8
Elements o f Differential Calculus
5.5.4 Derivatives of Implicit Functions It is not always necessary to express y explicitly in terms of x (as in y = f(x)) to find its derivative. We shall now see how to differentiate a function defined implicitly by a relation in x and y (such as, g(x, y) = 0). dy Example 11 Let us find  if x and y are related by dx ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Differentiating throughout with respect to x, we get
or
dy dx

 (ax + hy =
,(hx+by+f)
See if you can find
E
+ g) dy for the following implicit functions. dx

dy dx
E 11) Find  if x and y are related as follows: a) b) c) d)
x: + y2 = I y =4ax . x~~~+ x2y2+ xy + 1 = 0 cosx cosy  y2sin'x + 2x2 tanx
=
0
Derivatives of Some Standard Functiens
SUMMARY
5.6
In this unit we have 1.
obtained derivatives of the exponential and logarithmic functions, hyperbolic functions and their inverses. We give them in the following table.
Function
Derivative
ex
ex
lnx
1 
Function
1
sinh' x
aXlna
log, X
 log, e
cosh' x ,
1
cosh x
I
2. 3.
cosh x
sinh x
tanh x
sech2x
cotb x
cosech2x
sech x
sech x tanh x
cosech x
cosech x coth x
d extended the result  (xr ) dx
= rxr' to


3
1 1x 2 , I x l < 1
coth'x
1 Ix
sech'x
1 xJm?"<x<l
cosech'x
all x
/
tanh'x
X
sinh x
dx'tl 1
X
ax
Derivative
E
R and x > 0.
illustrated logarithmic differentiation, differentiation of functions involving parameters and differentiation of functions given by implicit relations.
2
,lx('>l
1
Elements o f Differential calculus
5.7
SDLUTIONS AND ANSWERS
E2)
e) 22X+ 1n2 f) 7wSX ( sinx) In7 f(x)=2"1n2f(O)=In2
'
Hence f increases
times faster at x
=
112 than at x = 0.


1
G  54 sin3 x cos x
I
)
X
cos ec h2
(i]
d)  sec h lnx tanh lnx \
e) ex (sinhx + coshx)
1
b,
 [sec h' (cos2 ~ ) 1  ~ ' ~
3
1
1
1
. X
2 c o s x sin x
C) Then f (x) = sin xX(ln sinx + x cotx ) and gf(x) = cosxh
Derivatives of Some Standard Functions
(sec2x ln cos x  tan. 2 x) \dY  = f (x) + gf(x) dx ~ , = x("'), x > o d) Let f(x) = ( x ~ )g(x) If y = xX,lny = xln
(x) = x ( ~[xX' ) + xXln x (1 + ln x)] Answer = f ' (x) + gf(x) = (xX)"[ h X+ x(1 + h ) ] + x(=) [$I + xx' 1m (1 + lnx)] d In sin x e) (sin x)InX= (sin x)InX(ln x cot x + dx x d (xX)= xX(l+Inx) dx In sin x Answer = (sin x)'"" (I" X Cot x + X + x X ( l+ l n x ) dx dx =  a sin 8,  = a cos 8 a) d8 de 3 g'
)
E 10)
b) c) d)
zdy
2a 1 2at t dy 3bsin28cos8 =   tang  dx 3acos2 8 sin8 a d~ a sin8  sin 8  dx a(1 cos 8) (1  cos8) 
dy . dy d)  cosx sin y   sin x cosy  2y  sin'x dx dx + 2x2 sec2x = o
Y \rT
sinx cosy + 3
dy  = dx
YL
 4x tanx  2x2 sec2 x
(1x
 (cos x sin y + 2y sin' x))
+
4x tan x
UNIT 1 Structure 1.1
Introduction Objectives
1.2 1.3 1.4 1.5 1.6 1.7
Second and Third Order Derivatives nth Order Derivatives Leibniz Theorem Taylors Series and Maclaurins Series Summary Solutions and Answers
1.1
INTRODUCTION
In the first block you have found the derivatives of a number of functions. You know that the derivative f of a differentiable function f is again a function and is called the derived function off. We have already seen in Unit 3 that the concept of differentiation was motivated by some physical concepts (I~kethe velocity of a moving particle) and also by geometrical notions (like the slope of a tangent to a curve). The second and higher order derivatives are also sunilarly motivated by some physical considerations (like the acceleration) and some geometrical ideas (like the curvature of a curve), which we shall study in the remaining units of this block. We shall introduce higher order derivatives in Sec. 1 and 2. Leibniz Theorem which is given in Sec. 4 gives us a formula for finding the higher derivatives of a product of two fufil~cjnsIn the later sections, we shall study some useful formulas, called series expansions. The sigir~ficanceof these expansions will become clearer in Unit 1 of Block 4.
Objectives After reading this unit you should be able to : calculate higher order derivatives of a given h c t i o n f use the Leibniz formula to find the nth derivatives of products of functions expand a function using Taylor's Maclaurin's series.
1.2 I
I
HIGHER ORDER DERIVATIVES
SECOND AND THIRD ORDER DERIVATIVES 
Consider the fhnction f(x) = x4. You know that f (x) = 4x3. Now, this f is again a polynomial fhnction and hence, can be differentiated (see Example 5, Unit 3). We shall denote the derivative o f f by f '. Thus.
f' (x) = 12x2.This f"' (x) is calledthe second derivative of the fhnction fat the point x. It is also denoted by
I
d 2~
dx
(read as d square y by d x square ) of yz or
e2)or ~ ' y .
Let us differentiate f'. We get f" (x) = 24x, where Y'denotes the derivative off', or the third
d3y derivative off. Other notations o f f " (x) are 7or y, or e3),or D3y.Differeiitiating f"', dx d4y we get the fourth derivative off, P4)(x) = 7 = y4 = 24. . dx
Thus, differentiating (if possible) a given function f, then differentiating its derivative, and then derivative's derivative and so on, we get the first derivative, the second derivative, the third, fourth, ........derivatives of the function f. If n is any positive integer, then the nh derivative o f f is denoted by f(")or by
dny (read as d. dxn
nybydxn)orbyy,orDny.

.
Drawing C u r v e s
Note that in the notation 6")the bracket is necessary to distinguish it from f", that is, f raised to the power n. This process of differentiating again and again, in succession, is called sllccessive differentiation. We have already seen that there are functions f that are not differentiable. In other words f need not always exist. Similarly even when f exists it is possible that f' does not exist (see ~ x a m p l 3e near the end of this section). In general, for each positive integer n there are functions f such that 6") exists, but itn+')does not exist. However, many functions that we consider in these sections possess all higher derivatives.
n
A twice differentiable function is a function f such that f' exists. ~ e tbe a positive integer. A function f such that ff") exists is called an ntimes differentiable function. If $3exists for every positive integer n, then f is said to be an infinitely differentiable function. Now we give some simple examplesof higher derivatives. Example 1 If we are given that the third derivative of the function f(x) = ax3+ bx + c has the value 6 at the point x = 1, can we find the value of a? Here, f(x) = ax3+ bx + c Differentiation this we get f (x) = 3axZ+ b
Differentiating this again, we get P(x)=6ax Differentiating once again, we get
P)(x) = 6a Taking the value at x = 1,

P3)(1) 6a It is given that P ) ( l ) = 6. Thus 6a = 6. Therefore a = 1. Such equaiions involving the derivatives are known as differential equations .
Example 2 If y = 2 sin x + 3 cos x + 5, let us prove that yz + y = 5. Now,y=2sinx+3cosx+5 y, =2cosx3sinx yz=2sinx3cosx yz+y=2sinx3cosx+ 2 s i n x + 3 c o s x + 5 = 5
:.
The example below gives a function f for which f exists but f' does not exist.
Recall (Unit I ) that
Example 3 Consider,thefunction f(x) = x 1 x 1 for all x in R The furaction f(x) can be rewritten as
At points other than 0 we have f (x)=2xifx>O f (x)=2xifx<O At x = 0, the right derivative off, h2  o2 R f (0) = lim+  lim h = 0, and hbo h h+o+ the left derivative off,

h2  o2 Lf (0) = lim  lim h=O. hboh hboTherefore f (0) = 0. Thus,f (x)=2 IxIforallxinR We have already seen in Example 7, Unit 3 that the absolute value furaction I x 1 fails to be differentiable at 0. Therefore, f is not differentiableat x = 0. That is, fc2) (0) does not exist.
Higher Order Derivatives
some exercises before going any further. Find the second derivatives of the following functions. a) f(x)=x34 b) y = e2X
I
,
;.':
i.;:
&
E2) Find f13) (7~4)for the following functions, a) fTx)=sec x b) f(x)=sin2x+cos2x
> t, 1. 5.
,
C.
E E3) 'Prove that the following functions satisfy the differential equations shown against them. a) y =sinx; b) y = c o s x ;

Y4 = Y (y2)*+ ( ~ 3 ) 1~
a
Drawing Curves
E E4)
Find the value of integer k in each of the following a) f(x) = sink x where fi2)(d6)= 2 J? b) f(x)=xk+kx2+1wherefi2)(1)=12
.1.3
nth ORDER DERIVATIVES
Let n be a natural number. We have already defined the n'h derivative of a hnction in Sec. 2.
When a function f is given by a formula, it is often necessary to express its nh derivative also by a formula using f and n. Usually, one can guess P")after working out $I), fi2) and P). However, a rigorous proof would require an application of the principle of mathematical induction. In the example below we shall derive formulas for the nfi derivative of various functions. Study them carefully as we shall be using them in later sections. But, first let us recall the principle of mathematical induction. If {P,) is a sequence of propositions (statements) satisfying. 9 P, is true (usually N = 1). ii) The truth of Pi implies the truth of Pi+,,i 2 N,
1,
then Pn is true for all n 2 N. We shall apply this principle in the examples that follow.
Example 4 We shall prove here that the n"' derivative of the polynomial hnction xn', m 2 0, w.r.t. x is dnxm m(m1) ....(mn+l)xm",if nSm,and dx" 0, if n > m Let us denote by Pn the statement
Note that the product m (m 1) ,.....(m n + 1) has n factok. When n = 1, we have already shown that
Thusthcstatement P, namely.
(5
mm*ml)is true.
Next, suppose we have proved for some n that the statement Pn is true. This means that the nh derivative of xmis m(m  1) ....m(m  n + I), xm". if n I m, and is 0, if n > m
1
f
Higher Order U e r i v a t i b c *
Then the (n + 1)" derivative of xmis = the derivative of the nh derivative of xm
When
m (m  1) ..... (m  n + I) (m  n: xm"I if n < m
11
. m. the
nlh
C~~IIV.I!I\,
IS cor~stdnt,because
u" '.  x I , therefore the (n r 1 ) " d e ~ ~ v a t ~1sv/em e
Oif n > m. !
m(m1) .....( m  n + I ) ( m  n ) x Oif n+I>m.
m(n1)
,if n + l < m
This means that the truth of P, implies the truth of P,,,. Therefore, by the principle of mathematical induction, P, is true for all n > I. Hence our result is true for all natural numbers n. Remark 1 When n = m, the nthderivative of x'" is =m(m 1) ....... (mn+ 1)xn'"=m(m 1).........3.2.1.Thisisthesarne1sm!.
Example 5 If f(x)= In (1 + x), let us find e")(x). 1 Differentiating f ( x ) = In (1 + x), we get f (x) = 1+ x I Differentiating again, fc2)(x)= (1 + x ) ~ 2 Differentiating once again, E3)(x) = (1 + Can you guess fin) (x) now? If you have guessed correctly, you must have arrived at these conclusions. i)
The denominator of fin) (x) is ( 1 + x)".
ii)
Its sign is positive or negative according as n is odd or even.
Its numerator has (n  I)! Do not think that it is merely (n  1). There is a factorial symbol too. To be convinced of this, calculate E4)(x) and see. % (I),' x (n  I)! ...( I) 'Therefore our guess is fir#) (x) = (1 + x)" This guess remains to be proved. A proof is necessary because there could exist many other formulas for P")(x) that coincide with the correct answer when n = 1.2 or 3. For example, if we omit the factorial symbol, we get one such formula. But we have already mentioned that this formula does not hold for E4) (x). So, let's try to prove (I). iii)
We first note that it is true for n = 1,2 and 3 as we have seen in the beginning. Assume that it is true for n = m, that is,
fi'") (x) =
(I)"'' x (m  I)! (1 + x)"'
Differentiating this we get, fil'l+l)
(x) = (1 y*1(m I)!
m  (I)"'' (1) x m (m  I)! (1 + x)"'+~ (1 + X ) ~ +I
Thls proves the guess for n = m t 1. Thus, assuming the truth of the formula (1) for n = m,we arrive at the huth of this formula for n = m + 1 . Therefore by the principle of mathematical inductioc, our guessed answer w correct for all positive integers n. Thus, when f(x) = In (I + x). fin'(x)=
(I)"' (11  I ) !
(I + X)n
Check !hill IIIC i o l ~ t l ~ ~ l <'I]; n snl or 13 > rn " and "n r i 2 n;" are equlvalenL.
Drawing Curves
E
E 5) If y = (1 + x)'; where r is a real number, find y,, where n is natural number (n < r).
Example 6 If f(x) = cos 2x, let us use the principle of mathematical induction to find a formula for 6") (0). We first find fln)(x) when n = 1,2,3,4. We have f(x) = cos 2x. Differentiating this successively, we get fi') (x) =  2 sin 2x fi2' (x) =  4 cos 2x fi3)(x) = 8 sin 2x fi4)(x)=16c0s2x We see that in the formula for fin)(x), we have to have i)
a sign (positive or negative),
ii)
a coefficient (some power of 2), and
iii)
a trigonometric function (sin 2x or cos 2x)
We observe that the first two terms cany negative sign, the next two carry positive sign, the next two negative and so on. We also observe that sin and cos occur alternately. Therefore our guess is
I
2" sin 2x if n is the form 4k + 1 2"cos 2x if n is of the form 4k + 2
f(") (x) =
2"ssin 2x (2"cos 2x
+3
Wecanalso write this in a compact form as #@(x) = 211cos (2x + nn / 2)
Recall that cos (9 + 112) =  sin 8 cos (0 + I)=  co* 8cos ( + 3 ~ ~ =2 SIP ) 9
10
if n is of the form 4k jf n is of the form 4k
You can easily check that (2) and (3) are equivalent by putting n = 4k + 1,4k + 2,4k + 3 and 4k in (3). We shall now prove formula (3). ..
...(2)
...(3)

We have already seen that it is true for n = 1,2,3 and 4. Suppose it is true for n = m, that is, P)(x) = 2"' cos (2x + d 2 ) . Differentiatingthis we get, P+') (x) = 2"'.2.( sin 2x + m d 2 ) sin (2x + m 7~12) or, Pnl+')(x) = 2"'+' cos [2x + (m + 1)d 2 ] So here again we see that the truth of the formula (3) for n = m implies its truth for
I
Therefore by the principle of mathematical induction, we see that the guessed formula for fin) (x) is true for all natural numbers n. Now substitute x = 0. We obtain f(")(0) = 2" cos nn12
I
/
T h ~ sr the required answer. We can also use this method to prove a general result for the nthderivative of a sum of two functions. Example 7 Iff and g are two functions from R to R and ifboth of them are ntimes differentiable, we can prove that
We shall prove this result by induction. When n = 1, this means (f + g)' = f + g'. This has already been proved in Unit 3. Suppose (f + g)(ln)= fin') + g(") is true. Differentiating this we get [(f + g)(m)]'= [f("'J + g(m)]r= [f("')lp+ [g(m)]r
I

This is the same as (f+g)(m+l)=fI*I)+g(nl+l)
i
Thus the result is true for n = m + 1. Therefore by the principle of mathematical Induction (f + E)(n) = P"+g(ll)holds for all natural numbers n. Remark 3 Similarly one can prove that (cf)(")= c.Pn)holds for all natural numbers n and all scalars c. This fact combined with Example 7 can be restated in the6'linear algebra terminology" as :
' I
The collection of ntimes differentiate functions is a vector space under usual operations. Try to solve these exercise now.
E
E6) Find the n"' derivative of the following functions a) f(x) = (ax + b)3 b) f(x)=(ax + b)"'
b
c) f ( x ) = ~
Higher Order Derivatives
cos (2x + (m + 1 ) x12) = cos (2x + n1d2 +d2) =  sin (2x + rnd2)
Drawing Curves
E E 7 ) If f(x) = sin x, prove that fin) (x) = sin [x + nn/2] holds for every natural number n.
E
E8) If y = sin (ax + b), prove that for every positive integer n. we have y, = ansin [(nn/2) + ax t b]
E
E9) Prove that the n'" derivative of the polynomial function f (x) = a, + alx + a,x2 + .....+ anxl'is the constant function n! a,, .
E E 10) If y = cos x and if n is any positive integer, prove that [y,I2 + [yn+,I2= 1
1.4
LEIBNIZ THEOREM
In Unit 3 we have proved some rules regarding the derivatives of the sum, scalar multip!e, product and quotient of two differentiable functions. These were (f+g)'=f +g' (cf)' = c f (fg) = fg' + g f gf'  fg' (flg)' = , 2(g (x) # 0 anywhere in the domain) b
! i
In the last section we have seen. (Example 7 and Remark 3) that the first two rules can be extended to the n"' derivatives i f f and g are ntimes d~fferentiablefunctions. In this section we are going to extend the product rule of differentiation. We shall give a formula for the nLhderivative of the product of two functions. The product rule for two functions u and v can also be written as (uv)] = U I V + U V I Now we look for a similar formula for (uv)?, (uv),, etc. But first let us recall the meaning of the notation C(n, r), where n and r E Z+and r 5 n. C(n, r) stand for the number of ways of choosing r objects from n objects. Socletinles it is also denoted by I1Cror
( :) .
Drawing C u r v e s
Also recall the formulas
ii)
C(n, 0) = C(n, n) = 1
iii)
C(n, r) = C(n, n  r)
iv)
C(n,r)+C(n,r+ l ) = C ( n + l , r + 1)
These are combinatorial identities, true for all positive integers r and n with r 5 n. Theorem 1(Leibniz Theorem) Let n be a positive integer. If u and v are n times differentiable functions, then (uv), = C(n, 0) u,v + C(n, 1) u,,v, Leibniz had stated this result in his first article on differential which was published in 1684.
+ C(n, 2) uW2v2+ ....... + C (n, n) u v,
The pattern in the formula for (uv), can be compared with the expansion of (x + y)". The coefficients are binomial coefficients and they appear in the same order as those in the expansion of (x + y)". The order of the derivative of u goes on decreasing one at a time, and the order of the derivative of v goes on increasing one at a time. The number of terms is n + 1. Remark 4 We omit the proof of this theorem and merely indicate how this can be proved by induction on n. Firstly, when n = 1, the above formula is the same as the already known product formula, and therefore is true. Assuming that it is true for n = m, we can prove it for n = m + 1, by applying the product rule for each term of the expansion of (uv), and by using the combinatorialidentities mentioned. (See E 17) for more details. We start with a simple and direct application of the formula. Example 8 If f(x) = x sin x, let us find the fourth derivative off, using Leibniz Theorem. We first observe that far n = 4, the Leibniz Theorem states
In this problem we take u = x and v = sin x, so that f = uv We have u = x u, = 1 %=o=ujr.u4
v=sinx v1 = COS X v2=sinx v3=COSX v, = sin x
Substituting these in the above formula, we get
What happens if we attack the same problem directly without the use of Leibniz Theorem? We have f(x)=x sin x Differentiating this, we get f (x) = x cos x + sin x (by product rule) Differentiating once again, we get r ( x ) =x(sinx)+ 1cosx+cosx =2cosxxsinx Differentiating once again, we get fi3)(x) =2 sin x  (x cos x + sin x) =3sinxxcosx Differentiating once again, fi4)(x) =3cosx[x(sinx)+cosx] =xsinx4cosx
We notice that we obtain the same answer. In this direct method, we had to apply the product formula four times, once for each differentiation. It is clear that when we want the n" derivative for bigger values of n, Leibniz theorem provides an easier method to write down the answer, avoiding the difficulty of repeatedly applying the product formula. Example 9 If y =(~in'x)~, prove that (1  X2)Y,+~  (2n + 1) xy,+,  n2y, = 0 for each positive integer n. Differentiating both sides of y = sin'^)^, we get
,/=
2 sin' x y1=
Squaring and crossmultiplying we get (1~~)~~~=4(~1I1I~)~=4y Differentiating once again, we get 2(1 x2) yly22xyI24y, = o Dividing throughout by 2 y, gives us (1x2)y2xyl2=0 Differentiating n times, using Leibniz Theorem for each of the first two terms we get (1 x2) y,, C(n, 1) 2xy,,,  C(n, 2) 2 Y,, {xY,+,+C(n, 1) yn) =O That is, ( I  x2)Y,,+~  (2n + 1) xy,,,  n2y, = 0 The following exercises will give you some practice in applying Leibniz Theorem.
E 11) State Leibniz Theorem when n = 5. That is, (U.V)~ =?
E 12) Prove that when n = 1, Leibniz Theorem reduces to the product rule of differentiation.
E 13) Find the third derivative of sin2xusing Leibniz Theorem. Find the same directly also and verify that you obtain the same answer.
Hlgher Order Derivatives
Drawing Curves
E E 14) If f(x) = x eX, find the sixth derivative of f, using Leibniz fonnula.
'
E E 15) Find thc nn derivative of x3lnx
E E16) If y = F x2prove that y(")=eU[a"x2+2na*' x+n(n 1)aW2]
I
E
Higher O r d e r Derivatives
E 17) a) Write down Leibniz formula for (u.v),
b) Differentiate it term by term and obtain (~v),~=C(m,O)u,~ v+C(m, 1)(u,,,vI+u,,v2)+
........+C(m,rn)uv,,.
c) Deduce that (UV),, =c(m,O)u,, v + [C(111,0)+C(m, 1)1u,"v1 =LC( I)+ C(m,2)]~,~v2+.........+[C(m,m 1)+C(m,rn)]u1v,+C(m,m)uv,,.
1
d) Deduce from (c) the Leibniz foxmula for (uv),
I.
I
I
.
t
i
i 1
E
E 18) Using Leibniz Theorem and induction, prove that (xn)(")= n! for all natural numbers n.
1.5
TAYLOR'S SERIES AND MRCLAURIN'S SERIES
In this section we obtain aeries expansions for many important functions. For this, we use higher derivatives. You muat have coma across tho following series : i)
Exponential Series :
ii)
Logarithmic Series :
Drawing Curves
iii)
Geometric Series :
ii)
Binomial Series :
r (r  1) (1 +x)'= 1 +rx+  x2+ ............. providedIxI< 1 1.2 We observe that each of them is of the form f(x) = a, + a, x + a2x2+ a3x3+ ..........+ anxl1+ .........,where a,, a,, a2, .......... all,.............are some constants. We ask ourselves the questions. Is there anything else common to these four examples? Is it possible to express a,, a,, .......,a,, ........ in terms of the function f? Our answer is : Yes. In all these examples,
f'"' (0) an=n!
In other words, the series is of the form f ' (0) f'2' (0) f'"' (0) f(x) = f(0) + x+ x2+ ........ + xn+ .......... l! 2! n! Brook Taylor (1685173 1) and Colin Maclaurin (1698 1746) were both d~sc~ples of Newton. Taylor first published his series in a paper In 1715. Maclaurin used Taylor's series PS a fundamental tool in his work on calculus.
.
We shall prove this for the above four instances, in the examples worked out below. .This expansion is called Taylor's series for f around zero. It is also known as Maclaurin's series for f. The name "Taylor's Series for f around zero" suggests that there npy be a Taylor's series for f around x,,(x,,# 0). But in this course we shall restrict ourselves only to the series around zero. This series expansion makes sense only when f is infinitely differentiable at zero. It is valid for many important functions (though not for all functions). You will learn mo,re about the validity of these series in the course on real analysis. In this section, you should train yourselves to write down Maclaurin's series for many functions. We have said above that the function f should be infinitely differentiable, that is, it should have derivatives of all orders. How do we check this condition? For some functions it is not difficult. For example, we have
'cos hx, If n is a w n rlln hx. if n iu odd
In these cases, we can say that the derivatives of all orders exist for all values of x and we can confidently use Taylor's series. Example 10 Let us verify that the k n o w series expansion of exis the same as its Maclaurin's series.
Maclaurin's series for exis
Where f(x) = ex Now, f(0) = e0= 1 :. f (0) = 1. Also, f (x) = ex
e")
&fact, we know that (x) =ex for all natural numbers n, which means fit') (0) = e0= 1 for all natural numbers n. Substituting these values of f(O), f (0), .......... en)(0)inMaciaurinls series we get
which is the known expansion for ex. Example 11 Obtain Taylor's series for In (1 + x) around zero. Let fix) = ln (1 + x) Then we have already seen 1n Example 5, that (n  I)! 1)lll (1 + x)"
Therefore, en)(0)= ( 1)"' (n l)! ... f'"' (0)  (l)"l . n! n Therefore Taylor's series around zero is (1 ) X + (1)2' xln (x) = 7
1
2
+ .....+(l)"I n
xI1 + ....
Hence, ~ a c l i u r i n ' sseries for ln (1 + x) is
We note that this is the same as the already known logarithmic series. Example 12 Next, let us write down Maclaurin's series (or Taylor's series around zero) for ll(1 x).
1
f (x) = ;f (0)= 1 (1  x ) ~ 2
fi2) (x) = O];e2)(0)=2 We can prove by induction that n! ( 4= n+l and therefore (1 x) Therefore Maclaurin's series is
en)

f'"' (0)
en)(0) = n! and hence = 1. n!
Note that this agrees with what we already know, namely that the sum of the geometric series 1 + x + x 2 +.........+ x n + .........is 1/(1x).
Example 13 Suppoee we want to wtite down Taylor's series for (1 + x)' around zero, where r is a fixed real number. Let f(x) = (1 + x)'. Then f(0).= 1 i'(x)=r(l +x)'' ; f (O)=t e2J(x)=r(r 1)(1 +x)'~:fi2)(0)=r(r 1)
Higher Order Derivatives
Drrwlng Curves
We can prove by induction that fln)(x)=r(r 1) ......(rn+ 1)(1 +x)+" fin)(0)=r(r 1) .......(rn+ 1) Therefore Taylor's series around zero is
Note :This is the same as the binomial series that we already know. This expansion is valid only when I x I < 1. The reason for this will be clear when you study the course "Real Analysis". When r = 1, this binomial series becomes
Note that Example 12 follows from this on replacing x by  x throughout. Further, we note that if r is a natural number than the series terminates after finite number of terms. So far we have seen that the four known series occur as Taylor's series. In the next two examples, we find that we can write down similar series even for hnctions like sin x and cos x. Example 14 Let us write down Maclaurin's series for the h c t i o n sin x.
Let f(x) = sin x. Then we have already seen in E 7) 0 if n is even 1 if n is of the form (4m + 1) 1 otherwise, i.e., n is of the form (4m + 3) nn 2
fin)(0) = sin We see that, as n varies over 0, 1,2,3,4,5,6,7 ............ fen) (0) takes the values 0,1,0, 1,0,1,0,  1, ......... Therefore Maclaurin's series for sin x is
Example 15 To fmd Taylor's series for cos 2x around zero, let us write f(x) = cos 2x. We have
already seen in Example 6 that nx 2
fin)(0) = 2" cos Therefore, Taylor's series around zero is
Example 16 Suppow we want to a) write downthe flnt four term of M.chrin'a d e a for tm x. b) write down the fint thm non m o tennr of thh nrioa.
Therefore the first four t e r n of Maclaurin's series for tan x are given by 1 0 2 :x2 3, 0,x,x I ! 2! 3! Hence Maclaurin's series for tan x is x x++ 3
.......
Now, we want the next nonzero term.
I j1
I L
We have f(*)(x) = 16 sec4x tan x + 8 sec2x tan3x and .: tan (0) = 0. 64)(0) = 0. ~ e x t f ( ~ ) ( ~ ) = 1 6 s e c ~ x + 6 4 s e c ~ x t a n ~ x + 2 4 s e c ~16sec2xtan4x xtan~x+ This m a n s fiq (0) = 16.
Thus we have
Example 17If in Maclaurin's series for sin b , the coefficient of x3 is given to be  6 k,let us find all possible values of k. Maclaurin series for sin b is given by kx k3x3 + .......
3!
l!
since qx) =k cos b
f12)(x) = sin b f13) (x) =k3cos b , where qx) = sin kx
The coefficient of x3 is  (k3/6) Therefore (k3/6)=  6k This gives the equation k (k2 36) = 0 The roots are k = 0 , 6 or  6.

:
Thus, 0,6 and  6 are all the possible values of k such that the coefficient of x3 in Maclaurin's series for sin b is  6k.
E E 19) Wri@down Maclaurin's series for the following : 1
a) (l+x)z b) ( ~  2 ) ' + 1 c) COSX d) l / ( l  2 ~ )
Higher O r d e r Derivatives
Drawing Curves
I
E
E20) Write down the first three nonzero terms in Maclaurin's series of the following : a) sin3 x b) h(1X)
I
I 1 j
E
E21) Find the coefficient of x9in Maclaurin's series for the function. a) cos 2x b) sin
):
(X +
t
1
E
E22) If Maclaurin's series for sin x is differentiated term by term, do you get Maclaurin's series for cos x ?
E
E 23) If Maclaurin's series of ex is differentiated tern by term, we get the same series again. Prove this.
E
E 24) Consider the function y = a + tan' bx where a and b are fixed real numbers. We are given that its Taylor's series arond zero is 2 + 3x  9 x3+ ..... Find the values of a and b.
Higher O r d e r Derivatives
Drawing Chrves
E
E25) Find the coefficient of x3 in Taylor's series around zero for the function sin' x.
Some General Remarks on Taylor's and Maclaurin's Series Though we have obtained infinite series for many functions, it is necessary to give a note of caution. These Infinite series need not be valid for all values of x, and as such, these have to be used with care. In the course on real analysis, you will be able to study the conditions under which these series are valid.
1.6
SUMMARY
In this unit, we have 1)
introduced higher order derivatives,
2)
derived a formula (Leibniz's Theorem) for the nIhderivative of a product of two functions. (uv), = C(n, 0) u, v + C(n, 1) u, v, + C(n, 2) u, v2+ ........+ C(n, n) u v,.
3)
written Taylor's series around zeroiMaclaurin's series of a number of functions by using the formula
1.7
SOLUTIONS AND ANSWERS
E l ) a) 6x E2) a) 11
k
#
0, since
k=O*O=qJ?;, which is impossible.
JZ
* f12)(x) = k2sin kx * c2)( d 6 ) = k2sin knl6
E4) a) f(x)= sin kx
Now,k2sinkd6=2&*sinkd6=2&lk2 Since 1 ~ s i n k ~ l 6 < O ,  n < k x 1 6 < 0 * k =  1 or.2or3or4or5 Out of these, k =  2 is the value which satisfies sin k A 16 =  21 & lk2
Higher Order 1)crivatives
E6) a)
3!an ((3  n)!
m!an ((m  n)!
b,
c)
ex
(ax + b)3n,n I3
(ax + b)mn,if n Im
d) knekx
cos x if n = 4 k + 1 E7) f ( x ) = s i n x ~ f ( x ) = c o s x , f ' ( x ) ~  s i n x , sinxifn=4k+2 f(3) (x) =  cos x, f(4) (x) = sin x and so on. So Our guess is that f (") (x) cosxifn=4k+3 Now use the principle of mathematical induction to prove that f tn) (x) = sin Isin x if n = 4k (X
y)
+
i
as in Example 6.
E10) y =cosx*y,=sinx,y2=cosx,y3=sinx,y,=cosxandsoon. yn = cos (x + nd2) 3 yn+,=  sin (x + nd2) 3 yn2+ yn+,2 = cos2(X+ nd2) + sin2(x + nd2) = 1. E l l ) (uv),=u,v+5u4vl + 10u3v2+10u,v3+5ulv4+vS E 12) (u v), = u, v + u v, which is the pribduct rule of differentiation. d3(sin2X) d3 =7 (sinxsinx)=cosxsinx3 sinxcosx dx3 dx d dx
 (sin2x)= 2 sin x cos x
(I)"+, x3
E 15)
xn
[(nl)!3C(n,
l)(n2)!+6C(n,2)(n3)!
 6C (n, 3) (n  4)!]
1
E21) a) 0
b,
9'JZ
I E 22) Yes
E24) We take that tan' bx always takes values between x/2 and d 2 . Then, a = 2, b=3
,
UNIT 2
THE UPS AND DOWNS
Structure 21
Introduction Objectives
22
MaximaMinima of Functions 2 2.1 Definitions and Examples 2 2 2 A Necessary Condition for the Existence of Extreme Points
23
Mean Value Theorems 2.3.1 Rolle's Theorem 2.3 2 Lagrange's Mean Value Theorem
2.4
Sufficient Conditions for the Existence of Extreme Poiqts 2.4.1 First Derivative Test 2.4.2 Second Der~vativeTest
2.5
More Information from the Second Derivative 2 5 1 Concavity/Convex~ty 2.5.2 Polnts of lnflect~ons
2.6 2.7
2.1
Summary Solutions and Answers
INTRODUCTION
Why do drops of oil on the surface of water coalesce? Why do honeycombs have hexagonal cells? Why is a drop of water spherical? Why does a red corpuscle in blood have the shape of a biconcave disc? The answers to these questions are closely related to minimum and maximum values of some functions. Drops of oils tend to coalesce so as to mnimise total surface tension. The scheme of hexagonal cells enables bees to store a fixed amount of honey by using the mlnunum amount of wax for sealing. A drop of water is spherical because a sphere is the shape which encloses a given volume with minimum surface area. The oxygen carrier red corpuscle, on the other hand, is in the shape of a biconcave disc so as to maximise the surface area. It enables our system to carry the maximum amount of oxygen on the surface of a fixed amount of blood. In this unit we shall discuss an important technique involved in solving the problem of maximising or minimising various functions. This technique, as you will soon see, involves the use of derivatives which you studied in Units 36. We shall also discuss Rolle's theorem and the mean value theorem which have very important applications as you will see further in your study of calculus.
Objectives After having gone through this unit, you should be able to obtain the maximum and minimum values of some functions solve practical problems of maximaminima state Rolle's theorem and the mean value theorem find the points of inflection and the curvature of a curve determine whether a given function is concave or convex or neither in a given interval.
2.2
MAXIMAMINIMA OF FUNCTIONS
Look at the points P and Q in the graphs in Fig. 1. How are they different from other points on the graphs? We could describe Q's as the peaks or hilltops and the P's as the,.valleybottoms. Using the language of mathematics, we could say that each P has the property that the value of the function fat P is smaller than the value off at neighbouring points. Similarly, the peaks Q are distinguished by f having a maximum value there when compared to the values at nearbypoints. But before proceeding any further, let us define maximum and minimum points of a fbnction precisely.
The Ups and Downs
t
Fig. 1
2.2.1
Definitions and Examples
Definition 1 :Mpoint c is said to be a maximum point for a function f if there exists a 6 > 0, such that f(c) 2 f(x) for all x such that I x  c 1 <6. f is said to have a maximum at c, and f(c) is called a maximum value off.
We can similarly define a minimum value of f. f is said to have a minimum at a point c of its domain provided there exists a 6 > 0 such that f(c) 5 f(x) whenever ( x  c 1 < 6. A maximum or a minimumvalue is known as an extreme value.
Example 1 The fimction f(x) = ( x I + 1 has a rninimum value atx = 0. You can see from Fig. 2 that f(0) = 1 and f(x) > 1 for all x in any every neighbourhood 10 6,0 + 6[ of 0. The mimimum value off is thus f(0) = 1. Do you agree that this function has no maximum value?
Fig. 2 : Grrph of f(x)

( ' rI + I
111 rlris exaniple we see that f(x) 2 f(0) in every neighbourhood of 0. But to prove that f(0) is a ~r~i~iim value, i ~ nwe ~ need to find only one such neighbourhood.
Our next example shows that Definition 1 cannot be applied to functions defined on closed ~ntervalsif their extrema occur at either of the end points of the interval. Example 2 Consider the function qx) = x + 2, v x E [O, 23. From its graph (Fig. 3), it is clear that x = 0 is a minimum point and x = 2 is a maximum poilit of this function. In fact, you will see that z = f ( o ) s q x ) s f ( 2 ) = 4 tX~E [0.2]
]x  6, x + 6[, that IS, the set of points whose distance from x is less than a pos~tivcnumber 6 is called a neighbourhood ol' x.
Xote that here we cannot find a 6 > 0 such that f(0) I f(x) v x E ]  6 , 6 [ for the simple reason that f7x)is not defined for x < 0. The same argument holds good for the rnaxlmum. Since f7x) is not defined for x > 2, we cannot find any 6 > 0, such that f7x) I f72) v x E 12  6,2 + 6[. This means f does not have extrema at 0 or 2 in the sense of Definition 1. How do we resolve this paradox? We modify Definition 1 to suit such cases.
Drawing Curves
Y t
Let f be defined on [a, b]. We shall say that f has a maximum (minimum) at a if we can f u d 6 > 0, such that f7a)2 f7x)(f7a)< f7x))for all x E [a, a + 6[.
4
/ ,
/
Similarly,we shall say that f has a maximum (minimum) at b, if we can find 6 > 0 such that f(b) 2 f7x) (f(b) I f7x))for all x E ]b  6, b]. I
2 /j
Example 3 The function f7x) sin x has a maximum value at several points. Fig. 4 shows the graph of this function. Can you see that these points are d 2 , 5 d 2 , ....... 3d2, 7d2, .......?
I
I I I I I
0
2
E
Fig. 3
Fig. 4 : Graph of f(x) = sin I
In general, the sine fimction has a maximum at each of the points 2nrc + (7c/2), n being any integer. The maximum value is sin [2nn + (d2)] = 1. This function has a minimum value too at several points. What are these points? Do you agree that the minimum value at these points is l? We can now say that this function has an extermum at all points m + d 2 , n being any integer. The functions in the examples considered so far were all continuous. For such functions, a valley in the graph of the function indicates a minimum and a peak points to a maximum.But we can talk about maximum and minimum values of noncontinuous functions too. Here, we may find extreme values which are neither at a peak nor at a valleybottom, as in Example 4. Example 4 Consider the function f defined follows :
This funition has three extreme points, P, Q and R (see Fig. 5). You should be able to see for yourself that P is a maximum and R is a minimum point of f. The h c t i o n also has a minimum at Q, that is, at x = 1. Let us see why. Consider the neighbourhood
] 5 1 [ 1
defined as follows :
1
9
+
9
that is;]
1 3
5
9
[ of
1. The function in E s interval is
The Ups and Uo~vns
I ' 9 :
' 1: '
I
4I ;r
a5 !
f
Fig. 5
[ ,<
b'
~hu~f(l)=l1'=2 For
j ik
i4 b
1
< x < 1,x2<1 =$x2>2 =$
/
:i i' \
=$
1
1x2>2 flx)>fll)
Now, when 1 < x < 312, f(x) is positive, and therefore is greater than f( 1). Hence f(x)2 f (1) whenever x E ] l a , 312 [.
I
I
Thus f( 1) =  2 is a minimum value of the function. Remark 1We found in the above example that f(0) =  1 was a maximum value off whereas f(3) = 0 was a minimum value. Did you find it hard to swallow that a maximum value of the function is smaller than a minimum value? If yes, recall our defdtion of an extreme value. We were concerned with the values of the function only in a neighbowhood (that is, points nearby) of the extreme point. Thus, the concept of maximais essentially a local phenomenon. What happens globally or elsewhere was not under consideration. For this reason, some people use the tenns local (or relative) maximum and local (or relative) minimum insteadofmaximumandminimua
!, ?
A value qc) such that f(c)2 f(x) for all x in the domain of the *tion,
is then called an absolute (or global) maximum. Similarly, if 0 ) 5 f(x)for all x in the domain, then 0 ) is called the absolute (or global) miuimum. Therefore, a function may have many local minima or maxima, but it can have only one absolute minimum or maximum. In the light of this we see that f(0) =1 is the absolute minimum value for the function in Example 1.
E
E 1) Find the maxima, minima of each of the following functions. If a function has no lmixhdminimumsay so.
iii) flx)=x,forO<x< 1. [Be carell! 0 and 1 are NOT in the domain of the function!] iv) flx)=x2,for all x E R
D r a w i n g Curves
2.2.2 A Necessary Condition for the Existence of Extreme Points So far, we have used the graph of a function as an aid to discovering extreme points. Clearly, this is not going to be feasible in all circumstances. Drawing graphs may be timeconsuming and cumbersome. Certainly there must exist quicker and neater techniques. In this subsection we describe an analytic method of f~ndingextrema. But fust, a definition : Definition 2 A point c of a set A G R is said to be an interior point of A if for some 6>0,]~6,~+6[~A. The point 1 E A = [O, 21 is an interior point, since ] 1  6 , 1 + 6 [ c [O, 21, 1
if we choose 6 =  .But neither 0, nor 2 is an interior point of A. 2 The fallowing theorem gives us a necessary condition for the existence of an extremum. Theorem 1Let f be a function which is derivable at an interior point c of its domain D. Iff has an extremum at c, then f (c) = 0. Proof :Since f is derivable at c, f (c) exists. This means that Lf (c) and R f (c) exist, and are equal. Suppose, fust that f has a maximum at c. This means. f(x) 5 f(c), t/ x E ]C  6, c + 61, where 6 is some positive number. *f(x)f(c)<O v X E ] C  6 , ~ + 6 [
Nowifc6<x<c,thenxc<Oand
f(x)  f(c) XC
Similarly,if c < x < c + 6, then x  c >0, and Therefore, Lf (c)=
2 [f(x)  f (c)] XC
,and
f(x)  f(c) XC
But we know that Lf (c) = Rf (c), Hence the above inequalitiesyield Lf (c) = R f (c) = 0. that is, f (c) = 0. Proceeding exactly as above, we can prove that f (c) = 0 even when f has a minimumat c (see E 2). Remark 2 The condition f (c) = 0 is only a necessary condition for f to have an extremum at c. This means that iff has an extremum at c, then we must necessarily have f (c) = 0. But it is by no means a sufficient condition. In other words, if we are given that f (c) = 0, this information is not sufficient for us to conclude that f has an extremum at c. That is, a function may not have an extreme value at c even though f (c) is zero. For example, take the function fix) = x3 shown in Fig. 6. We know that f (0) =0. Now, qO) =0, whereas qx) <0 for  6 <x <0, and qx) > 0 for 0 < x < 6, whatever positive number 6 may be. Thus there is no 6 for which either fix)<f(O),foraUx~16,6[orfix)2fiO),forallx~] 4 , 6 [ . This indicates that f has neither a maximum, nor a minimum at x = 0, even though f(0) = 0. Remark 3 The condition f (c) = 0 applies in the case when f (c) exists. But a function may have an extremum at x = c even though it is not derivable at c. For instance, the function f of Example 4 is not derivable at x = 1. Yet it has a minimum there. Similarly, k t i o n fix) =  I x I is not derivable at x = 0 but has a maximum thm. Can you point out another function fiom the examples above, which is not derivable but has an extremum? From this discussion we arrive at the following corollary: Corollary 1 If a ~ k t i o fnhas an extreme value at a point x = c, then one of the following conditions is satisfied :
3 ii)
f is not derivable at c. f is derivable at c, and f (c) =0.
'
Definition 3 A point at which either f (x) does not exist.or is zero is brown as'a critical point. If you study Definition 3 and Cor. 1 carefully, you will realise that any extreme point is a critical point. But the converse is not true. That is, a critical point need not be an extreme point. You will find an example to illustrate this in Remark 2 above. Example 5 Consider the function qx) = xY3for all x E R. 2
f (x) =  x'I3. Thus, this function is not derivable at x = 0. 3 This means that f has a critical point at x = 0. Further, at other points f exists, but f (x) =(2/3)x'" # 0. Hence, the function has only one critical point, namely 0. For every 6 > 0, and every x in ]6,6[,~x)=xm=(~'")22Oandfi~)=~.'Ihus~=~isafninimumpointandf(~)=~isthe minimum value (also see Fig. 7).
Fig. 7 : Graph of f(r)

1"'.
Drawing C u r v e s
Example 6 A farmer has a certain length of fencing wire. We wants to make a fence in a rectangular shape to keep his goats. Now what dimensions should he choose to ensure maximum enclosed area? It is clear that in order to get a maximum enclosed area, the farmer should use the whole length of wire. Suppose it is 4k metres. Then 4k is the perimeter of the enclosure. Thus if x and y are the length and the breadth, respectively, of the enclosure (in metres), then 2(x + y) = 4k, or y=2kx,O<x<2k. . Thus, we can regard the enclosed area xy (in m2)as a function f of its length x. Thus, we have .the function f defined as f(x) = x(2k  x), for all x > 0 (length is always positive). Our task is to find that value of x, for which Rx) becomes maximum. Now f(x) is a polynomial in x, and therefore f is derivable for all x. Hence its critical points (among which we should seek the extreme values) are obtained from the equation f (x) = 0, which in this case gives 2k  2x = 0, that is x = k.
I
This means that the function qx) = x (2k  x) has a critical point at x = k. Let us see whether it has an extremum at x = k. Now, f(k) = k2,and f(x)=2kx~~=k~(xk)~
Since there is only one maximum value, namely f(k), of qx), f(k) 1 f(x) for all x s.t. o < x < 2k.
...(1)
It is obvious from ( 1) that for values of x less than or greater than k, (x  k)2 > 0 and therefore, f(x)< k2= f(k). Hence, for any 6 > 0, f(x) 5 f(k)for all x E ]k 6, k + 6[. This means Qk) is a local maximum. But since there is only one extreme point, f(k) is the maximum value ofthe area. Now, when x = k, y = 2k x = k. So, the farmer should have a square fence made, taking the entire length of his wire.
In this example it was easy to compare the neighbouring function values with f(k). Are you wondering how you would tackle this problem when the functions under consideration would not be so nice? There is no need for speculation. We have suficient condition which tell us whether f(x) is an extreme value or not, and if it is, whether it is a minimum or a maximum value, without actually having to compare values. Section 6 is, in fact, devoted to such explorations. Before that, in the next section we shall discuss the mean value theorems. But now it's time to do some exercises.
1
1 I f
I
I
Find the critical points of each of the following functions. a) f(x)=(x3)(x5),v
XE
R.
b) f43=d+13x2+Sx+7,v X E R c) fTx)=sinx+3,t/ x e R. d) f(x)=ex,t/
XE
R.
e) f(x)=2lxl,v X E R f) f(x)=lxl + 2 , v X E R .
g) f(x)=(xl+lx1 ) , v X E R.
h) f(x)=x+ l/x,x>O
2.3 t
I
L
MEAN VALUE THEOREMS
,In this section we shall study the means value theorems. These theorems have proved to be very handy tools in proving other theorems not only in calculus, but also in other branches of mathematics, such as Numerical Analysis. Their importance lies in their wide applicability and tremendous usefulness. We shall first consider Rolle's Theorem, which is a special case of Lagrange's mean value theorem. We shall not attempt the proofs of these theorems here, but you will agree that both are intuitively obvious. We shall discuss their geometrical significanceand illustrate their usefulness through some examples.
2.3.1 Rolle's Theorem Rolle's Theorem was not actually proved by Rolle. He had only stated it as a remark. In fact, Michel Rolle (1652 1719) was known to be a critic of the newly founded theory of Newton and
4
The Ups and Downs
Drawing Curves
Leibniz. It is ironical then, that one of the most important theorems of this theory is known after him. Now let us see what this theorem is.
Fig. 8
In Fig. 8 we see the graphs of two continuous functions defined on the closed interval [a, b]. Here we observe the following features common to both of them. Precise Statement.
Rough S t a t e m e n t
1. The curve is drawn without breaks or gaps.
The function f is continuous on [a, b].
2. There are no comers in the curve, except possibly at the end points.
The function is differentiable in the open rnterval ]a, b[.
3. The two end points of the curve lie on the same horizontal line.
f(a) = f(b)
4. The curve admits a horizontal tangent (drawn as a dotted Irne) at some point.
f(c) = 0 for some c in ]a, b[.
i
The line joining the two end points may be imagined to be pushed upward or downward, keeping it always horizontal, and keeping the curve unmoved. Then there is a position, shown by the dotted line, where it touches the curve. This makes us believe that the fourth property holds for all the functions satisfying the first three properties. This is what Rolle's Theorem states.
Theorem 2 (Rolle's Theorem) Let f be a function continuous on the closed interval [a,b] and differentiablein the open interval ]a, b[. Further, letfja) = f(b). Then there is some c in ]a, b[ such that f (c) = 0. We give some examples below to illustrate this theorem.
Example 7 Consider f(x) = sin x on the interval [O, 2x3. All the assumptions ofRolle's theorem are satisfied here. f(O) = 0 = q2x)
1.1
1 1
:
I
I
Therefore according to Ro!le's theorem, there should exist c in ]0,2x[ such that f (c) = 0. Here f(c) = cos c.
Can we find an element c such that cos c = O? Yes. In fact there are two such points c in ]0,21t[, namely n12 and 31~12.
I
~
~
. .
The Ups and Downs
A
0 1
I Fig. 9
At d 2 , the function sin x attains its.maximumvalue. At 3d2, the functionsin x attainsits minimumvalue. Both these belong to the iat'e~al]O,2x[. Rolle's theorem asserts that there is at least one c in ]a, b[ such that f (c) = 0. Example 7 shows us that there may be more than one points m ]a, b[, at which f (x) = 0. In Rolle's theorem, a function f on [a, b] has to satisfy three conditions. i)
f is continuous on [a, b]
ii)
f is differentiable on ]a, b[
'iii)
f(a)= f@)
.
.
Now, we shall see through some examples that each of these conditions is essential. We cannot drop any one of them and still prove the theorem.
Example 8 Let fix) = x  [x] = fractional part of x, be defined on [0, I.].This can also be described. y as f (x) =
I
xifOSx<l. 0 if x = 1.
T
I,
Here 40) = f( 1) = 0. f is differentiable in the open interval 10.1 [.Thus, two of the three conditions af Rolle's theorem are satisfied by f. The derivative off is 1 at every pomt of 10.1 [. There is no point of ]O,1[ where the derivative is zero. What happens to Rolle's Theorem in this example? Obviously, its conclusion does not hold here.
/" +b
The reason is that f is not continuous on the closed interval [O, I], since it fails &be continuous at 1. Fig. 10
In the next example, we see that the assumption of differentiability in ]a, b[ cannot be omitted. Example 9 Consider f(x)= Ix I on [I, 11. There is no c in ] , 1,1[ such that f (c) = 0. Actual computation shows that
i
lif  1 < x < O f'= lifO<x<l does not exist at x = 0
I
I
i
f is continuous on [  1,1]. Also,f(l)=f(l). But f is not differentiable hE l , 1[. Our next example shows that the assumption f(a) = fib) is essential in Rolle's Theorem.
Example 10Let f(x)= x3on [O,l], Then f is continuous on [0, 11, and is differentiablein 10, 11. But fiO)# fil).
Y 9
k,
I\
/
0
/
~lg.11
b
X
Drawing Curves
In this case f (x) = 3x2# 0 for any x E 10, 1[. Thus, we see that the conclusion of Rolle's theorem may not hold when f(a) # f(b)
fY I
\
Lastly, we give an example where Rolle's Theorem is applicable and yields a unique c.
1
I
i
Example 11Let f(x) = x2on [I, I]. Then f (x) = 2x. Here all the three conditions of Rolle's Theorem are satisfied. There is only one c, namely c = 0, such that f (c) = 0.
\
4
You will now be able to solve these exercises.
I
E
Fig. 12
E4) Can Rolle's Theorem be applied to each of the following functions? Find 'c' in case it can be applied. a) y = sin2x on the interval [0, n]. b) f(x)=x2+1on[2,2]. C) f(x) = x3+ x on [0, 11. d) f(x) = sinx + cos x on [0, d2]. e) f(x) = sin x  cos x on [O,2n].
E
E5) Consider the function f(x) = x2 3x + 2. Prove that f(  1) = f(4). Find apoint c between 1 and 4 such that the derivative off vanishes at c. Is this point the midpoint of  1 and 4?
The Ups and Downs
I:LE E6) For the same function f(x) =x2 3x + 2, verify Rolle's Theorem
E
011 the interval [I, 21.
E 7) Let f(x) = ax2+ bx + c be the given function. If p and q are two real numbers such that
f ( ~= ) qd,prove ~ l a t
f1
=0 .
: E8) Consider the curve y = ax2+ bx + c. Let x, be the unique real number such that the
I
L
tangent at (x,, yo) to this curve is horizontal. Prove that the function y is oneone on the interval [x,, m [ (Hint :If f(p) = f(@, apply Rolle's theorem in the interval [p, ql).
Drewing Curves
E E9)
Let I be an open interval of R. Let f :I +R be a differentiable function such that f does not vanish on I. Prove that f is oneone on I.
2.3.2 Lagrange's Mean Value Theorem Now we shall discuss the mean value theorem. It was proved by Joseph Louis Lagrange, a towering mathematician of the eighteenth century.
We have already mentioned that Rolle's theorem is a special case of the mean value theorem. Let us recall the statement of Rolle's Theorem in the following f o m Let f be a continuous function on the closed interval [a, b], and differentiable in the open interval ]a, b[. The graph off is a curve in the plane. If the endpoints of this c w e lie in the same horizontal line, (that is, f(a) = 0)) there is a point c on the curve where the tangent to the curve is horizontal (f (c) = 0). . The last sentence' can be restated as follows. If the endpoints of the curve lie in the same horizontal line, there is a point on the c w e , where the tangwt to the curve is parallel to the line joining its endpoints. The mean value theorem asserts the same conclusion, even without the assumption of horizontality of the line joining the endpoints of the curve. Fig. 13 illustrates this difference. H m P and Q are the end points of the curve. The line PQ is horizontal m Fig. 13 (a), but not in Fig. 13(b). But in both the cases the point R on the curve has the property that the tangent to the cunle at R is parallel to the line PQ.The number c is the xcoordinate of R
Fig. 13
Fig. 13(a) illustrates Rolle's theorem, whereas Fig. 13(b) illustrates Lagrange's mean value theorem, The two end points of the curve are (a, f(a)) and (b,f(b)). The line joining these two points has the slope [f(b)  f(a)]/(b  a). Any line parallel to this line will a l s have ~ the same slope. Therefore, the conclusion of the mean value theorem is f (c) = [f(b) f(a)]/(b  a) for some a < c < b. This is because, we already know that f (c) is the slope of the tangent to the curve at (c, f(c)). Now we are ready to give the precise statement of the theorem. Theorem 3 (Lagrange's Mean Value Theorem) Let f be a continuous function on a closed interval [a, b]. Let f be differentiable in the open interval ]a, b[. Then there is a point c in the
open interval ]a, b[ such that f (c) =
f(b)  f (a) a
Rolle's Theorem has three assumptions : a continuity assumption, a differentiability assumption, and the assumption f(a) = f(b).
.
The mean value theorem has only two assumptions. These are the same as the fust two assumptions of Rolle's Theorem. Suppose in addition to the two assumptions of the mean value theorem, f(a) = f(b) also holds. Then what does the mean value theorem yield? It says that f(b)  f(a) for some a < c < b. But f(b)  f(a) = 0. f ( ~ ) = ba Therefore, we get f (c) = 0 for some a < c < b. This is the same as the conclusion of Rolle's theorem. This proves our contention that Rolle's theorem can be deduced from the mean value theorem. But why the name mean value theorem? What is the mean value here? f(a) is the initial value off. f(b) is the final value off. Therefore f(b)  f(a) is the total change in the value off. This change has occurred when the xcoordinate has changed from a to b. For a change of b  a in the domain, there is a change of f(b)  f (a) in the value of. Therefore, the mean value, that is, the average value of the rate of change is [@)  f(a)]/(b  a). The mean value theorem asserts that this average value of the rate of change o f f is assumed at some point c by derivative f . We shall illustrate the same thing by meansof an example. Consider a car moving from h i e a to time b; let f(t) be the position of the car at time t. Then the average speed of the car is distanceltime= [@)  f(a)]l(b a) According to mean value theorem, the speedometer of the car would have shown this [f(b)  f (a)]/ (b  a) at some time between a and b. For instance, if the car has travelled 100 krrrs. in two hours, at sbme point of time, its speed would have been actually 50 kmph. (which is its average velocity over the span of two hours).
Example 12 Let us veriQ the truth of Lagrange's mean value theorem for ~e hnction f(x)=x22xontheinte~al[1,2]. This is a polynomial function. Therefore it is continuous on [I, 21 and differentiable in ]1,2[. (We have seen this in Unit 2 and 3). Here
We want to check that f ' (c) = 1 for some c such that 1 < c < 2. Now f '(x) =2x  2. For what value of x will it be l?
The Ups and Downs
Drawing Curves
Now. 2x  2 = 1 when x + 312 and 312 E ] 1,2 [. Thus, we see that
Eow consider the function f : [a, b] + R which satisfies the assumptions of mean value theorem. Let p and q be any two points such that a I p < q I b. Is there some c between p and q such that
f (c)  [f(q)  f(p)j/(q  p)? To answer this, consider the restriction off to the interval [p.q]. It satisfies the assumptions of the meanvaluetheorem. Therefore such a point c exists. This result can be geometrically interpreted as follows. (p, f(p)) and (q, f(q)) are two points on the curve y = f(x). f(q). f(p) is qP the slope of this chord. What we have shown is that the slope of this chord is the same as the slope of the tangent at the point (c, f(c)). This means, that the tangent at (c, f(c)) is parallel to the chord (see Fig. 14). Thus, for any chord of the curve, there is a point on the curve where the tangent is parallel to the chord. The line joining them is called a chord of the curve.
Fig. 14
Example 13 i) Let us find the point c in ]  d 4 , n/4[ such that the tangent to f(x) = f(x) = cos x at (c, f(c)) is parallel to the chord joining ( 4 4 , f(d4)) and (d4, f (d4)). ii) We shall prove further that for the same c, the tangent at (c, g (c)) to the curve g(x) = cos x + x2+ x is parallel to the chordjoining (d4, g( ~ 1 4 )and ) (nl4, g(d4)).
9
The slope of the chord joining ( n/4, f( ~ 1 4 )and ) (n4, f(d4)) is f(nl4)f(n/4)  11431/43 =0 n/4(n/4) n/2 Therefore we seek c such that f (c) = 0. We have f (x) =sin x. The only point in 1 4 4 , d 4 , [ where this vanishes is at c = 0. The corresponding point on the curve is (0, f(0)) = (0,l).
The slope of the chord joining (( n/4), g( d4) and (~14,g(d4)) is
When c = 0, we want to prove that the tqngent at (c, g(c)) to the curve g(x) also has the same slope 1. In other words, we must prove that g' (0) = 1.
This proves that, for both the functions f(x) and g(x) over ]  nI4, n!4[, it is the same point c where the conclusion of the mean value theorem holds. ~xarnple14 For the curve y = In x, suppose we want to find a point on the curve where the tangent is parallel to the chord joining the points (1,O) and (e, 1).
Since In 1 = 0 and In e = 1, these two points (1,O) and (e, 1) lie on the curve y = In x. Consider this function on the closed interval [l, el (see Fig. 15). It is continuous there. It is also differentiable on 11,e[. I
Fig. 15
Therefore by the mean value theorem, there is a point c between 1 and e such that the tangent at (c, It! c) is parallel to the chord joining (1,O) and (e, 1). We have to find this point. Now, y' = l/x. Its value at c is l/c.
The Ups and Downs
The required point is given by
The required point on the c w e is (e  1, In (e  1)).
Remark 4 Let f : [a, b] ,R satisfy the assumptions of the mean value theorem. Then 9 8 , 0 < 8 < 1, suchthat@)=f(a)+(ba)f(a+O(ba)). This is because any point c between a and b is of the form a + 8(b Notethata=a+O(ba)andb=a+ 1 (ba).
 a) for some 0 < 8 <
Are you ready for these exercise?
E 10) Verify the mean value theorem for f(x)= x2+ 1 on the following intervals a) [I, 11 b) 11,21
E 11) Verify the mean value theorem on the interval [O, 21 for the following functions a) f(x)=sinm
Drawing Curves
E E 12) a)
Let f(x) = x3on [0,1]. Find a point c in ]O,1[ as in the mean value theorem.
b) Let f(x) = x3on [I, 01. Find a point c in 1 1,0[ as in the mean value theorem c) Let f(x) = x30n [I, 11. Show that there are two points c hi ]  1,1[ such that
E
E 13) Lef f be a function on [a, b] satisfying the assumptions of the mean value theorem. Let c be a point guaranteed by the mean value theorem. Prove that if g,(x)=flx)+ 1 and g, (x) = f(x) + x for all x in [a, b], then the same point c satisfies g, tb)  g, (a) ba
= g,'(c)
g2 (b)  gz (a) and ba
= g,'
(c) also.
.
Drawing Curves
The mean value theorem assures us that the equation 3x2 cos x = 25 n2 has atleast one solution, c. But it does not enable us to find the value or values of c. You can study methods of solving such equations iu the course in numerical analysis.
In the next section we shall see how the mean value theorem helps us to derive sufficient conditions for the existence of extreme points of a fimction.
2.4
SUFFICIENT CONDITIONS FOR THE EXISTENCE OF EXTREME POINTS
In Sec. 2 we have seen that a necessary condition for the existence of an extreme point of a given derivable function is that the derivative is zero at that point. We have also seen that the condition is not a sufficient one. In this section we shall discuss some tests which give suff~cientconditions for the existence of extreme points. Lagrange's mean value theorem which we have studied in Sec. 3 is used in deriving these tests.
2.4.1
First Derivative Test
The following theorem gives a sufficient condition for a function f to have an extreme value at an interior point c of its domain. It also tells us whether the extreme value is a minimum or a lnaximm
Theorem 4 (FIRSTDERIVATIVE TEST) Let c be an interior point of the domain of a func'tion f. Suppose f is derivable on ]c  6, c + 6[ for some 6 > 0 and that f (c) = 0. Then a)
i f f (x)>Owhenc6<x<cand f (x) iO when c < x <c + 6, then f has a maximum at c, and
b)
i f f (x)<Owhenc6<x<c and f (x) > 0 when c < x < c + 6, then f has a minimum at c.
Proof a) We have to prove that f has a maximum at c. In other words, we have to show that f(c) 1 f(x) for all x in some neighbourhood of c. Now, let x E ]c  6, c[. Then f is differentiable onlx, c[ (in fact on [x,c]), and continuous on [x, c] because differentiability at a point implies continuity there (Theorem 6, Unit 3). By Lagrange's theorem, there exists a E ]x, c[ such that f(c)  f (x) CX
=f
(a). That is,
f(c)  f(x) =(c  x) f (a).
...(1)
therefore, f (a) > 0. Also, c x > 0 since x E ] c  6, c[. Hence, from (1) f(c) f (x) > 0 or f(c) > f(x).
...(2)
Similarly, if x E ]c, c + 6[, then by Lagrange's theorem, there exists a point b E ] c, x[ such that f(x)f(c)=(xc) f (b).
...(3)
Since f * (x) < 0, when x E ]c, c + 6[, f*(b)<0. Further, since x E ]c, c + 6[, x  c > 0.
(3) now gives

...(4)
f(x) f(c) <0 or f(c)> qx).

Thus,it follows from (2) and (4) that whether x e ]c 6, c[ or to ]c, c + 6[, f(c) > f(x). In other words, f(c) 2 f(x) y x E ]C  6, c + S[. Hence, f(x) is a maximum value off and f has a maximum at C. The proof of part (b) proceeds on similar lines. See if you can write it yourself.
E E 15)Prove part (b)ofTheonm4.
Tbc Ups and Downs
We can also interpret Theorem 4 as follows. If while crossing a point, moving fiom left to right, the derivative changes sign fiom positive ' to negative, then that point must be a maximum.Similarly, while crossing a point, if the derivative changes sign from negative to positive, then that point must be a minimum.
Remark 5 From Theorem 4 we can deduce that when f (x) > 0 in an interval, the function f is strictly increasing in that interval. Similarly, f (x) < 0 in an interval would imply that f is a strictly decreasing function in that interval f = 0 in an interval, likewise, means that the function is a constant function in the interval. VJe shall talk about this in Block 4. You should note that the conditions stated in Theorem 4 are sufficient and are by no means necessary for the existence of extreme poins. Thus, a function may have an extreme value at c even though the conditio~wgiven in the theorem are not satisfied as shown in the following example.
f(0) = lim h+O
=
lim
f(0 + h) f(0) h
*
h [5 + sin(1 / h)] h
Thus f (x) exist for all x. However, neither in the interval ] 6,0[, nor in ]0,6[ does f (x) keeps the same sign, no matter how small 6 is (you may check this yourself). Thus, the question o f f (x) being positive or negative on two sides of 0 docs not arise.
Example 18Among all rectangles having a given area, is there one which makes the perimeter aminimum? Suppose that the length and the breadth of the rectangle are x and y, respectively. Then x > 0, y > 0. The area xy being constant, k2say, we have y = k2x' .Then, the perimeter of this rectangle I
,
!
= 2 (length +breadth) = 2 (x + k2 XI), x >0. This is a fhction of x alone. Let us denote it by fix). Thus,
f(x)=2(x+lr'X'),x>O.
Drawing Carves
we want that value of x which makes f(x)a minimum.The function f is derivable at all points of its domain. The derivative is given by f (x) = 2 [l (lC2Ix2)] Now,f(x)=0~x2=kZ~x=fk.1fweassumethatk>0,~enkisnotinthedomainoff.' Thus the only critical point is x = k. Let 6 be any positive number such that k  6 >0. Then f is derivable at all points of ]k  6, k + &. 1)
xâ‚ŹJK&k[*x<k, x2 <k=, k2ht2> 1, *f(x)<O
*
Hence by Theorem 4, f has amhimum at x =k. Also, when x = k,
y=kzxl =k.
This means that a square shapeyields a minimum perimeter among all rectangles with a fured area.
You should be able to solve the followidg exercisenow.
E E 16) Find all possible extreme values of each of the following functionsby applying the fmt derivative test a) f l ~ ) ~ ~ ~  5 d + 5 ~ ~  l f o b d x â‚Ź R [It may be helpful to fktorize f(x)]
2.4.2 Second Derivative Test
be up8 and Downs
We now investigate another condition which, if satisfied, does away with the need to examine the sign o f f (x) in ]c  6, c[ and ]c, c + S[ as in the first derivative test. This condition also is only sufficient, but very functional and hence, useful.
Theorem 7 (Second Derivative Test) Let f be derivable in ]c  6, c + 6[ for some 6 > 0 and suppose f (c) = 0. Then a)
f has a maximum at x = c provided f' (c) exists and is negative.
b)
f has a minimum at x =c proided f' (c) exists and is positive.
Proof :a) Since f"(c) < 0, f is a strictly decreasing function in the neighbourhood of c (see Remark 5). Thus there exists an & > 0 (and we may take it smaller than 6 ) such that f (x) > f:(c) when x E ]C  â‚Ź, C[ (sincex <c) and f (x) < f (c) when x E ]c, c + &[(since,here, x >c) Since f (c) =0, this means that f (x) > 0 when x E ] c  E, C[and f (x) < 0 when X E ]c,c+E[. By the first derivativetest it follows that f has a maximum at c. 'lhe proof for (b) follows along similar lines.
Remark 6 Ybu must have obsaved that this theorem says nothing about the cam when f"(c) isaim zero.In this case the function may have a maximumor a minimum value or neither as the followingexamples show.
3
f(x)=x4,forallx~R Here f (0) =0 = f' (O), but the functionbas a maximum at 0. (see Fig. 17 (a)).
ii)
f(x)=x4,f~allxeR H m f '(0) = 0 =f '(0). but the function has a minimum at 0. (see Fig. 17 (b))
Here f (0)= 0 = f'(0) and the functionhas neither a maximumnor a minimumat 0 (see Fig. 17 (c)). 'lhus, the first derivative test does have some merit.
S
Example 19 Let us find tbe extmne vrlues of the W o n f defined by f(x)=2x+3/x,forallx#O Here, f (x) = 2  3/x2,and therefore,
J37Z
f (x)=0*x=f Also, f' (x) =6xj, This means f'(&Z)>~andf'(
,/m)<0.
Drawing Curves
rn ), it follows from the second derivativetest that f has a minimum at rn and a maximum at  rn .The minimumvalue is Since, f (f(m)=
) = 0 =f (
2,k ,andthemaximurnvalueisf( m ) =  2 , k .
Example 20 From each comer of a square paper of side 24 cm, suppose we remove a square of side x cm and fold the edges upward to f o m an open box. Let us try to find that value of x which willgive us a box with maximum capacity.
Fig. 18
Clearly, 0 < x < 12 for a box to be formed. Also, the box thus formed has dimensions (24  2x), (24 2x) and x (see Fig. 18). The volume fix) is a function of x given by
Since 12 is hot in the domain of our function f, 4 is the only critical point. Also, f"(x)=24x 192.
Hence x = 4 is a maximum point off. The maximum value f(4) off (that is, the maximum capacity of thp box) is 1024cm3.
Are you surprised that the box is not a cube for maximum capacity? But, had it been a cube, four 414areseach of side 8 cm (the removed portions) would have been wasted, whereas, now four*sqmseach of side only 4 cm have been thrown away. There had to be a compromise between the waste mataial and making the box as near a cube as possible. Moral : Mathematics does not fail even though intuition may! Here are some exercises for you to solve.
E ,817) Find the extretne points for each of the followingfunctions. Point out which of them are maximum, which are minimumand which lue neither. Also find the extreme values off.
d)
qx) =a,+ a,x2 + q x 4+ ....+ anx2., where x E R and eacha, is positive. D o not get bogged down by the degree, of qx)!]
(Hint :A maximumpoint off will be the minimumpoint of llf aid vice versa).
The Upr and Downs
E
E 18) Show that d3 is a critical point off, whe~e qx) = sin x (1 + cos x), x E R Does f have a maximum or minimum at this point?
E E 19) Show that the rectangle of maximum area which can be inscribed in a ctcle, is a square. [Notice that the diagonals of the rectangle must be diameter of the circle]
E E20) A man wants a nameplate with &splay area equal to 48 cm2bordered by a white strip 2 cm along top and bottom and 1 cm along each of the two remaining sides. What dimensions should the plate have so that the total area of the plate is a minimum?
2.5
MORE INFORMATION FROM THE SECOND DERIVATIVE
In the previous section we used the sign of the second derivative at a critical point to discover wh& the function baa a maximumor a minimum at that point. For &wing the necessary conclusion we regarded f" as the derivative of the anction f , and then applied the first derivative test. We shall now use the monotonicity considerations o f f to draw another conclusion, which would give us an idea about the shape of the graph of the function f. Given that a function f increases in [a, b], its graph may be anything like (a), @) or (c) inFig. 19.
In this section we shall use the second derivative to determine the type of graph the given function f has. We shall assume that the function is twice differentiable in its domain.
The Ups and Downs
Fig. 19
2
Iff' >0 in some interval [a, b], thm f increases in this interval (Remark 5 applied to f). In other words, as x increases, the slope of the tangent to the graph at (x, qx)) increases, so that the tangent hvns counterclockwise. This results in the graph bending upward or bulging &wnward.Such graphs and functions are known as convex. We also use the terms concave upward or convex downward. Such graphs lie below their chord (linesegment joining the points on graph which correspond to x = a and x = b) and above their tangents. See Fig. 20 (a). Similarly i f f ' < 0 in [a, b] so that f decreases, and hence slope decreases, the tangent tums clockwise. Such functions are known as ccncave (equivalently, concave downward and convex upward). The graph in this case lies below the tangents and above the chord. (see Fig. 20 (b)).
We say that a function f is concave (convex) at a point, if it is concave (convex) in a neighbourhod of the point.
II
Rcmuk 7: i) The can&vity is towards the chord. If the chord lies above the graph, the graph is concave upward. When the chord lies below the graph, the graph is concave downward.
hi
Every convex function is of the form f, where if is concave.
hi Only concave and convex hctions have the property that each of their tangent lines intersect their graph, exactly once. Till now we.weh concerned only with the manner of bending of a graph. Let us now &s.souss the measure of the bending of a graph at a point also known as the curvature at that pomt. We measure the curvature at (c, qc)) by the ratio.
Drawing Curves
k(c) =
f "(c) [l + f ' ( ~ ) ~ ] ~ ' ~
1 The radius of curvature at (c, qc)) is denoted by p (c) and is defined by p(c) = k(c) ,if y c ) # 0. You will note that the radius of curvature is positive if the function is convex at that point and i s negative if the function is concave there. How does one get the ratio? Oh well! Surely, you don't thlnk this course covers the why of everything about functions? /'
.'I,< /. Points of Inflection f.. . /
/
LC,f(c))
,(/
2.5.2
Suppose f' (c) = 0 for some c E ]a, b[, and f changes sign in passing through c". That is, f" has one sign on the left of c, in ]c  c[ say, and the other on the right of c, in I*, c + 6 [ say. Then on one side of (c, qc)) the tangents are above the graph and on the other side of (c, qc)) these are below the graph. This means that the tangent at (c, qc)) must be crossing the graph (Also see Fig. 21).
//
,
Such points (c, qc)) are known as the points of inflection of (the graph of) f. Note that we did not really use the fact that f' (c) is zero. In fact f' (c) = 0 is neither a necessary nor a sufficient condition as you will find out from Examples 2 1 and 22. A function may have a point of inflection at x = c without Q' (c) existing at all. For the tangent to exist, the existence of f (c) is enough. However, note that iff' (c) exists in ]c  6, c + 6[ and changes sign in passing through c, then f' (c) must be zero.
~4
Example 21 Consider the function qx) = x 1 x 1, for all x E R shown in Fig. 22. This function can be rewritten as
Hence,
I '

But f' (0) does not exist and
Fir. 22
f "(x) =
2ifx>O 2ifxe0
Thus f' changes sign at 0 and the tangent crosses the graph at 0. x =O, is thus a point of inflection even though f' (0) # 0 (it does not exist!). Hence f' (c) = 0 is NOT a necessary condition for f to have apoint of inflection at x = c.
4
Example 22 Let qx) = x4.The first derivativetest shows that f has a minimum at x = 0. Thus f
das not have a point of inflection at x = 0 even though f'(0) = 0 (also see Fig. 17 (b)).This happens because f' (x) = 12x2> 0 for all x # 0, and accordingly it does not change sign in passing 0. Thus, f' (c) = 0 (by itself) is not sufficient either, for f to have a point of inflection atx=c.
~ x a m ~23 l ; Suppose we want b fmd the values of x for which the graph of the function qx) = llx, x E R\{O) is convex (i.e., concaveupward) and concave (i.e., concave downward). We shall also find if the graph (Fig. 23) has any points of inflection, and the curvature at x= 1. Here, f (x) = 11x2,f' (x) = 2/x3. Pig. 23
nT f'(x)+OfbranyxcR\{O),and
It follows fiom (i) that the graph in question is convex (i:e., concave upward) in 10, [. From (ii) and the fact that f' exists for all x in the domain off, we conclude that the graph has no points of inflection. From (iii) we deduce that the graph is concave (i.e., concave downward) in]  w, O[.
The Ups and Downs
Further, curvature at x = 1, is
Example 24 Suppose we want to examine the function f for points of inflection if f(x)=xZn+',forallx~ R,ne N. Therefore, f'(x) =0 +x =0. Since f' (x) exist for all x, there can be at most onepoint of inflection, namely, (0, @)), or (40). Now to the left of x 30, &at is, for x <0, f' (x) <0, and to the right of x = 0, f' (x) > 0. Hence, f' changes sign (fromnegative to positive) in passing through the origin. Thus the origin is a point of inflection of fix) = xw'.
E E21) Examine each of the following functions for concavity, convexity and points of inflection.
If f ' (c) exists, theh f '(c) = 0) is a necessary condition for (c, f (c)) to be a point of inflection.
D y w i n g Curves
I
I
E 22)
2.6
Find the curvature at an arbitrary point of the graph of the function
SUMMARY
In tbis unit we have discussedthe following points. 1) A function f is said to have a maximum (resp.,minimum) value at a point c of its domain if
+,
thereexists a positive number 6 such that for all x E ]C 6, c + f(x)5 f(c)(nsp. f(x)2 qc)). Maximum and minimum values are known as the extreme values of the function. 2) At an extreme point c of a h t i o n f, either the derivative does not exist or is zero.
3) Critical points for a h t i o n are those where either the derivative does not exist or else has the value zero. All extreme points are critical points. A critical point may fail to be an extreme point.
4) Rolle's and Lagrange's mean value theorems, and their geometrical inerpretation.
5) A sdiicienf condition for a function f to have an extreme value at x = c is that f is continuous at c and the derivative f changes sign in passing through c. If the change is from positive to negative, c is a maximum point. In the other event, c is a minimum point. This test is known as the first derivative test. 6) Second derivative test, another sufficient condition for the existence of extreme points asserts that i f f (c) = 0 then f' (c) > 0 implies f has a minimum at x = c and f' (c) < 0 guarantees a maximum value at c. 7) I f f ' (x) > 0 on some interval then f is convex on it. I f f ''(x) < 0 then f is concave on it. 8) I f f ' (c) = 0 or does not exist and f' changes sign in passing through c, then f has a point of inflection at x = c. This means the tangent at (c, f(c)) crosses the graph off at this point.
9) The radius of curvature =
2.7
.
,,(c)
SOLUTIONS AND ANSWERS
E 1) a) all point of R are maxima as well as minima:
b) no maxima or minima on R c) no maxima or minima on ]O,l{. d) 0 E R is a minimum. No maxima.
E2) f has a minimumat c a 3 6 such that qx)2f(c) VXE ]cS,c+S[
+~ ( x )  ~ ( c ) ~ O ~ / X E ] C  ~ , C + ~ [  f(c) * f(x) 2 O,c<x<c+6 (xc) f(x)  f(c) l0,cs<x<c (x c) 3 Rf (c) 2 0 and Lf (c) < 0. But since f is differentiable at c, f (c) exists and Lf (c) = Rf (c) a each is equal to zero. *f (c)=O.
and
E3) a) f (x)=(x3)+(x5)=2x8 . f(x)=Ox=4 x = 4 is a critical point.
..
d) no critical points e) x=O
f)x=O
g) All pohts x for which 0 5 x 5 1 are critical points because the function is &fwd ae
[I2x,if x < O
The Ups and Downs
Drawing Curves
E4) a) Y q y ; = 2 s i n x c o s x = s i n 2 ~ = 0 i f x = d 2 [O,IC] ~ b) Yes,f (x)=2x=0 ifxaO E [2,2] c) No. f (x) = 3x2+ 1 # 0 for x E [0, 11. Rolle's t h e ~ r ~ d onot e shold as 40)#4l)
E 5) c = 3 0 . Yes.
E8) Suppose f(x) is not (1  1) on [boo [ *p,q[xo,=[,suchthatp*qandf@)=f(q)
P*9 = x,,as x,, 2
is the unique point with f (x,,)~ 0 Therefore either p c x,,or q c x, since p and q both cannot be equal to x,,. This is a contradiction as we have taken, p,q E [x, a [. E9) Suppose p,q e I s.t. p # q and f@)= f(q) If p < q we have [p, ql c I, f is differentiableon [p, ql and f@)= f(q). Thus f satisfiesthe conditions of Rolle's theorem on [p, ql ~f(x,,)=Oforsomex,,e[p,qlcl. But this is a contradiction Therefore f is oneone
E l l ) a) qO)=O=f12) f (x)=~cosmaf(1/2)=0.
The Ups and Downs
E 13) f (c) =
f(b) f(a) =I ba
g2(b)  g2(a)  f(b)  f(a) + 1= f 9 ( c ) +1zg2.(c) ba b a E 14) a) y, = nxb'. Slope of the chord from (0,O) to (2,2") is
b) Slope of the chord from (0,O) to (1,l) is 1
1" point: piai' piT) ' ( I E 15) Let x E ]c  6, c[. Applying the mean value theorem to [x, c], 3a E ]x, c[, such that f(c)f(x)=f (a)(cx)<O,sincef (a)<Oandcx>.O.
Hence flc) >flx). Similarly, if x E ]c, c + 6[, then by the mean value theorem 3b E ]c, x[, such that f(x)f(c)=f (b)(xc)>Osincef (b)>Oandxc>O. a flx)>f(c),or f (c) <flx). *cisaminknun. E16) a) Max.atx=l,min.atx =3.Ext~emevaluesat.eOand28.Butx=Oisnotanextremum because there is a neighburhood of 0 in which f (x) has the same sign on either side of 0. tai ,' b) Min.atx= 1,x = 3 Max.x=1 Extreme values a~e3,285,29
E 17) a) x = 0; min. point. extreme value =0 b) There are no extreme points.
d) f (x) = xg(x) where g(x) is polynomial in x2with all coefficients positive. Hence g(x) >0 for all x # 0. Therefore the only extreme point off is x =0. Clearly, f(O)=a,andf(x)>a,,x#O.HenceOisarnin.andE.Visg
Drawing Curves
E 19) If a and b are the sides of the inscribed rectangle, a2+b2=d2*b=
J
/
Area=A=ab=a A8=Oifa=dJZ A"=<Ofora=dJZaa=#JZ
isamax.
a = d J Z a b = d J Z atherectangleisasquare. E20) Suppose the display area is a rectangle with sides a cm and b cm. Then the dimensions of the name plate are a + 2 cm and b + 4 cm. ab=48*b=48/a. A=(a+2)@+4)=(a+2)(48/~1+4)
Dimensions of the plate :2(1+
& ), 4(1+ & )
E21) a) Convex in 10, m[; concave in ]  , O[; point of inflection (P.1) = (0,o). b) Convex in ]  m, O[; concave in 10, m[; P.I. :0,O) c) Convex in],
l [ u ] 2 , m[; concave in] l,2[, P.1.: 1, (1,12),(2,47)
d) Convex if x > 3, concave if x <3; no P.I. e) Concave everywhere; no P.I. f ) Convex in]d2,3d2[; concave in 10, d 2 [ u] 3d2,2x[; P.I.; (d2,Q)and
(3fl, 0).
UNIT 3
GEOMETRICAL PROPERTIES OF CURVES
Structure 3.1
Introduction Objectives
Equations of Tangents and Nonnals Angle of Intersection of Two Curves Singular Points
32 3.3 3.4
3.4.1 Tangents at the Origin 3.4.2 Classifying Singular Points
Asymptotes
3.5
3.5.1 Asymptotes Parallel to the Axes 3.5.2 Obique Asymptotes
3.6 3.7
3.1
Summary Solutions and Answers
.
INTRODUCTION
. We started our study of Calculus by stating two problems. One of them was the problem of fmding a tangent to a curve at a given point. In Unit 3 we have seen that the solution of this problem was instrumental in the development of differential calculus. Now having studied various techniques of differentiation, we shall once again take up#&pioblem. The study of the tangents of a curve will then lead us to nonnals and asymptotes of curves, which we shall study in Sec. 2 and Sec. 5, respectively. In the last unit we discussed some other geometric featuresof functions, like maxima, minima, points of inflection and m a t u r e . You will see that all these will prove very useful when we tackle curve tracing in the next unit.
Objectives After studying this unit, you should be able to obtain the equations of the tangent and the nonnal to a given curve at a given point calculate the angle of +tersection of two curves at a given point of intersection obtain the angle between the radius vector and the tangent at a point on a given curve define the identify a singular point, a node and a cusp defme asymptotes and obtain their equations.
3.2
EQUATIONS OF TANGENTS AND NORMALS bD
In Unit 3 we have seen how a tangent can be defined precisely with the help of derivatives. We have noted that the slope of a tangent to the curve y = f(x) at (x,, yo) is given by f (xo), whenever it exists. In fact, we had also obtained the equations of the tangents of some simple curves. Once we know how to find the equation of a tangent, it is easy to find one for a normal too. A nonnal to a curve, y = f(x) at (x, yo) is a line which passes through (4yo) and is perpendicular to the tangent at that point. This means that the slope of this nonnal will be
Now, what happens when f (XO) = O? f (%) = 0 implies that the slope of the tangent at (x,, yo) is zero, that is, this tangent is parallel to the xaxis. In this case, the normal (which is perpendicular to'the tangent) would be parallel to the yaxis. The equation of this normal, would then be x = x,. Now let us look at various curves and try to obtain the equations of their tangents and normals.
A line L, is perpend~cularto a l~neL, iff rnl% = I, where rn, and % are the slopes of L, and L,, respect~vely.
Drawing Curves
a = 0 gives us the bivial case when the curve is the line y = 0, or the xaxis
Recall that the equation of a line through (x,,, y,,) having a slope rn is y  yo = rn(x  x,,).
a,
Example 1 Consider the'curve y = 2 a # 0, shown in Fig. 1. 2a d~ . Thus, exists and is nonzero for all y # 0. Now y will be zero only if dx Y dx x is zero. Thus, we can fmd the equations of tangents and normals to this curve at any point, except the origin (0,O). We know that the slope of the tangent at any point (x,,, y& will be 2aly,. The slope of the n o d will, therefore, be yd2a. Thus, the equation of the tangent at (%Y&is 2a YYo(x*
y,
The equation of the normal at (x,,,y& is
Now let us see an example where the equation of the curve is given in the parametric form. In Unit 4 we have already seen what a parameter is.
Example 2 To find the equations of the tangent and the normal at the point 0 = d 4 to the curve given by x = a cos30,(see Fig. 2), we first note that dy dy/d0 3asinZ0cos0 =  tan0 dx dx / d0 3a cosZ0 sin 0 =X
Fig. 1
Hence, the slope of the tangent at 0 = d 4 is  tan d 4 = 1. The slope of the normal at this point thus comes out to be 1. Now, if 0 = d 4 , cos 0 = 11fi and sine= 1 / f i Thus, x = a/2& and y = a12 fi . The equation of the tangent at (d2 fi ,d 2 fi ) is
The equation of the normal at (aD&
,a/2 f i ) is
Example 3 illustrates the method of fmding the equations of tangents and normals when the equation of the c w e is given in the implicit form.
Example 3 Let us find the equations of the b e n t and the normal to the curve defined by x3+f6xy=Oatapoitlt(x,,,y&onit. Fig. 2
Fig. 3 shows this curve.
In Unit 4 we have seen how to calculate the derivative when the relation between x and y is expressed implicitly. We shall follow the same procedure again. Differentiatingthe given equation throughout with respect to x, we get dy dy 3x2+ 3y2   6y6x  =0, which means dx
dx
Thus, the slope at the point ( x , yo) is
2 ~ 0xo2  2x0
y02
1
Hence, the equation of the tangent at (xo, yo) is
L
Simplifying, and using the relation x , , ~+ :y = 6x,,y0, this reduces to (2y0x,2)x+(23,~,2)~+ 2 w , = o Now the normal at (s,yo) is a line passing through (x,, yo) and having slope
p
 (yo2 2 ~ 0 ) 2 ~ 0X o
2
. Hence, the equation of the normal at (x,,,yo) is
If you have followed these examples, you should have no problem in solving the following exercises.
E E 1)
Find the equations of the tangent and the normal to each of the following at the specified point.
Geometrical Properties of Curves
Drawlng Curves
Vertical Tangents By now, you are quite familiar with the fact that f (x) or dy!dx may not exist at some points. At such points either the tangent does not exist, or else, is parallel to the yaxis, that is, vertical. To examine the existence of vertical tangents at (x,, yo), we examine = 0 , then, we conclude that there is a vertical tangent at
(x,, yo). In such cases the equation of the tangent can be written as x = %. The normal corresponding to a verticalgkngent will obviously be horizontal or parallel to the xaxis. This means we can write its equation as y = yo, as its passes through (%, yo).
If you take the curve in Example 2, you will find that
dy does not exist when 9 = d 2 . dx

dx dx Let us examine at this point.  =cot 8 = 0 if 8 = d 2 . dy dy This means that the curve has a vertical tangent and, consequently, a horizontal normal at this point. Now, when 9 = d 2 , x = 0 and y = a. Thus the equation of the tangent at (0, a) is x = 0 and that of the normal is y = a. See if you can solve this exercise now.
E E2)
Are there any points on the following curves where the tangent is parallel to either axis? If yes, find all such point..
Let us now look at mother example.

Example 4 To find the equations of those tangents to the c w e y x3,which are parallel to the line 12x  y  3 = 0, we f m observe that the slope of the line 12x  y  3 = 0 is 12. Thus, the slope of any line parallel to this line should also be 12. Now, the slope of the tangent to the c w e y = x3at m y point (x, y) is f (x) = 3x2.
If we equate f (x) to 12, we will get those points on the curve where the tangent is parallel to 12~~3=0.
' I
I I
Thus, the points in question are (2,8) and (2, 8). The equations of the tangents at these points are y  8 = 12(x 2) and y + 8 = 12(x + 2), respectively. The following exercises will give you some more practice in applying the concepts learnkd in this section.
E E3)
Find the equations of the tangent and the normal to each of the fdllowing curves at the point 't' :
E E4)
Find the equation of the tangent to each of the following curves at the point (%, yo) a) xz+y2+4x+6y1=0
b) xya
Geometrical Properties or Curves
Drawimg Curves
E E5)
Prove that the line 2x + 3y = 1 touches the curve 3y = eZX at a point whose xcoordinate is zero.
E E6) Rove th.1 the equation of the nonnal to the hyperbola x2 y2 = ~ a t a p o i m ( a f i , b ) i s u + b f i y = ( a ~ + bf~ i) . a2
b2
Drawing Curves
Fig. 4 shows Y  @ to be an acute angle. But if the curves f and g were as in Fig. 5, then angle 8 = x  (Y  0 ), since we take the acute angle as the angle of intersection.
But we are not in a position to decide whether we should take tan 8 as tan (Y  0 ) or as
 tan (Y  a ) , unless we have drawn the curves. Since it would be tedious to fust draw the curves and then decide, we think of an alternate scheme. We observe that since 8 lies between 0 and d 2 , tan 0 is nonnegative. Thus,we take tan 0 to be [tan (Y  a)].
Having proved this theorem, we can easily deduce the following corollaries.
Corollary 1 Two curves y = qx) and y = g(x) touch each other at (x,, y,), that is, have common tangents at (x,, y,) iff 8 = 0, that is, iff f (x,) = g'(x,). Corollary 2 Two curves cut each other at right angles, or orthogonally, at (x,, y,) iff f (x,) g' (x,) =I. If you study Example 5 cawhlly, you will have no difficulty in solving the exercises later.
Example 5 Let us find the angle of intersection of the parabola 9 = 2x and the circle x2+9=8. First we fmd the points of intersection of these curves, if there are any. The coordinates of these points will satisfy both the equation to the parabola and the equation to the circle. 2x in x2+ 9 = 8 ,we get So substituting 9'= x2+2x=8,0rx=4or2. It is clear from 9 = 2x that the abscissa x (= y2/2)of any common point must be nonnegative. So we reject the value  4 of x. When x = 2, y = f 2. Hence the common points are P (2,2) and Q (2, 2). Since both curves are symmetric about the xaxis (see Fig. 6) and since P and Q are reflections of each other w.r.t. the xaxis, it is sufficient to find the angle at one point, say P; the angle at Q being equal to the angle at P. Differentiating the two equations wr..to x, we get
Ceometricrl Properties of Curves
Pig. 6
Hence the values o f f (x) and g' (x), that is, the slopes of the tangents to the two curves at (x, y) are liy and x/y. Hence the s+s of the tangents at (2,2) to the two curves are 112 and 1. Hence if 9 is the required angle, then
Hence, 8 = tan' 3 = 71.56' You can by these exercises now.
E E 7) Find the angle of intersection of the parabolas 9 =4x and x2= 4y.,'
E E8)
Show that a) the ellipse x2+ 4y2= 8 and the byperbola x2 2y2= 4 cut each other orthogonally (at right angles) at four points. b) the curves xy = a2and x2+ yJ = 2a2 touch each other (have a common tangent) at two.points.
tan 8 = ylx. The equation of a curve is sometimes expressed in polar coordinates by an equation r = f(8). For example, the equation of a circle with centre 0 and radius r is r = a. Now let us turn once again to the problem of finding the angle of intersection of two curves.
Geometrical Properties of Curves
This method that we have been following till now, cannot be used if the equation of the curve is given in the polar form. In this case we follow a somewhat indirect method.
Fig. 8
Take a look at Fig. 8. It shows a curve whose equation is given in the polar form as r = f(8). P(r, 0) and Q(r + 8r, 0 + 88) are two points on this c w e . PT is the tangent at P at OPR is the line through the origin and the point P. We shall now try to find 8, the angle between PT and OR. We note here, that the tangent PT is the limiting position of the secant PQ. If we denote the angle between PQ and OR by a,then we can similarly say that I$ is the limit of a as. _ , Q + P along the curve. Now from AOPQ we have OQ OP
=
sinLOPQ siil LOQP
Remember the sine rule for a AABC ? sin A sin B ==
A
R
sinC
C
r + Sr sin (n  a ) or = r sin ( a 60) 6r or1+= r
sin (n  a ) sin (a68)
6r sin a sin(a  68) or  = (since sin ( x a)= sin a ) r sin ( a  68) 1 Sr' or= r 68
sin A
 sin B
= 2 sin
(?)A  B
cos
A +R
2cos(a68/2)sin(68/2) sin (a  68). 68
 2 cos ( a  68 12) sin (60 I 2)

sin (a 60)
60 1'2
Recall lim
1 dr cos 4 == r d0 sin 4
m,o
cot 4
de
sothat tan$= r.
dr
This formula helps us to find the angle between OP and the tangent at the point P on the curve defined by the equation r = f(0).
sin (68/ 2) 6812 ='
Drrwlng Curves
We shall use this result to find the angle between two curves L, and C, which intersect at P (say). If the angles between OP and the tangents to C, to C, and P are $, and d, respectively, the angle of intersection of C, and C, will be I $,  d I (see Fig. 9).
x Fig. 9
This can be easily calculated as we know tan 8, and tan +,.
Further, if the curves intersect orthogonally, tan $, .tan $2 =  1. The following examples will help clarify this discussion.
Example 6 Suppose we want bo find the angle of intersection of the curves r = a sin 28 and r = a cos 28 at fhe point P (a/ JZ ,d8).The coordinates of P satisfy both the equations r=asin28andr=acos20. If 8, is the angle between OP and the tangent to r = a sin 28, then 1 de = a sin 28  a sin 28 =tan 1 r28 = dr drId8 2acos28 2 2 Similarly, if $, is the angle between OP and the tangent to r = a cos 28, then
nus9M(qlq2)=
tan41  t a n 9 2 l+mgI
=
112+112 4 =1114 3
Thus, g,  q2= M  I (413) = 53.13O, which is the required angle. Now try to do a few exercises on you own.
E E9)
Find the angle between the line joining a point P(r, 8) on the curve to the origin and the tangent for each of the following curves.
a) ? = a2cos 28 c)
.F= amcos me
Drawing Curves
3.4
SINGULAR POINTS
In this section we shall study a category of points on curves, called singular points. But to properly classify singular points we have to find the nature of tangents at these points. So let us first study on easy method of finding the tangents to a curve at the origin. This knowledge will then help us to find the tangents at any point of the c w e easily.
3.4.1
Tangents at the Origin
We shall now give you a simple method of finding the tangent to a curve at the origin, when the equation of the curve is given by a polynomial equation. That is, the equation is of the form f(x, y) = 0 where f is a polynomial in x and y. You will agree that the constant term in this polynomial is zero since the curve passes through the origin. For such a curve the equation of the tangent at the origin can be found out by equating to zero the lowest degree terms in x and y (we shall not prove the here). Thus, if x3 + 3xy + 2x + y = 0 is the equation of a curve, the equation of the tangent at the origin is2x+y=O. Similarly, if the equation of a curve is x4 + x2 y2 = 0, then the equation of the tangent to this cweat ( o , o ) ~ ~ x ~  ~ ~ = o , ~ ~ x ~ = ~ ~ , o ~ x = + ~ .
When the origin is shifted to (h, k) the coordinates of a point P(x, y ) in the new coordinate system are given by
x'=xh y' = y  kr
Hence we get two equations x = y and x =  y. This means that the curve x4 + x2  y2 = 0 has two tangents at the origin. We shall consider such eventualities in the next subsection. Now, consider a curve given by g(x, y) = 0, where g (x, y) is some polynomial in x and y. Suppose we want to find the equation of the tangent to this curve at some point, say (h, k), on it. What we do is, we shift the origin to (h, k). Then with respect to this new origin, the equation of the curve will be g(x + h, y + k) = 0. We can also express it by G(x, y) = 0, where G(x', y') = g(x + h, y + k). Note that this change of origin does not change the shape of the curve. Now the tangent to the curve G(x, y) = 0 at the origin will be the tangent to the curve g(x, y) = 0 at the point (h, k) (see Fig. 10(a) m d (b)).
The method for finding the equation of the tangent at any point of a curve will be clear to you when you read our next example.
Example 7 Consider the curve defined by the equation ay2= x(x + a)2. Let us find the equation of the tangent to this curve at the point (a, 0). For this, we first shin the origin to (a, 0). The equation of the curve then becomes ay2=(xa)(xa+a)' or a? = x2 (x  a).
Geometrical Properties of Curves
We can also write this as a(x2 + y2)= x3 Now, the equation of the tangent to this curve at the origin will be given by ~ ( x ~ + ~ ~ ) = O =3~~+~2=0 a x 2 LY2
This is impossible, since the square of any real number has to be nonnegative. But we can write this as x =
I
+ iy, where i = ais an imaginary number.
Thus the equations of the tangents to the given curve at the point (a, 0) are x + a = iy (shifting back the origin).
+
I
In such cases we say that the curve has imaginary tangents at the point ( a , 0). Now that you have seen how to find the tangents to curves given by polynomial equations, let us try and categories the points on a given curve with the help of the tangents at those points. Applying the procedure used in the Example 7, you should be able to solve this exercise.
E
E 11) Find the equations of the tangents at the origin to each of the following curves
3.4.2 Classifying Singular Points An equation of the type y = qx) determines a unique value of y for a given value of x. This means, every straight line parallel to the yaxis meets the curve y = qx) in a unique point. However the equation of a curve is often given as f(x, y) = 0. If flx, y) is not a linear expression in y, then it may not be possible to write qx, y) = 0 in the form y = qx) uniquely. For example, if f(x, y) = xZ+ yZ a2,then qx, y) = 0 gives y = + ,/= This gives us two relations y =
and y = 
Jmof the type y
= F(x).
Each one of these two relations determines a certain portion or branch of the curve x2+y2=a2,asyoucanseefrom~ig. 11.
Points A and B are common to the two branches. Put, differently we can say that two branches of the circle x2+ yZ= a2pass through points A and B. We have a generic name, singular points, for points like A and B. A precise defmition is as follows : Definition 1If k branches of a curve pass through a point P on the curve f(x, y) = 0 and k > 1, then P is said to be a singular point or a multiple point of order k. Singular points of order two are known as double points. Thus the points A and B in Fig. 1l(c) are double points. Obviously, a curve will have more than one tangent at a singular point (one corresponding to each branch). Depending upon whether tangents at double points are distinct, coincident or imaginary, we shall give special names to such points. Definition 2 A double point is known as
3
a node if the two tangents at that point are real and distinct,
ii)
a cusp if the two tangents are real and coincident,
ii]
a conjugate (or isolated) point if the two tangents are imaginary.
In Fig. 12 we show an example of each. For the curve f(x, y) = 0, the origin is a node. For the curve g(x, y) = 0, the points P,, P2,P, and P, are cusps, while the point Q on the curve h(x, y) = 0 is a conjugate point.
In Example 7 we have seen one more example of a conjugate point. See if you can solve this exercise now.
E E 12) Show that ( 1, 2) is a singular point on each of the following curves.
Also determine the nature (cusp, node etc.) of this singular point in both cases. (Hint :Shift the origin to ( 1,  2) and check the tangents at the new origin.
In this section we shall study another feature of curves which will prove very useful in tracing curves as you will see in the next unit. This involves taking limits as x = +f or y + f =. You have come across such limits in Unit 2. Let us define an asymptote now.
Definition 3 A straight line is said to be an asymptote to a curve, if as a point P moves to infinity along the curve, the perpendicular distance of P from the straight line tends to zero. Example 8 Consider the rectangular hyperbola xy = c shown in Fig. 13,xy = c implies y = CIXand this implies that as x + = or  , y + 0. Now ( Y I is the distance of a point P(x. y) on the hyperbola from the xaxis. So, we can say, that as x + m or  the distance 00,
Geometrical Properties of Curves
Drawing Curves
of a point, P(x, y) on the hyperbola from the xaxis approaches zero. In other words, this means that the xaxis is an asymptote of the hyperbola.
Writing xy = c as x = cly, and repeating the arguments exactly as above, we can prove that h e yaxis is also an asymptote of the hyperbola. 10

\
Example 9 Let us prove that the xaxis is an asymptote of the curve y = shown in I+x Fig. 14. From the equation of the curve, it is quite clear that y 4 0 as x , or  a.Again, this means that the distance of the point P(x, y) on the curve from the xaxis tends to zero as x + or  a.This proves that the xaxis is an asymptote of the curve.
3.5.1 Asymptotes Parallel to the Axes ~ip.14
Here we shall derive tests to decide whether a given curve has asymptotes parallel to the x and y axes. For this we shall consider a curve given by qx, y) = 0, where f(x, y) is a polynomial in x and y.
Theorem 2 A straight line y = c is an asymptote of a curve qx, y) = 0 iffy  c is a factor of the coefficient of the highest power of x in fix, y). Proof Arrange qx, y) in descending powers of x so that the equation of the curve is written as
P(X, y) tends to infinity means atleast one of x and y must
tend to infinity.
76
The perpendicular distance PM of P(x, y) from the line y = c is I y  c 1. (Check this by drawing a suitable figure). Now according to Definition 3, y = c is an asymptote iff PM tends to zero as P tends to infinity, that is iffy ,c as P tends to infinity. If the y coordinate y of P + c (a finite number) as P tends to infinity, then its xcoordinate x must tend to infnity. Now since P is a point on the curve, its coordinates satisfy the equation 4%Y)=0.
Geometrical Properties o f Curves
So, as P tends to infinity along the curve, we get lim f(x, y) = O p'm

From this we can say that y = c is an asymptote iff lim f(x, y) = 0 x+m YrC
Hence, y = c is an asymptote of (1) iff

lim [g,,(y) + g,(y)
x+m
1 1 ; + ........+ gn (y) p] = 0
Y'C
&(c) = 0, y  c is a factor of g,,(y), the coefficient of the highest power of x in f(x, y).
This theorem can also be interpreted as follows. Asymptotes parallel to the xaxis are obtained by equating to zero the real linear factors of the coefficient of the highest power of x in the equation of the curve. We can also state a theorem, similar to Theorem 2, giving a test to decide whether a given curve has an asymptote parallel to the yaxis or not. Theorem 3 Asymptotes parallel to the yaxis are obtained by equating to zero the real linear
factors ax + b of the coefficient of the highest power of y in the equation of the curve. Proof :Similar to that of Theorem 2
1 Example 10 Let us find the asymptotesparallel to either axis for the curve y = x +  . X
Writing the given equation in the form f(x, y) = 0, we have x2 xy + 1 = 0. You can see the graph of this curve in Fig. 15. The coefficient of the highest power of x is 1. It has no factors of the form y  c. Hence there are no asymptotes parallel to the xaxis. The coefficient of the highest power of y when equated to zero gives x = 0. Hence there is one asymptote parallel to the yaxis and moreover, it is the yaxis itself. See if you can do these exercises on your own.
E E 13) For each of the following curves, find asymptotes parallel to either axis, if there are any. a) x2y=2+y
b) xy2=16x2+2W
~ i g 15 .
Drawing Curves
3.5.2 Oblique Asymptotes You may be wondering whether an asymptote must always be parallel to a coordinate axis. No, there are many curves having asymptotes which are not parallel to either axis. Such asymptotes are generally referred to as oblique asymptotes. We shall now learn how to find oblique asymptotes y = mx + c to rational algebraic curves f(x, y) = 0. The problem is to determine m and c so that y = mx + c may be an asymptote to f(x, y) = 0. The perpendicular distance of a point P(x,, y,) from the line ax+by+c=Ois
Suppose that the line y = mx + c is an oblique asymptote to the curve f(x, y) = 0. This means that m = 0. The perpendicular distance PM of a point P(x, y) on this curve from I ymxc ( y=mx+cisgivenbyPM= J T .Now, since m + 0. P can be at infmity on the (l+m curve only when x (as also y), tends to m. Thus, as x + m, PM + 0. This means that asx+m,(ymxc)+O. or, lim (ymxc)=O X+Q
That is, c = lim (y  m). xPm
Thus c would be known as soon as m is known. Now,
= lim (ymx) lirn X+Q
X00
=c.O=O,
Hence, tn
%! ):(
using (1)
. Y
Thus, given any c w e f(x, y) = 0, we first find lim  = m and then use this m to XbQ x calculate c = lirn (y  mx). The followin~~x"arnp1e will clarify this procedure. Example 11Let us examine the curve x3 y3 = 3xy for oblique asymptotes. Suppose that the given curve has an oblique asymptote y = mx + c. The equation of the curve can be written as x3y33xy=o.
Dividing throughout by x3 we get
Iy3 x3
3~ 1= o x'x
[
..I
33 Thus, lim 1  y3  x+m x
k
y =o
1 0,since lirn  = 0 x+m x
x+m
I
*
m3= 1 m = 1, the other roots of m3 1 = 0 being complex numbers. Rewriting the equation of the curve as (x  y) (x2+ x i + 9)= 3xy, we have c = lim (y  x)
=
x+m
lim x+
=
lim x+m
  3
[
I
x 2+~Y+Y2 3 2
X
,since lim  = lim
1+1+1
I
Geometrical Properties of Curves
x+m
y
= 1
. b
;[
:C
i
1
Hence the required asymptote is y = x  1. Try to solve this exercixes now.
x+m
.'
E E 14) Find oblique asymptotes to each of the following curves. a) x3+y3=3ax2
b) k4f+xy=O
1
Drawing Curves
SUMMARY
3.6
In this unit we have covered the following points. 1)
2) 3)
The equation of the tangent at (x,, yo) to the curve y = f(x) is Yy,=f (x,)(x%J dx The curve has a vertical tangent at (x,, yo) if  = 0 at this point. dy The angle 8 of intersection of two curves y = f(x), y = g(x) is the acute angle between the tangents at that point to the curves. It is given by the relation tan 8 =
1 ll?o(~~!~~) 1
4)
y = fix) and y = g(x) cut each other orthogonally at (%, yo) i f f (x,) . g' (x,)  1.
5)
The angle $ between the tangent and the radius vector of the curve r = f (8) at the do point 8 is given by tan $ = r  . dr The tangents at the origin to any curve (which passes through the origin) are obtained by equating to zero the lowest degree terms in the equation of the curve.
6)
7)
If k branches of a curve pass through a point P on the curve fix, y) = 0 and k > 1, then P is said to be a singular point or a multiple point of order k. Singular points of order two are known as double points. A double point is known as a node, a cusp or a conjugate (isolated) point according as the two tangents at that point are real and distinct, real but coincident, or imaginary.
8)
A straight line is said to be an asymptote to an infinite branch of a curve, if as a point P on the curve moves to infinity along the curve, the perpendicular distance of P from the straight line tends to zero.
9)
Asymptotes parallel to the coordinate axes are obtained by equating to zero the real linear factors in the coefficients of the highest power of x and the highest power of y in the equation of the curve.
Y lirn 10) If y = rnx + c is an oblique asymptote of the curve fix, y) = 0, xPw X lirn (y  mx) = c. X+Q
::lx=,
E 1) a) 
.
= 4 Equation of the tangent at (1,4) is
(y4)=4(x 1) Slopeofthenonnalat(1,4)=114 Equation of the normal at ( l,4) is (y  4) = ( 114)(x  1). b) Slope of the tangent =  bla Slope of the normil = alb at t = ~ 1 1x, = a/ ,5 ,y b/ ,/Z . Equation of the tangent : (Yb/&)= bla(xdfi)

c) Slope of the tangent = 314
= m and
Geometrical Properties o f Curves
Slope of the normal =  413 Equation of the tangent : y  4 = (314) (x + 3) Equation of the normal :y  4 = ( 413) (x + 3) E2) a) Tangents are parallel to the xaxis at x
= (1 f
f i )/3
b) Tangents are parallel to the xaxis at all points where x = nx: + x:/2 for some
integer n. There are no tangents parallel to the yaxis. E 3) a) Tangent : ty = x + at2 Normal :y + tx =at(2 + tZ) b) Tangent : (1 + cos tjy = sin t(x at). Equivalently, sin (t/2)x  cos (tl2)y = at sin (V2) Normal : sin (t/2)y + cos (t/2)x = 2a sin (d2) + at cos (d2)
I
2
=   (0, 113) is a point on the curve. The tangent E 5) 3y = e2xzdy x = o dx 3
at (0,113) is given by
a Slope of the normal =  a/b .,
a a Equation of the normal is y  b = (X a f i ) bJZ
E7) yZ =4x~x=y2/4~x2=y4/16=4yatthepointofintersection. + ~ ~  6 4 ~ = 0 ay(y364)=0 a y ( y  4 ) ( y 2 + 4 y + 16)=O + y = 0,4 (other roots are complex) a x=Oor4. Slope of the tangent to y2 = 4x at (4,4) = 112 Slope of the tangent to x2 = 4y at (4,4) = 2 a angle of intersection = tan' (314) The tangent at (0,O) to y2 = 4x is vertical, and the tangent at (0,O) to x2= 4y is horizo~ltal. Hence the angle of intersection at (0,O) is ~ 1 2 . E8) a) Thefourpointsare ( 4 / & , & & / 3 ) , (  4 / & , *
:.
They cut orthogonally.
a13)
Drawing Curves
E 9) a) 2r
=
d0 2a2 sin 20 d;
a angle = tan' (r
a tan
+, tan g2=  1
d0
r =
2
2
r 2 sin 20 = tan' ( cot 28)
d0 z) = ='an'

~
the curves cut orthogonally.
b) The curves cur orthogonally.
El 1) a) 16?= 16x2x4
The equation of the tangent is 16y2  16x2=0.a?xZ=O*y=f
x.
b) y22xzy+x4=x4+3x3. The equation of the tangent is y2 = 0 * y = 0.
c) Equation : y2= 0 or y = 0.
,. 
E 12'
2)
Change the crigin to (1, 2). Then the equation of the curve is ( ~  1 ) ~ + 2 i( )~2 + 2 ( ~  i ) ( y  2 )  ( y  2 ) 2 + 5 ( ~ 1)2(y2)=0. ~ X ~  X ~ + ~ X ~  ~ ~ = O The equation of the tangents at the origin is x22xy+ y2=o e(~y)~=oox=~. There are two real and coincident tangents at this point. Hence it is a cusp.
b) After shifting the origin we get the equation : (x1)3+(~1)(y2)2+2[(x1)2+(y~2)2]+4(x1)(y2)+5(x1) +8(y2)+8=0. . The equation of the tangents at the origin is y2 x2 = 0, that is, y2=x20ry= + x There are two real anddistinct tangents at this point. Hence it is a node.
E13)a) x Z y = 2 + ye x z y  y  2 = 0 Highest power of x is 2. The coefficient of x2 is y. Heece y = 0 i s an asymptate. Highest power of y is 1.The coefficient of y is x2 1= (x  1) (x + 1). Hence x =  1 and x = 1 are two asymptotes. b) No asymptotes parallel to the xaxis. x = 20 is an asymptote.
c) No asymptotes parallel to the yaxis. y = 0 is an asymptote.
+
d) y = 3 are asymptotes. x = f 3 are asymptotes.
Geometrical Properties o f Curves
e) y = 0 is an asymptote. f ) y = 0 is an asymptote.
I
+
c= =
lim
x+m
= lim x+m
3a x

lim (y  mx) = lim (y + x)
x,m
xPm
3ax2
Jim x+m

3
):(
X
2
3a 1+1+1
= lim xm
3a I   ~ / X + ( ~ / ~ ) ~
a
Hence the equation o f the asymptote is y + x = a.
UNIT 4
CURVE TRACING
Structure 4.1
Introduction Objectives
42 4.3 4.4 4.5 4.6 4.7
Graphing a Function and Curve Tracing Tracing a Curve :Cartesian Equation Tracing a Curve :Parametric Equation Tracing a Curve :Polar Equation Summary Solutions and Answers
4.1
INTRODUCTION
A pichue is worth a thousand words. A curve which is the visual image of a functional relation gives us a whole l d of infonnation about the relation. Of course, we can also obtain this infonnation by analysing the equation which defines the functional relation. But studying the associated curve is otten easier and quicker. In addition to this, a curve which represents a relation between two quantities also helps us to easily find the value of one quantity corresponding to a specific value of the other. In this unit we shall try to understand what is meant by the pichue or the graph of a relation like f(x, y) = 0, and how to draw it. We shall be using many results from the earlier units here. With this unit we come to the end of Block 2, in which we have studied various geometrical features of functional relations with the help of differentialcalculus.
Objectives After studying this unit, you should be able to list the properties which can be used for tracing a c w e trace some simple curves whose equations are given in Cartesian, parametric or polar forms.
4.2
GRAPHING A FUNCTION AND CURVE TRACING
Recall that by the graph of a function f :D 4 R we mean the set of points {(x,f(x)) :x E D) . Similarly, the set of points {(x, y) :f(x,y) = 0) is known as the graph of the functional relation qx, y) = 0. Graphing a function or a functional relation means showing the points of the corresponding set in a plane. Thus, essentially curve tracing means plotting the points which satisfy a given relation. However, there are some difficulties involved in this. Let's see what these are and how to overcome them. It is often not possible to plot all the points on a curve. The standard technique is to plot some suitable points and to get a general idea of the shape of the curve by considering tangents, asymptotes, singular points,extreme points, inflection points, concavity, monotonicity, periodicity etc. Then we draw a free hand curve as nearly satisfying the various properties as is possible. The curves or graphs that we draw have a limitation. If the range of values of either (or both) variable is not fmite, then it is not possible to draw the complete graph. In such cases the graph is not only approximate, but is also incomplete. For example, consider the simplest curve, a stpight line. Suppose we want to draw the graph off : R 4 R such that qx) = c. We know that this is in line parallel to the xaxis. But it is not possible to draw a
I I
Curve Tracing
Fig. 1
In the next section we shall take up the problem of tracing of curves when the equation is given in the Cartesian form.
4.3
TRACING A CURVE : CARTESIAN EQUATION
Suppose the equation of a curve is f(x, y) = 0. We shall now list some steps which, when taken, will simplify our job of tracing this curve. 1) The first step is to determine the extent of the curve. In other words we try to find a region or regions of the plane which cannot any point of the curve. For example, no point on the curve y2 = X, lies in the second or the third quadrant, as the xcoordinate of any point on the curve has to be nonnegative. This means that our curve lies entirely in the first and the fourth quadrants.
A curve is symmetrical about a line if, when we fold the curve on the line, the two portions of the curve exactly coincide.
A point to note here is that it is easier to determine the extent of a curve if its equation can be written explicitly as y = f(x) or x =fly).
2)
The second step is to find out if the curve is symmetrical about any line, or about the origin. We have already discussed symmetry of curves in Unit 1. Fig. 2, shows yo6 some examples of symmetric curves.

Here we give you rorne hinta to help you determine the rymmetry of r curve. a) If all the powem of x occurring in f(x, y) 0 are even, then f(x, y) ir rymmetrical about the y ax^.
f(
x, Y)mdthe curve
A curve is symmetrical about the origin if we get the same curve after rotating it through 180"
Drawing Curves
In this case we need to draw the portion of the graph on only one side of the yaxis. Then we can take its reflection in the yaxis to get the complete graph. We can similarly text the symmetry of a curve about the xaxis.

b) If f(x, y) = 0 f(x,  y) = 0, then the curve is symmetrical about the origin is such cases, it is enough to draw the part of the graph above the xaxis and rotate it through 180' to get the complete graph. c) If the equation of the curve does not change when we interchange x and y, then the curve is symmetrical about the line y = x. Table 1 illustrates the applicatioilof these criteria for different curves. Table 1
Equation
SY
x3+y+)r=o
About the xaxis (even powers of y)
X4+f'+y=O
About the yaxis (even powers of x)
X4 + x2g + f
x2+f=10
=0
About the origin (6(%~)=06(&~)=0) About both axes (4% Y)= 4 % Y),4% Y)= 4% Y)) Abouttheliney=x (4% Y)= 4~94 ) About both axes, (even powers of x and y) but not about y = x 4% Y)+ 4~3x)
3) The next step is to detennine the points where the curve intersects the axes. If we put y = 0 in f(x, y) = 0, and solve the resulting equation for x, we get the points of inersection with the xaxis. Similarly,putting x = 0 and solving the resulting equation for y, we can find the points of intersection with the yaxis. 4)
Try to locate the points where the function is discontinuous.
5) Calculate dyldx. This will help you in locating the portions where the curve is rising (dyldx > 0) or falling (dyldx < 0) or the points where it has a comer (dyldx does not exitst).
6) Calculate d2y/dx2.This will help you in locating maxima (dyldx = 0, d2y/dx2<0) and minima (dyldx = 0, d2y/dx2> 0). You will also be able to determine the points of inflection (d2y/dx2= 0). These will give you a good idea about the shape of the curve.
7) The next step is to find the asymptotes, if there are any. They indicate the trend of the branches of the curve extending to infmity.
8) Another important step is to detennine the singular points. The shape of the curve at these points is, generally, more complex, as more than one branch of the curve passes through them. 9) Finally, plot as many points as you can, around the points already plotted. Also try to draw tangents to the curve at some of these plotted points. For this you will have to calculate the derivative as these points. Now join the plotted points by a smooth curve (except at pointa of diecontinuity). The migents will guide you in this, re they give you the direction of the curve.
We shall now illustrate this procedure through a number of exiunples. You wlll notice, that It may not be necessary to take all the nine steps mentioned above, in each case. We begin by tracing some functions which were introduced in Unit 1.
Example 1Consider the h c t i o n y = I x 1. Here y can take only poeitive values. Thus, the
graph lies above the xaxis. Furthet, the fundtion y = I x 1 is symmetric about the yaxis. On the tight of the yaxis, x > 0 and ao ( x 1 = x, Thus the graph reduces to that of y = x and you know that thitl is a straight line equally inclined to the axes (Fig. 3(a) below).
The curve meets the yaxis only at the origin. Taking its reflection in the yaxis, we get the complete graph as shown in Fig. 3@). We have drawn arrows at the end of the line segment to indicate that the graph extends indefinitely.
Fig. 3 : (a) Graph to the right o f the yaxis.
(B) Complete
graph.
Example 2 The greatest integer function y = [x] is discontiwous at every integer point. Hence there is a break in the graph at every integer point n. In every interval [n, n + 1[ its value is constant, namely n. Hence the graph is as shown in Fig. 4. Note that a hollow circle around a point indicates that the point is not included in the graph.
Fig. 4 : Graph of (I) = 1x1.
Example 3 Consider the curve y = x3. Now (x, y) lies on the curve a y = x3 w  y = (  x ) e ~ ( X,  y) is on the curve. This means that the curve is symmetric about the origin. Thus, it is sufficient is draw the graph above the xaxis and join to it the portion obtained by rotating is through 180째.
3fi
Above the xaxis, y is positive. Hence x = must be positive. Thus, there is no portion of the graph in the second quadrant. The curve meets the axes of coordinates only at the origin and the tangent there, is the xaxis. dy = 3x2which is always nonnegative. This means that as x increases, so does y. Thus the dx graph keeps on rising.
This implies that there ate no extreme paints, and that (0,O) is a polfit of inflectiorl. The gtaph Y has no asymptotes paballel to the axes. Furthst lim  = lim x' and x i s , x %+a
Curve Tracing
Drawing Curves
obviously, this does not exist. This means that the curve does not have any oblique asymptotes.You can also verify that it has no singular points. The graph is shown in Fig. 5. 4
Fig. 5: Graph of y = x3
1 Example 4 Consider y = 3. The ycoordinates of any point on the curve cannot be negative. X .
So the curve must be above the xaxis. The curve is also symqetric about the yaxis. Hence we shall draw the graph of the right of the yaxis first. The curve does not intersect the axes of coordinates at all.
4
2 and dy < 0 for all x > 0,the function is = .Since dx x3 dx2 X dx nonincreasing in 10, [, that is, the graph keeps on falling as x increases. Further, since dy =
dy is nonzero for all x, there are no extreme points. dx
d2y Similarly, since 7is nonzero, there are no points of inflection. Writing the equation of the dx curve as x2y = 1, we see that both the axes are asymptotes of the curve. There are no singular points. Therefore, the curve does not fold upon itself. The curve is shown in Fig. 6.
Example S Let us try to trace the curve given by the equation xy = 1. Here we can nee that either x and y both will be positive or both will be negative. This means that the curve lies in the fmt and the third quadrants. Fwther, it is symmetric about the origin and hence, it is sufficient to trace it in the first quadrant m d rotate this through 180째 to get the portion of the curve in the third quadrant.
(1, 1) is a point on the curve and x = lly means that as x increases in the first quadrant, y decreases.

1 Now the distance of any points (x, y) on the curve from the xaxis = i y I y =  + 0 as X
x + m . This means that the xaxis is an asymptote. Arguing on the same lines we see that the yaxis is also an asymptote. dy dx
=
1
#
0 for any x. That is, there are no extrema.
x2
dy At the ~ o i n(t1 , l ) we have,  =  1, which implies that the tangent at (1.1) makes an angle dx
of 135" with the xaxis. Considering all these points we can trace the curve in the first quadrant (see Fig. 7 (a)). Fig. 7 (b) gives the complete curve.
(a) Fig. 7 (a) Graph of xy

(b) 1 in tbe first Quadrant (b) complete graph.
The curve traced in Example 5 is a hyperbola. Ifwe cut a double cone by a plane as in Fig. 8(a), we get a hyperbola. It is a section of a cone. For this reason, it is also called a conic section. Figs. 8(b), (c) (d) and (e) show some other conic sections. You are already familiarwith the circle in Fig. 8(d) and the pair of intersecting lines in Fig. 8 (e). The curve in Fig. 8(b) is called a parabola and that in Fig. 8(c) is called an ellipse.
(b)
(c) Fig. 8
The earliest mention of these curves is found in the works of a Greek mathematician Menaechmas (fourth century R.C.). Later Apol!onlus (third century B.C.) studied them extensively and gave them their current names. In the seventeenth century Rene' Descartes discovered that the conic sections can be characterised as curves which are governed by a second degree equation in two variables. Blaise Pascal (: 6231662) presented them as projections of a circle. (Why don't you try this? Throw the light of a torch on a wall at different angles and watch the different conic sections on the wall). Galileo (15641642) showed that the path of a projectile thrown obliquely
Curve Tracing
Drawing Curves
(Fig. 9) is a parabola. Paraboloid curves are also used in arches and suspe~isionbridges (Fig. 10). Paraboloids surfaces are used in telescopes, search lights, solar heaters and radar receivers.
Fig. 9
Fig. 10
In the seventeenth century Johannes Kepler discovered that planets move in elliptical orbits around the sun. Halley's comet is also known to move along a very elongated ellipse. A comet or meteorite coming into the solar system from a great distance moves in a hyperbolic path. Hyperbolas are also used in sound ranging and navigation systems.
Let's look at the next example now.
Example 6 Consider the curve y = x3+ x2. There is no symmetry and the curve meets the axes at (0,O) and ( 1,O). dy dy  3x2+ 2x. The xaxis is the tangent at the origin as  = 0, at x = 0. Since dx
dx
dy
 = 1 when x =  1, tangent at (1,O)
dx
makes an angle of 45" with the xaxis (Fig. 1l(a)).
I
d2y dy d2y Further 7= 6x + 2. This means (0,O) is a minimum point as  = 0 and > 0 at dx dx dx dy d2y x = 0. The point (213,4127) is a maximum point as  = 0 and 7< 0 at x = 213. Thus in dx dx Fig. 1 1 (b), 0 is a valley and P is a peak.
Fig. 11
d2y
 = 0 at x =
dx
Curve Tracing
 1 and changes sign from negative to positive as x passes through  113. 3
Hence (113,2127) is a point of inflection.
dy 2 If   < x < 0, then  < 0. Thus the graph rises in ]  ,  213 [ and 10, [, but 3 dx falls in ]  2/3,0[. As x tends to infinity, so does y. As x , , so does y. There are no asymptotes. Hence the graph is as shown in Fig. 1l(c). I'
I  i.
So far, all our curves were graphs of functions. We shall now trace some curves which are not the graphs of functions, but have more than one branch. Example 7 To trace the semi cubical parabola y2 = x3, we note that x3is always nonnegative for points on the curve. This means x is always nonnegative and no portion of the curve lies on the left on the yaxis. There is symmetry about the xaxis (even powers of y). The curve meets the axes only at the origin. The tangents at the origin are given by y2 = 0 so that the origin is a cusp. (see Sec. 4 in Unit 8).

g
2t 4
2
I
3
\h \
',
\ t
In the first quadrant y increases with x, and y + as x ,. There are no asymptotes, extreme points and points of inflection. Taking reflection in the xaxis we get the complete graph as shown in Fig. 12. Example 8 Suppose we want to trace the curve. y2=(x2)(x3)(x4).
,
If x < 2, we get a negative value for y2 which is impossible. So, no portion of the curve lies to the left of the line x = 2. For the same reason, no portion of the curve lies between the lines x=3andx=4. Since y occurs with even powers alone, the curve is syrnrnekical about the xaxis. We may thus trace it for points above the xaxis and then get a reflection in the xaxis to complete the graph. The curve meets the axes in points A(2.0). B(3,O) and C(4,O). At each of these points, the curve has a vertical tangent (see Sec. 2 of Unit 8). Combining these facts, the shape of the curve near A, B, C must be as shown in Fig. 13 (a).
Fig. I 2 : Seml cubical parabola, y2 = xz
I
'
Drawing Curves
Let us take y > 0 (i.e., consider points of the curve above the xaxis). Ihen
and p = 3  11J? then a lies between 3 and 4, and can therefore be ignored. Also, 3 x 2 '  1 8 ~ + 2 6 = 3 ( x  P ) ( x  a ) a n d 2 < P < 3 < a . ~ o r x]2,3[,xaremains ~ dy negative. Hence for 2 < x < $,  < 0 since (x  a ) and (x  p) are both negative. dx dy Similarly, for P < x < 3,  < 0. Hence the graph rise in.j2, P[and falls in ]P, 3[. Thus
dx
the shape of the curve is oval above the xaxis, and by symmetry about the xaxis: we can complete the graph between x = 2 and x = 3 as in Fig. 13(b). Now let us consider the portion of the graph to the right of x = 4. Shifting the origin to (4, O), the equation of the curve becomes y2=x(x+ l)(x+2)=x3+3x2+2x. As x increases, so does y. As x + m ,so does y (considering points above the xaxis). When x is very small, x3and 3x2are negligible as compared to 2x, so that near the (new) origin, the curve is approximatelyof the shape of y2 = 2x. For large values of x, 3x2and 2x are negligible as compared to x3, so that the curve shapes like y2 = x3 for large x. Thus, at some point the curve changes its convexity. This conclusion could also be drawn by showing the existence of a point of inflection. There are no asymptotes or multiple points. Considering the reflection in the xaxis, we have the complete graph as shown in Fig. 13 (c).
Example 9 Let us trace the curve (x'  1) ( 9  4) = 4. There is symmetry about both axes. We can therefore sketch the graph in the first quadrant only and then takes its nflection in the yaxis to get the graph above the xaxis. The reflection of this graph in the xaxis will give the complete graph. Notice that the origin is a point on the graph and the tangents there, are given by 4x2 + y2 = 0. These being imaginary, the origin is an isolated point on the graph. The curve does not meet the axes at any other points. For x > 0, y > 0, the equation (x2 1) ( 9  4) = 4 shows that x should be greater than 1 and y should be greater than 2.
Fig. 14
Equatmg to zero the coefficients of the highest powers of x and y, we get y = f 2 and x = f 1 as asymptotes of the curve. Thus, the pomon of the curve in the first quadrant ~pproaches the lines x = 1 and y = 2 in the region far away from the origm.
I
I
s
4 In the first quadrant, as x increases, so does x'  1, and since x2 1 =
(y2 4) '
y decreases as x increases.
There are no extreme points, singular points or points of inflection. As x + m ,y
+ 2 and as y + 0 0 , x +
Example 10 To trace the curve y"
r
I. Hence the graph is as shown in Fig. 14.
(x  1) (x  2)' we note that there is symmetry about the
xaxis.
No portion of the curve lies to the left of x = 1. Points of intersection with the axes are A(l, 0) and B(2,O) and the tangent at (1,O) is vertical. Shifting the origin to B(2, O), the curve transforms into y' = x2(x + 1). The tangents at the new origin B, are given by y2 = x2. This means that B is a node, and the tangents at B are equally inclined to the axes. Let us try to build up the graph above the xaxis between x = 1 and x = 2. Differentiating the equation of the curve with respect to x, we get.
when 1 < x < 2, (x  2) < 0. If y is positive, then y' > 0 provided 3x  4 < 0. Thus y' > 0 when x E ] 1,4/3[ and y' < 0 which x E ]413,2[. The tangent is parallel to the xaxis when 3x  4 = 0, that is, when x = 413 (see Fig. 15(a)).Hence, for 1 < x < 2, the curve shapes as in Fig. 15(b). Now for x > 2, As x + = , y + in the first quadrant. Note that when B(2,O) is taken as the origin, the equation of the curve reduces to Y 2 = x 2 ( ~1+) = ~ 3 + ~ 2 00
This shows that when x > 0 and y > 0, the curve lies above the line y = x (on which y2 = x2). Hence the final sketch (Fig. 15 (c)) shows the complete graph.
Curve Tracing
Drawing Curves
If you have gone through Examples 1  10 carefully, you should be able to do the following exercise.
E 1) Trace the curves given by a) y=x2 b) y 2 = ( ~  2 ) 3 c) y(l+x2)=x d) y 2 = ~ 2 ( 1  ~ 2 ) (Graph paper is provided at the end of this unit.)
TRACING A CURVE : PARAMETRIC EQUATION
4.4
Sometimes a functional relationship may be defined with the help of a parameter. In such cases we are given a pair of equations which relate x and y with the parameter. You have already come across such parametric equations in Unit 4. Now we shall see how to trace a curve whose equation is in the parametric form. We shall illustrate the process through an example.
Example 11 Let us trace the cycloid x = a(t + sin t), y = a(1 cos t) as t varies from  n ton. dx dt
 = a(l
d~
+ cos t), = a sin t, so that dt dx dt
dy dx
 = tan (tl2). Since  > 0 for all t E 1 n, n[, x increases with t from  a A (at
dy Also  is negative when t E ] n, 0[ and positive when t E 10, A[.Hence y decreases dx from 2a to 0 in [ n, 01 and increases from 0 to 2a in [0, x]. Let us tabulate this data. t E [O, XI
tE [n,O]
x increases from 0 to a
i ) x increases from  a to 0
i)
ii) y decreases from 2a to 0
ii) y increases from 0 to 2a
iii) Hence the curve falls
iii) Hence the curve rises
Also, at the terminal points n, 0 and n of the intervals [  n, 01 and [O, x], we have the following. t x
(x, Y) (ax,2a)
0
(0,o)
R
(are, 2a)
dx
dy 

dx
dy
not defined
0
0
not defined
not defined
0
Tangent vertical horizontal vertical
On the basis of the data tabulated above, the graph is dnwn in Fig. 16.
I
Fig. 16
I
I
Curve Tracing
Remark 1If t is increased by 2x, x is increased by 2xa and y does not change. Thus the complete graph can be obtained in intervals....... [  5x,3x3, [ 3x, x],[x,3x1,[3n,5x1 ..... by mere translation through a proper distance. The cycloid is known as the Helen of geometry because it was the cause of many disputes among mathematicians. It has many interesting properties. We shall describejust one of them here. Coilsider this question : What shape should be given to a trough connecting two points A and B, so that a ball rolls from A to B in the shortest possible time? Now, we know that the shorter distance between A and B would be along the line AB (Fig. 17). But since we are interested in the shortest time rather than distance, we must also consider the fact that the ball will roll quicker, if the trough is steeper at A. The Swiss mathematicians Jakob and Johann Bernoulli proved by exact calculations that the trough should be made in the form of an arc of a cycloid. Because of this, a cycloid is also called the curve of the quickest descent. The cycloid is used in clocks and in teeth for gear wheels. It can be obtained as the locus of a fixed point on a circle rolls along a straight line. <
A
\(\:?:.
See if you can do this exercise now.
,>,. = ..__ ::. _ .__  _ .. :: '.>.
,,, ,> :\
E2) Trace the following curves on the graph paper given at the end of this unit.
\,I
.
_  __ _ _ _ _  :  .:,,      _  /: :::_I
.._ __.Fig. 17
4.5
TRACING A CURVE : POLAR EQUATION
In this section we shall consider the problem of tracing those curves, whose equations are given in the polar form. The following considerations can be useful in this connection. Symmetry :If the equation remains unchanged when 8 is replaced by  8, then the curve is symmetric with respect to the initial line. If the equation does not change when r is replaced by  r, then the curve is symmetric about the pole (or the origin). Finally, if the equation does not change when 8 is replaced by x  8, then the curve is symmetric with respect to the line 8 = d 2 .
Extent : (i) Find the limits within which r must lie for the permissible values of 8. If r < a (r >a) for some a > 0, then the curve lies entirely within (outside) the circle r = a. (ii) If $ is negative for some values of 8, then the curve has no portion in the corresponding region.
Angle between the line joining a point of the cuwe to the origin and the tangent :At suitable points, this angle can be determined easily. It helps in knowing the shape of the curve at these points. Recall that angle $ is given by the relation tan $ = r
de dr'
We shall illustrate the procedure through some examples. Study them carefully, so that you can trace some curves on your own later.
Example 12 Suppose we want to trace the cardioid r = a (1 + cos 0). We can make the followihg observations. Since cos 8 = cos (  â‚Ź9, the curve is symmetric with respect to the initial line. Since  1 5 cos 8 5 1, the curve lies inside the circle r = 2a. .'!It.
,,
 ~.
dr dr  =  a sin 8. Hence  < O when 0 < 8 < x . Thus r decreases as 0 increases in the interval d0 de
.
Drawing Curves
10, R[. Similarly, r increases with 8 i n ] ~ / 2R[. , Some corresponding values of r and 8 are tabulated below.
'
d0  a(l + cos 0)  =  cot (0 12) = tan . This shows that the angle between the r dr  a sin 0 line joining a pont (r, 0) on the curve to the origin and the tangent is 0 or n/2 according to 6 = x or 0. Hence the line joining a point on the curve to the origin is orthogonal to the tangent when 0 = 0 and coincides with it 8 = x. Combining the above facts, we can easily draw the graph above the initial line. By reflecting this portion in the initial line we can completely draw the curve as shown in Fig. 18. Notice the decreasing radii 2a, c,, c,, c, etc.
L/'
This curve is called a cardioid'since it resembles a heart.
Example 13Let us trace the equiangular spiral r = aeeCota. We proceed as follows.
Fig. 18
dr
= r cot a, which is positive, assuming cot a > 0. Hence as 8 increases so does r. d0 de r  = tan a. Thus, at every point, the angle between the line joining a point on the curve to dr the origin and the tangent is the same, namely a. Hence the name.
y4 1
/
113
Combining these facts, we get the shape of the curve as shown in Fig. 19. The equiangular (or logarithmic) spiral r = aeeCO'a is also known as the curve of pursuit. Suppose four dogs start from the four comers of a square, each pursues the dog In front with the same uniform velocity (always follow~ngthe dog in front in a sbalght line), then each will describe an equiangular spiral. Several shells and fossils have forms which are qulte close to equiangular spirals (Fig. 20). Seeds in the sunflower or blades of pine cones are also arranged in this form.
Fig. 19
This spiral was first studied by Descartes in 1638. John Bernoulli rectified this curve and was so fascinated by it that he willed that an equiangular spiral be carved on his tomb with the words 'Though changed, I rise unchanged' inscribed below it.
,
I '
, '
Fig. 20
4y
I
The spiral r = a0 is known as the Archimedean spiral. Its study was, however, initiated by Conan. Archimedes used this spiral to square the circle, that is, to find a square of area equal to that of a given circle. This spiral is w~delyused as a cam to produce uniform linear motlon. It is also used as casings of centrifugal pumps to allow air which increases uniformly in volume with each degree of rotat~onof the k n blades to be conducted to the outlet without creating backpressure. The spiral r0 = a, due to Varignon, is known as the reciprocal or hyperbolic (recall that xy a is a hyperbola) spiral. It is the path of a particle under a central force which varies as the cube of the distance. Now let's consider one last example.
Examplq 14 To trace the curve r = a sin 38, a > 0, we note that there is symmetry about the line 8 = n12, since the equation is unchanged if 8 is replaced by n  8.
 .. . j
The curve lies inside thr circle r = a, because sin 38 5 1. The origin lies on the curve and this is the only point where the initial line meets the curve.

r = 0 6 nnI3. where n is any integer. Hence the origin is a multiple point, the l~nes 8 = 0, d 3 , 2 ~ . / 3n,, 4n/3,5n/3,2n etc. being tallgents at the pole.
I (0 a) Pig. 21
96
dr
= 3 cos 38. Hence r increases in the intervals 10, d 6 , [ I n!2,5d6[, de
and ]7x/6,3x/2[, and
decreases in the intervals ]n/6, ~ / 2 []566,7n/6[ , and]3n/2,5n/3[. Notice that r is negative when 8 â‚Ź]x/3,27~/3[or8 E In, 4n/3$r 8 E ]5n/3,2x[.
Hence the curve consists of three loops as shown in Fig. 2 1. The function is periodic and the curve traces itself as 9 increases from 2n on. Now try to trace a few curves on your own.
E 3) Trace the following curves on the graph paper provided. a) r = a ( l cosO),a>O. b) r = 2 + 4 c o s 0 . c) r=acos39,a>O. d) r=asin29,a>O (Graph paper is provided at the end of this unit.)
4.6
SUMMARY
In this unit we have covered the following points. 1)
Tracing a curve y = f(x) or f(x, y) = 0 means plotting the points which satisfy this relation.
2)
Criteria for symmetry and monotonicity, equations of tangents, asymptotes and points of inflection are used in curve tracing.
3)
Curve tracing is illustrated by some examples when the equation of the curve is given in a) Cartesian f o m b) Parametric f o m c) Polar form
4.7
SOLUTIONS AND ANSWERS
Dotted lines represent tangents or asymptotes throughout. E l ) a)
b) Shifting the origin to (2,O) we get y2= x3 which you know how to draw.
't
C u r v e Tracing
Drawing Curves
c) y = x is the tangent at the origin. Origin is a point of inflection. xaxis is an asymptote. Either x, y are both positive or both negative. Function rises in 1 1,1[ and falls elsewhere. Graph is shown alongside.
d) 5  1  x
shows that the entire curve lies within the lines x = f 1. Tangents at the
JZ ,1/4), symmetry
origin are y = f x. Tangents at x = f 1 are vertical. Maxima at (f 11 about both axes.
curve Tracing
Drawing Curves
UNIT I
DEFINITE INTEGRAL
Structure 1.1
Introduction Objectfves
1.2
Preliminaries 1.2.1 Partitions of Closed'Interval 1.2.2 Upper and Lower Product Sums 1.2.3 Upper and Lower Integrals
1.3 1.4 1.5 1.6
Defrnite Integral Fundamental Theorem of Calculus Summary Solutions and Answers
1.1
INTRODUCTION
We have seen in Unit 3 of Block 1 that one of the problems which motivated the concept of a derivative was a geometrical onethat of finding a tangent to a curve at a point. The concept of integration was also similarly motivated by a geometrical problemthat of finding the areas of plane regions enclosed by curves. Some recently discovered Egyptian manuscripts reveal that the formulas for finding the areas of triangles and rectangles were known even in 1800 B.C. Using these one could also find the area of any figure bounded by straight line segments. But no method for finding the area of figures bounded by curves had evolved till much later. In the third century B.C. Archimedes was successful in rigorously proving the formula for the area of a circle. His solution contahed the seeds of the present day integral calculus. But it was only later, in the seventeenth century, that Newton and Leibniz were able to generalise Archimedes' method and also to establish the link between differential and integral calculus. The definition of the definite integral of a function, which we shall give in this unit was fist given by Riemann in 1854. In Unit 2 of this block, we will acquaint you with various methods of integration. You have probably studies integration before. But in this unit we shall adopt a new approach towards integration. When you have finished the unit, you should be able to tie in our treatment with your previous knowledge.
Objectives After reading this unit you should be able to : define and calculate the lower and upper sums of some simple functions defined on [a,b], correspondmg to a partition of [a,b], define the upper and lower integrals of a function, define the definite integral of a given function and check whether a given function is integrable or not, state and prove the Fundamental Theorem of Calculus, use the Fundamental Theorem to calculate the definite integral of an integrable function. '
1.2
PRELIMINARIES
We have mentioned in the introduction that Archimedes was able to find the formula for the area of a circle. For this he approximated a circle by an inscribed regular polygon (See Fig. 1 (a)). Further, we can see from Fig. l(b) that this approximation.becomesbetter and better as we increase the number of sides of the polygon. Archimedes also tried to approximate the area of the circle by a gumber of circumscribed polygons as in Fig. l(c). The area of the circle was thus rnmnressed hetween the inscrihed and the circumscribed nolv~ons.
Fig. 1
We shall follow a similar procedure for finding the area of the shaded region shown in Fig. 2. We begin with the concept of a partition. 1.2.1
't I
Partition of a Closed Interval
Let us consider the closed interval [a, b] c R Then we have the following definition. Defintion 1 Let x,, x,, x, .......,xn, ,X,..be numbers in [a,b] such that 
a=x,<x,
~~

<x,< ...<x,,Cxn=b.
Then the ordered set P = {x,, x, x,
............. x,)
is called a partition of [a,b]
Example1 PI = {0,1~4,1/2,314,1)and=P,{O, 1/3,1/2,2/3,6/7,~l),botharepartitionsof[O, 11.
I
Moreover, Fig. 2
P, UP,= {0,1/4,1/3,1/2,2/3,3/4,6/7,1) and=P,nP,= {0,1/2,1) arealsopartitionsof[0,1]. See Fig. 3 (a), (b), (c) and (d).
By an ordered set we mean a set. in which, the order in which its elements occur is \ fixed. I
I
I
1
(b)
(a)
1
:
:
I
:
Fig. 3
(c)
I
I
(4
A set J is called a subinterval of an interval I, if i) J is an interval, and ii) j s 1
Apartition of P = {%, x,, x,...........x,) of [a,b] divides [a,b] into n closed subintervals, [%, x,], [xI9xz],........... '[~n,,Ih],
'A' is rcad as delta.
It follows that
with the n + 1 partitioningpoints as endpoints. The interval [xi,, xi] is called the ith subinterval of the partition. The length of the ith subinterval, denoted by A xi, is defined by
We call partition P regular if every subinterval has the same length, that is, if x,  %, "2X~,..........,xn x,,, h e all equal.' In this case, the length of [a, b], that is b  a, is equally divided into n parts and we get
Thus a regular partition of [a, b] . m y be written as {a, a + h, a + 2h, ...............,a + nh), where a + nh = b: We shall denote this partition by {a + ih) ni,. ForP={1,3/2,2,5/2,3,7/2,4),Ax1= x,xo=3121 =1/2, A x,=x,x,= ,2 312 = 112. If you calculate A x3, A x4,Ax5and Ax,, you will see'that P is a regular partition of [1,41.
E E 1) See Example 1. Which partitions among P,, P,.
Definite Integral
PI uP, and PI n P, are regular '? What are the lengths of the third subintervals in PI and in P, ?
E E2)
Write down a regular partition for each of the following intervals a) [O, 21 with 7 partitioning points. 'b) [2,9] with 11 partitioning points.
Definition 2 Given two partition PI and P, of [a,b], we say that P, is a refinement of PI (or P, is finer than P I )i f P 2 ~ P l . In other words, P, is a refinement of PI if each subinterval of P, is contained in some subinterval of PI. Example 2 Consider the partitions PI = {1,5/4,3/2,7/4,2), P,= {1,6/5,5/4,3/2,19/10,2), P, = {I, 5/4,3/2,2) PI and P, are both finer than P,, as PI 2 P, and P 2 2 P,. However, neither is P I a refinement of P, nor is P, a refinement of P I . If P, and P, are partitions of [a,b], then from Definition 2 it follows that i)
PI u P2is a refinement of both P, and P,.
ii)
P I and P2are both finer than P, n P,.
Now, suppose for every n e N we define P, as
This means PI,has 2" + 1 elements. We can see that PI,is a regular partition, with each subinterval having length
j
b  a  1 ba Now?n;i2(2n
A x, = x,  x,+,
ba =a+i2" ba
This means that the length of the subintervals corresponding to PI,, is half the length of 2" those corresponding to PI,. We can also see that Pn+l2 PI,. In other words, P,+, is finer than Pn (also see E3)). Thus we have defined a sequence of partitions {Pn) of [a,b], such that P,,, is a refinement of Pnfor all n. Such a sequence (P, } is called a sequence of refinements of partitionsof la,bl.
E E3) From the sequence of partitions {PI,) defined above, L
a) Find P, and P, . b) Yerify that P, 2 P, 2 P,. c) . What are the lengths of the subintewals in each of these partitions?
Integral Calculus
,
E
E4) Let {P, )=:, be a sequence of partitions of [a,b], and let P; = PI, P,'=P,UP,,P;=P,,UP, {P,' }
UP,,andingeneral, P:=P,
U P,',.Showthat
:,, is a sequence of refinements of [qb]
Upper and Lower Product Sums i.2.2 By now, we suppose you are quite familiar with partitions. Here we shall introduce the concept of product sums. It is through this that we shall be in a position to'probe the more subtle concept of a definite integral in the next section. Let f: [qb] +R be a bounded function, and let we have defined (supremum) and g.1.b. (infimum of a bounded set numbers in Unit I .
NOWfor my subinterval [xi,,xi],consider the set Si = {qx) :x E [xi],xi]}. Since f is a bounded function, S,must be a bounded subset of R This means, it has a supremum (or least upper bound) and infimum ( ~greatest r lower bound). We write
We now define the upper product sumU (PJ)and the lower product sum L (P, f) by
You must have come across this
L(P,f)=m, (x,x,,)+m,(3 x,)
notation earlier. But let us state clearly what (1) means :
+ ..........
+
mn(%xn,)
Thus,to get U(p,f) we have multiplied the supremum in each subintervalby the length of that
C.F.B. Riernrnn (18261866)
subinterval, and have taken the sum of all such products. Similarly, L(P,f) is obtained by summing the products obtained by multiplying the infimum in each subinterval by the length of that subinterval. U(P,f) and L(P,f)are also called Riemann sums after the mathematician George Friedrich BemhardRiemam.
R i m gave a definition of definite integral that, to this day, remains the most convenient
(b) Fig.
and useful one. We started this unit saying that we wanted to find the aiea of the shaded region in Fig. 2. Then what are we doing with partitions, U(P,f) and L(P,f)? Fig. 4 will give you a clue tothe path which we are going to follow to achieve our aim. Fig. 4 (a) and 4(b) give the geometric view of M,A x, and m,A x,as areas of rectangles with base A x, and heights M, and m, respectively. The shaded rectangles in Fig 4(a) are termed as outer rectangles, while the shaded rectangles in Fig. 4(b) are called inner rectangles. Thus, when f is a nonnegative valued fimction (qx) 2 0
t, x),
U(P,f) = sum of the areas of outer rectangles as in Fig. 4(a). L(P,f) = sum of the areas of inner rectangles as in Fig. 4(b), and
,

U(P,f) L(P,f) = difference of the ateas of the shaded rectangles along the graph of f as shown in Fig. 4(c). As you see fiom Fig. 5, U(P,f) and L(P,f) depend upon the fimction
f:{a,b] ,R (compareFig. 5(a) and (b)), and the partition P of [a,b] (compare Fig. 5(c) and (d)).
Fig. 5:
(a) U(P,f) where y = x (c) U(P,f) whenP = (0, l,2,3}
(b) U(P,f) where 9 = x
(d) U(P,f) when P = {0,1/2,1,3/2,2,5/2,3}
If we denote the area between the curve given by y = f(x), the xaxis and the lines x = a and x = b, (the shaded area in Fig. 2) by A, then it is also quite clear fiom Fig. 4(a) and (b), that L(P,Q IA I U(P,f).. The geometric view suggests the following theorem : Theorem 1 Let f: [a,b] R be a bounded function, and let P be a partition of [a.b]. If M and m are the supremum and the infimum off, respectively in [a,b], then
If X c Y: then sup X 5 sup Y and inf X 2 inf Y.
Integral Calculus
m (b  a) 5 L(P,F) 5U(P,f) 5 M (b  a).
Proof :Now M = Sup {f(x) :x E [a,b]}, and M,

sup (f (x): x
E
]
[xi,,xil .Hence Mi h M.
Further, m = inf {f(x): x E [a,b]), and mi = inf {f(x) : x E xi]). Thus, m 5mi. This m e w m5miIM,<M Once we ha:.:. the inequalities (2), we can complete our proof in easy steps. (2) implies that m A x i I m i Axi5MiA x i I M A x i This implies that if we take the sum over i = 1,2..........., n, we get .n
!
n
X m i=l C Ax,IL(P,f)SU(P,f)I Mi=l C Axi
*m(ba) < L(P,f) Iu(P,~)5M(ba), n
since m C A xi = the sum of the lengths of all subintervals i=l
the length of [a,b] = ba. Fig. 6 will help you understand this theorem better. Let us verify this theorem in the case of a given function. Example 3 Let f:[1,2] +R be a function defined by f(x) = x2, and let P = {1,5/4,3/2,5/3,2) be a partition of [1,2]. The subintervals associated with partition P are [l,5/4], [5/4,3/2], [3/2,5/3] and [5/3,2]. The function f is a bounded function on [I ,2]. In fact, the image set off is [1,4], which is obviously bounded. =
Fig. 7
Since f is an increasing function on each subinterval (see Fig. 7) the supremum off in [xi,,xi] will be attained at xiand the infimum will be attained at xi, That is, Mi = qxi)and m, = f ( ~ ~ Therefore, ~). we can m t e C Mi Axi = C f(xi ) Axi = C xi2 (xiU(P, f ) = X ~ ( X ~  ~ , , ) + ~ ( ~  X ~ ) + X ~ ~+xi(x4xJ ( X ~  ~ )
...
up, f )
=
(b) (;12 +
(a)
+
',(; (;)
+m2
(5)
Definite Integral
Now, the supremum of fIx) in [1,2] = M = f(2) = 2, =4, and M(ba)=(4)(2 1)=4,andm(ba)=(1)(2 1)= l.Thus, m(b  a) 5 L(P,f) 2 U(P,f) I M(& a). We have noted that the upper and lower product sums depend on the partition of the given interval. Here we have a theorem which gives us a relation betweenthe lower and upper sums corresponding to two partitions of an interval. Theorem 2. Let f:[a,b] + R be a bounded function and let P,and P, be partitions of [a,b]. If P2 is finer than PI, then L(P,,f)I L(P,,0 IU(P,,f) ~ U P , , f ) . Proof : For proving this theorem we look at Fig. 8(a) and (b). Let PI = {x,, x, x,,. ............xn)and P2 = {xg, vl, xi,x2,.............xn}be two partitions of [a,b]. P, contains one element more than P, ,Namely v,. Therefore, P, is finer than PI. In fact, P, can be rightly called a simple refinement of PI. We shall prove the theorem for this simple refinement here. PI divides [a,b] into n subintervals : [$, x,], [x1,x2],............[x*, xnl.
.
9
(b)
(a) Fig. 8
Fig. 8(a) clearly shows that L(P,,~) 5 L(P,,f) (by an amount represented by the area ofthe shaded rectangle). Similarly, Fig 8(b) shows that U(P,,f) lU(P,,f). Since L(P,,f) I U(P2,f),the conclusion of the theorem follows in this case. If P, is not a simple refinement of P,, and P2 has m elements more than PI.Then we can find (m  1) partitions p i ,:P , p23 .............
PT'
such that
Intcgral Calculus
PI c P: c P; c P; c.............c pZmIc p2 and each partition in this &pence is a e
simple refinement of the previous one. Theorem 2 then holds for each pair of successive refinements and we get
L(P,,f) 1 L(P&f ) 1 L ( P ~f ,) l ...................< L(P,"', f ) I L(P,,f) and U(P,,f) I u(p,"',f)
s ............... I lJ(~2,,f)1 u(p;,f) CU(P,,f)
Thus, YP,,f) 5 L(P,,f) IU(P,,f) IU(P,,f). From Theorem 2 we conclyde the following : Let f : [a,b] + R be a continuous and nonnegative valued function, and let
{P, )=:,
be a sequence of refinements of [a,b].
Then we have
L(P,,f) I L(P,,f) 5 ..........5 L(Pn,f) 1.........5 A 1.........IU(Pn,f) 5 ..........< u(p2,f) 5 u ~ l , f ) where A is the area bounded by the curve, the xaxis and the lines x = a and x = b.
E
E 5) Find the upper product sum and the lower product sum of the function f relative to the partition P, when a) f(x) = 1+ x2,P = {0,112,1,3/2,2)
b) f(x)
= llx, P = {1,2,3,4)
E E 6) Verify Theorem 2 for the function f(x) = llx, 2 I x 13, and the partitions P, = {2,512,3) and P,
= {2,914,5/2,1114,3)of [2,3]
Definite Integral
In this subsection we have seen that the area A in Fig. 2 can be approximated by means of the lower and upper sums corresponding to some partition of [a,b], further. Theorem 2 tells us that as we go on refining our partition, the lower and upper sums approach A from both sides. The lower sums underestimate A (L(P,f) IA), while the upper sums overestimate A, i.e., IJ(P,f)2 A. Let us go a step further in the next subsection, and definelpygr 'and ugper integrals. Upper and Lower Integrals
1.23
Let,f:[a,b] + R be a nonnegative bounded function. Then to each partition P of [a,b], there correspondsthe upper product sum U(P,f) and the lower product sum L(P,f). Let P b e the set of all partitions of [a,b], Then the set u = {U(P,f) :P E P ) is a subset of R and is bounded below since A I U(P,f) t/ P E P. Thus, it is possible to find the infimum of u. Sirmlarly the set u ' = {L(P,f): P E P ) is bounded above, since L(P,f) IA V PEP.
Recall (Unit I ) that every set which is bounded below has an infimum, and every set which IS bounded above has a supremum
Hence we can find the supremum of u '.The infimum of u and the supremum of u ' are given special names as you will see from this definition. Definition 3 If a function f is defined on [a,b] and if P denotes the set of all partitions of [a,b] then infmum of {U@,f):PEP) is called the upper integral off
on [a,b], and is denoted by
Jab
f(x)dx.
The symbol
The supremum of {L(P,f):PEP) is called the lower ihtegral off on [a,b], and is b
denoted by
[
f(x)dx.
From Theorem 2 it follows that
1:
f(x)dx t A and
Jab

f(x)dx i A.
I
Example 4. Let us find
f(x)dx and

f(x)dx.
0 if x is rational for the function f, defined by f(x)
1 if x is irrational
Suppose P = {xo,xl, xz,.. ...........,x,,,) i~a partition of [0,11. Each subinterval [xi, ,xi] contains both rational and irrational numbers, this means, Mi = 1 and m, = 0 for each i. Thus,
c~~ n
U(P,f)=
n
A 3 = c ( 1 ) (3xkl)=l0=l
iI
ill
and
Since P was any arbitrary partition of [O, 11, this means that u(P,f)= 1 andL(P,f)=O t/ PEP.
Thus,u= {U(P,f): PEP) = (1) and u'= {L.(P,f) :PEP) = (0) Hence inf u = 1 and sup u '= 0.That is,
f(x) dx = 1 and

See if you can do these exercise now.
'[I
is read as ~ntepral
Integral Calculus
E E8) If the fimction f and g are bounded nonnegative valued functions in [a,b] and if Qx) a g(x) in [a,b], provethat
jabf(x) d r i

1.3
Jab
Ia
b
f (x )dx i
jab g(x)dx and
g(x) dx.
DEFINITE INTEGRAL
.
In the last section we had restricted our discussion t ~ L. ~ O negative II valued functions. But we can easily extend our definition of L(P,f), U(P,f) and t b lc er and upper integrals to all bounded functions. However we shall have to modify om interpretation of these sums as areas. For this purpose, we introduce the concept of signed area. If R is any region, its signed area is defined to be the area of its portion lying above the xaxis, minus the area of its portiod lying below the xaxis (see Fig 9).
Definite integral
Fig. 9
With this definition then, we can interpret L(P,f) as the signed area of a polygon inscribed inside the given region, and U(P,f) as the signed area of a polygon circumscribed about the region. Thus, for any bounded function on a closed interval [a,b], we can define
jabf(x) dx = sup [L(P.f): P e P} and

Now we are in a position to discuss the definite integral for a bounded function on a closed interval. (The adjective 'definite' anticipates the study of indefinite integral later).
Definition 4. Let f:[a,b] only if,
R be a bounded function. f is said to be integrable over [a,b] if, and
b f(x)dx =
f(x) dx. a

This common value is called the definite integral off over the interval of integration [a,b], and is denoted by
f(x)dx.
In this notation for the definite integral, f(x) is called the integrand, a is called the lower limit and b is called the upper limit of integration. The symbol dx following f(x) indicates the independent variable. Here x is merely a dummy variable, and we may replace it by t or v, or any other letter. This means,
jabf(x)dx = j The symbol
I
b
b
f(t)dt =
f(v) dv.
reminds us of S which is appropriate, because a definite integral is in some
sense, the limit of a sum. In fact it is the common value (when it exists) of the lower and upper integrals which are themselves infimum and supremum sums. The use of f(x) dx reminds us that we do not take the sum of function values, rather we take the sum of terms, each of which is the product of the supremum or infimum of the function in an interval multiplied by the length of the subinterval. The definition of definite integral above, applies only if a < b, but it'would be appropriate to include the cases a = b and a > b as well. In such cases we define
and
jabf(x) dx =  J: f(x) dx
Provided the right hand iniegral exists.
Integral Calculus
In Example 4, we have seen that if Oifxisratid Xx) =
1 if x is h a t i d , then 1

f(x)dx = 0, and
I.
f(x) dx = 1
Since the lower and upper integrals for this function are not equal, we conclude that it Lo not integrable.
E
E 9) Check whether the function given in E7) is integrable or not.
Now we shall list some basic properties of detiaite integrals. I ) Integral of a constant function f(x) = c Jab
cdx = c(b  a)
This is intuitively obvious since the area represented by the integral is simply a rectangle with base b  a and height c, (see Fig. 10). Fig. 10
Now let us consider a function f which is integrable over [a,b]. 11)ConstantMultiple Property Jab
Lf(x) dr = lc lbf(x) dx.
III) IntervalUnion Property
If a'< c < b, then
(b) Fig. 11
Its geometrical interpretation is shown in Fig. 1l(a).
IV) ComparisonProperty If c and d are constants such that c 5 f(x) I d for all x in [a,b], then
Fig. 1l(b) makes this statement clearer. Note that c and d are not necessarily the minimum and maximum values of f(x) on [a,b]. The value carnaybe less than the minimum, and d may be greater than the maximum.
Definite Integral
The following theorem gives a criterion for a function to be integrable. Theorem 3 A bounded function f is integrable over [a,b] if and only if, for every E > 0, there exists a partition P of [a,b] such that 0 I U(P,f)  L(P,f) < E. Proof: Weknow that for any partition P of [a,b],
I
I
b
L(P,f) 2
3
0i
[ab
b
1
f(x) dx i
f(x) dx 2 U (P, f)
f(x) dx 
Iab
f(x) d r i U (P. f )  L (P, f).
If the function f has the property that for every E > 0 there exists a partition P of [a,b] such that U(P,f)  L(P,f) < E, we conclude that
From this it follows that
fab
f(x) dx 
jab f(x) dx = 0,

and hence f is integrable over [a,b].
On the other hand, iff is integrable over [a,b],
lab
f(x) dx = sup {L(P,f): P
E
P} = inf {U(P,f): P E P}.Thus, for every E > 0 we canBnd
partitions P' and P" of [a, b], spch that 02
fab
f(x)dxL(Pt.I.)<~/2,andOiU(P",f)
b
f ( x ) d x c ~ / 2(seeSec,2of
a
Unit 1). Taking some partition P which is finer than both P' and P", and adding the two inequalities, we have
This completes the proof. Now arises a natural question : Which are the functions which satisfy the above criterion? The following theorems provide an answer. Theorem 4 A function that is monotonic (increasing or decreasing) on [a,b], is integrable over [a,bl. Proof Let the function f:[a,b]
+ R be increasing. Then
ba For each positive integer n, let Pn = {a, a + h, ............., a + nh = b), where h = ,be a n regular partition of [a,b]. Then 11
u ( P , ~ ) = ~ AX^ M ~= i= l
ba " C Mh~= h C M~=  x f(a + ih) n
n
i=l
i=l
II
,=I
since the supremum of ax)in [a + (i  l)h, a + ih] is f(a + ih). ba " andL(~n,f)= h x m r = h x f ( a + ( i  l ) h ) n
i=l
"
iI
I
Integral Calculus
Therefore
u (Pn,f)L(Pn,f)
=
ba [.f(a+h)+f(a+2h)+ ..........+ f ( a n

ba n
=  [f (a + nh)
+ nh)
 f (a)]
Let E > 0.Can we choose an hwhich will make U(Pn,Q L(P,Q < E ? Yes, we c a , Try some n > (b a)[f(b) f(a)l . If we substitute this value of n in (I), r e gei E
Thus, applying Theorem 3, we can conclude that f is integrable. Theorem 4 leads us to the following useful result. 'Corollary 1 Iff is increasing or decreasing on [a,b], then
lab
f(x) dx = lirn h[f(a)
+ f(a+h) + .......f(a+(nl)h)]
h+O
= lim h [ f ( a + h ) ho
+
f(a+2h)
+..........+
f(a+nh)], where h
ba
= 
n
We shall illustrate the usefulness of Corollary 1 through some examples. But before that we state another theorem, which identifies one more class of integrable functions. Theorem 5 If a function f:[a,b) + R is continuous, then f isbintegrable. The proof of this theorem is beyond the scope of this course. We shall prove it in a later course on real analysis. In Sec. 5 in Unit 3, we have seen that differentiability implies continuity, Now we can write differentiability q continuity integrability Now, let us evaluate some definite integrals with the help of Corollary 1.
Example 5 To evaluate
Jab
cos x dx, 0 r a 5 b I
?c
/ 2, we observe that
f: x + cosx is a decreasing function of [a,b]. Therefore, by Corollary 1
jab cosx dx = h+O lim h[cos(a + h) + cos(a+2h) + ... ..... + cos(a+ nh)]. a + nh = b. Now 2 sin (W2) [cos(a + h) + cos(a +2h) + .......... + cos(a +nh)] = 2 sin (W2) cos (a +h) + 2sin (W2) cos (a + 2h) + ...... + 2 sin (W2) cos (a +nh).
2n+1 2111 [sin(a + () h)  sin (a + () 2 2
= [sin(b
cos(a+h)
+
h ) 2
h)]
h 2
 s i n ( a + ), s i n c e a + n h = b
+ cos(a+2h) +............+ cos(a+nh)
=
sin ( b + h / 2 )  sin ( a + h / 2 ) 2 sin h / 2
Thus
lab
cosx dx = lim [sin (b
= sin b
h+o
 sin a.
li h + )  sin ( a + )] 2 2
h/2 . sln ( h i 2
Example 6 Suppose we want to evaluate Here, f: x Therefore,
+
Defidte Integral
i2'
(X + x2) dx
x + x2is an increasing function on [1,2].
( x + x 2 ) d x = lim h T f ( l + i h ) . h = l / n h+O
lim h h+
O
x
[(l
i=l
+ i h ) + (1 + i h ) '
]
i=l
Recall that
lim [2h h+O
5 1+ 3h2 5 i + h3 x i2] =I
i=l
a=l
3 1 lim [2nh+ h2n(n+l) +  h3n(n+l) (2n+l)] 2 6
I~+O
3 lim [2 +  (1 + h ) o 2
h+
1
+ (1 + h ) 6
(2
+ h ) , since&=
1.
3 1 23 =2++=. 2 3 6 In this section we have noted that a continuous function is integrable. We have also proved that a monotone function is integrable. corollary 1 gives us a method of fmding the integral of a monotone function. One condition which is very essential for the integrability of a function in an interval, is its boundness in that interval. If a function is unbounded, it cannot be integrable. In fact, if a function is not bounded, we cannot talk of Mi or mi, and thus cannot form the upper or lower product sums. Now on the basis of the criteria discussed in this section you should be able to solve this exercise.
E E10) State whether or not each of the following functions is integrable in the given interval. Give reasons for each answer.
x + l when x < O f(x) =
in [I,]] 1x when XSO,
Integral Calculus
E Ell) Use corollary 1 to evaluate the following definite integral.
1.4
FUNDAMENTAL THEOREM OF CALCULUS
As you have already read in the introduction, the basic concepts of definite integral were used by the ancient Greeks, mainly Archimedes (287212 B.C.), more than 2000 years ago. This was long before calculus was invented. But in the seventeenth century Newton and Leibniz developed a procedure for evaluating a definite integral by antidifferentiation. This procedure is emboded in the Fundamental Theorem of Calculus (FTC). Before we state this theorem, we introduce the notions of the average value of a function and the antiderivative of a fuction. Definition 5 Let f be integrable over [a,b]. The average value y of y = f(x) over [a,b] is '
The following theorem tells us that every continuous hnction on a closed interval attains its average value at some point of the interval. We shall not give its proof here. Theorem 6 (Average Value Theorem) Iff: [a,b] 4 R is continuous,then b
1 f(x) = Jf(x) dx ba a
for some
x
E [a,b]
We shall now define the antiderivative of a function. Definition 6 Let f: [a,b]
R and F: [a,b] 4 R be two hnctions such that
d dx tive) of Rx).
F(x) = F' (x) = f(x) for each x â‚Ź]a$[, We call F(x) an antiderivative (for inverse deriva
x3 For example, is an antiderivative of x2,since 3 d cos x is an antiderivative of sin, since  (cos x) = sin x. dx x L  x 2 an antiderivative of x3 2x? Is 4 Consider the two hnctions f(x) = x2and g(x) = x2+ 5. Both these are antiderivatives of the function h(x) = 2x. This means that antiderivative of a function is not unique.
In fact, if F(x) is an antiderivative of f(x), then F(x) + c is also an antiderivative of f(x). This follows from the fact that d

d =f dx dx We can also say that any two antiderivatives of a function differ only by a constant. Because, if F(x) andG(x) are tsvo antiderivatives of Cx), then (F(x)) =
 (F(x) +c)
F '(x) = F '(x) = Cx). That is, F(x)  G(x)] ' = 0 We have noted in Unit 7 that if the derivative of a function is zero on an interval, then that function must be a constant. Thus F(x)  G(x) = c. Now having defined the average value and the antiderivative, we are in a position to state the Fundamental Theorem of Calculus. We shall give this theorem in two parts.
Theorem 7 (FTC): Let f: [a,b] + R be a continuous function. Part 1 If the function F : [a,b] + R is defined by
Then F is an antiderivative off, that is, F'(x) = f(x) for all x in ]a,b[. Part 2 If G is an antiderivative off in]a,b[, then b
Iabf(x)dx
=
G(x)] = G(b)  G(a). a
Proof of Part 1. By the definition of derivative, F'(x)
=
lim h+O
F(x + h)  f (x) h
by the interval union property of definite integrals. Gut, by the Average Value Theorem (Theorem 6 )
Therefore, FC(x)= lim f ( i ) . We know that h+O h
Definite Integral
+ 0, i +
i E [x, x + h]. This means that as
x Therefore,
F'(x) = lim f (t)= f (x), since f is a continuous function. t+x Hence, F as defined by (3), is an antiderivative off. Proof of Part 2 G is given as an antiderivative o f f in ]a,b[. Also, as shows in Part 1, F defined by (3) is an antiderivative off in ]a,b[. Therefore, G(x) = f(x) + c on ]a,b[ for some constant c. To evaluate c, we substitute x = a, and obtain c = G(a)  F(a) = G(a)  0 = G(a). Hence G(x) = F(x) + G(a), or F(x) =G(x) G(a) 1f we put x = b, wk get
The interval [a, b] on wh~chf and its antiderivative are defined, so t b t F' (x) = f(x) t/ x E ]a, b[, is implicit in our discussion here
Integral Calculus
The Fundamental Theorem of Calculus tells us that differentiation and integration are inverse processes, because Part 1 may be rewriker, as
r
fit) dt) = f(x), i f f is continuous.
&(
That is, if we first integrate the continuous function f with the variable x as the upper limit of integration and then differentiate with respect to x, the result is the function f again. So differentiation offsets the effect of integration. On the other hand, if we assume that G' is continuous, then Part 2 of FTC may be written as JaxG'(t) dt = G(x)  G(a)
Here we can say that if we first differentiate the function G and then integrate the result from a to x, the result can differ from the original function G(x) only by the constant G(a). If G is so chosen that G(a) = 0,then integration offsets the effect of differentiation. Till now we had evaluated the integrals of some functions by first finding the lower and upper sums, and then taking their supremum and infimum, respectively. This is a tedious procedure and we cannot apply it easily to all functions. But now. FTC gives us an easy method of evaluating definite integrals. We shall illustrate this through some examples.
h
3
Example 7 Suppose we want to evaluate
(a2
+
bx+cjdx.
Since f : x + ax2+ bx + c is continuous on [2,3], it is integrable over [2,3]. ax3 bx2 G(x) =  +  + cx is an antiderivative off (x). 3 2 Qence, by FTC (Part 2)
Example 8 Let us evaluate
1"
d Example 9 To evaluate dx
Now d du
du dx
=
lo
cos 2 x dx
lo
x2
sin t dt, we put x2 = u
2x. and using FTC (Part I), we get
d sint dt = sin u = sin x2. Thus. dx
x2
lo
sint dt = 2x. sin x2
Definite Integral
Example 9 suggests the following formula.
If you have followed these examples, you should be able to solve the exercises below. Remember that the main thing in evaluating a definite integral is to find an antiderivative of the given function.
E
E12) The second column in the table below consists of some functions which are antiderivatives of the functions given in column 1. Match a function with its antiderivative by pairing appropriate numbers. For example, we can match x" with
n+l
n+l
n+l
n +l
r since X is an antiderivative of xn.We shall
indicate this by iii) + viii)
Function
3 ii) iiii iv)
Antiderivative i)  Incosx
sinx cosx
n
Incoshx
4
hi
sechx
eaX
iv)
cosx
v)
sin
.v) vi)
a
vi)
vii)
tanh x
vii)
viii)
sech x tanhx
1 em a ax Il+I
E
viii)
.
n+l
E 13) Evaluate the following integrals by using FTC a
a)
JI3
zx3dx
C)
II2
e)
J:
g)
5,
b)
X(X+ I ) ~ ~ x
d)
x(x2 +
0
'
d~
(x  cos x)dx
h)
sinhxcoshx dx
,y
1
3
5,
3
(2x2+2x+l)dx
rf4
sec2 x dx
~
lrf2 L
+sin X r) dx
,504e2xdx 1
L,(sinh x cash x)dx
1
Integral Calculus
E 14) Find
1.5
d
p(x)] when F(x) is defined by the following definite integrals. dx
SUMMARY
In this unit we have conered the following points: 1) A partition P of a closed intewal [a,b] is a set {a = s,x,, x2..........,xn,,xn= b)] such that Xo<X,<X2<.... <Xn.
Definite Integral
Intdgrsl Calculus
2) 3) 4) 5)
A partition P, of [a,b] is finer than a padition P2,if P, =, P,, If M and m are the supremum and the infimum of a bounded function fin [a,b], then, given any partition P of [a,b], m(ba) l L (P,f) l U(P,f) l M(ba). The lower integral of a bounded function is less than or equal to its upper integral. A bounded function f is integrable over [a,b] if and only if its lower and upper integrals are equal. In such a situation the lower (or upper) integral is called the b
definite integral o f f over [a,b], denoted by
f(x)dx.
6) Iff is monotonic or continuous on [a,b], then f is integrable ovet[d,b].
1 b
7)
I f f is continuous on [a,b], then
f(x) dx represents the signed area of the region
bounded by the curve y = qx), the xaxis and the lines x = a and x = b. 8) I f f is monotonic on [a,b], then Jab
f(x)dx = lirn h h+O
2
f(a +ih), = lirn h
j=,
h+O
f:f ( a +(i  1)h). 1=1
b a where h = n The Fundamental Theorem of Calculus: 9) i) I f f is continuous on [a,b], then for x E ]a,b[
ii) Iff is continuous on [a,b] and F'(x) = f(x) for x E ] a,b[, then
lab f(x)dx F(b)  F(a). =
ba Ax inP3 is 8 GP ,:, 3 P:+, is a refinement of P: E 4) PI:+, = P,:, LJ pI: 3 E5) a) f(x) is an increasing function on [0,23.. Hence
5 1 1 1 3 1 and U(P,f) =.  + 2. + . 4 2 2 4 2
+ 5.1
2
b) f(x) is a decreasing function on [I ,4]. Hence
v n.
Definite Integral
1691 
3960 L(P1,f ) L(P2,f) I U(P2,f) U(P,,f). E 7) If P = {%, ............. x,) is a partition of (0, 1),
J0
lo 1
1
Hence
f(x) dx =
f(x) dx
=
2.
E 8) If P = {a = x,, x,,. .............x, =b) is any partition of [a,b],
thenL(P,f)=Cmi,,Axi I&,, Axi= L(P,g), where qVc = inâ‚Ź{f(x):x E [xi,,xi]) and m.'3g = inf {f(x):x E [xi,,xi]) and q I = q,, since f(x) Ig(x) for all x. Sirmlarly, U(P,f) l U(P,g) for all P. The result follows. E 9) f(x) = 2 is integrable E 10a), b), e) and g) are integrable as these are continuous. c), f ) are not integrable as they are not bounded. d) h) are integrable as these are increasing functions. ~ 1 1 )J 2 ( l + x ) d x
=
1
=
lim h[(l+l) h+O
lim h [2n h+O ,
+h
+ ( l + l + h ) + (l+1+2h) + ...........+ ( l + l + x ]
n(n  1) lim [2.1+ h2.I h+O 2 nh. (nh  h) = lim [2 + I h+O 2 =
E12)
3 ii) iii)
+ +
+
iv) v) viii) vi)

(O+l+ ..........+ n  1 as n h = 1
iv) + v) + 3 vi) + vii) vii) + ii) viii) + iii) 4 x . E 13 a) 71s an antiderivative of 2x3.Hence
Integral Calculus
=
d [ cost 2 dt = dr 2xcos x4  COS x2
JOx2
Cos t2dt  Id(ms t 2 dt]
UNIT 2
METHODS OF INTEGRATION
Structure 2.1
Introduction Objectives
2.2
Basic Definitions 2.2.1 Standard Integrals 2.2.2 Algebra of Integrals
2.3
Integration by Substitution 2.3.1 2.3.2 2.3.3 2.3.4
2.4
Method of Substitution Integrals using Trigonometric Forrnuias Trigonometric and Hyperbolic Substitutions Two Properties of Definite Integrals
Integration by Parts 2.41 ~
~
~of
2.4 2 Evaluation of
I 1\ih
l e a~x sin~ b x dt x a ni d
Summary
2.6
Solutions and Answers
2.1
~ s a x~c o s b x d x
1\/a'
2 4 . 3 Integoals of the type Iex[f(x)if.(x)l
2.5
I
h
&,
dx
INTRODUCTION
I b
In the last unit we have seen that the deffite integral
f(x) dx represents the signed
area bounded by the curve y = f(x), the xaxis and the lines x = a and x = b. The Fundamental Theorem of Calculus gives us an easy way of evaluating such an integral, by first finding the antiderivative of the given function, whenever it exists. Starting from this unit, we shall study various methods and techniques of integration. In this unit, we shall consider two main methods: the method of substitution and the method of integration by parts. The next two units will cover some special integrals, which can be evaluated using these two methods.
Objectives After reading this unit you should be able to define the indefinite integral of a funciton evaluate certain standard integrals by finding the antiderivatives of the integrands use the rules of the algebra of integrals to evaluate some integrals use the method of substitution to simplify and evaluate certain integrals integrate a product of two functions, by parts.
2.2
BASIC DEFINITIONS
We have seen .in Unit 1, that the antiderivative of a function is not unique. More precisely, we have seen that if a function F is an antiderivative of a function f, then F+c is also 311 antiderivative off, where c is any arbitrary constant. Now we shall introduce a notation here : We shall use the symbol
I f (x) dx to denote the class of all antiderivatives off. We call
11the
indefinite integral or just the integral off. You must have noticed that we use the samc sign
I,here that we have used for definite integrals in Unit 1.Thus, if F(x) is an antiderivative of
I
integral Calculus
qx), then we can write ] f(x) dx =F(x) + c. This c is called the constant of integration. As in the case of definite integrals, f(x) is called the integrand and dx indicates that f(x) is integrated with respect to the variable x. For example, in the equation (av + b) (av + b)4 dv = + C, 5a
I
(av + b)4 is the integrand, v is the variable of integration, and
(av + b) 5a
+c
is the integral of
the integrand (av + b)4. You will also agree that the indefinite integral of cosx is sinx + c, since we know that sin x is an antiderivative of cosx. Similarly, the indefinite integral of e2"is indefinite integral of x3
+ 1is
I
(x3
e2x
1 2
+ c, and the
& =  e'x
x + 1)dx = + x + c. You have seen in Unit 1 that 4
1 b
the definite integral
f(x) dx is a uniquely defined real number whose value depends on a,
b and the function f. On the other hand, the indefinite integral
I
f (x) dx is a cl'assof functions which differ from
one another by constants. It is not a definite number; it is not even a definite function. We say that the indefinite integral is unique upto an arbitrary constant. Unlike the definite integral which depends on a, b and f, the indefinite integral depends only * .
onf.
.
All the symbols in the notation
Jab
f (x) f(x) dx for the definite integral have an interpretation.
I
The symbol reminds us of summation, a and b give the limits for x for the summation. And f(x) dx shows that we are not considering the sum of just the function values, rather we are considering the sum of function values multiplied by small increments in the values of x.
I
In the case of an indefinite integral, however, the notation f(x) dx has no similar interpretation. The inspiration for this notation comes from the fundamental Theorem of Calculus. Thus, having defined an indefinite integral. let us get acquainted with the various techniques for evaluating integrals.
2.2.1 Standard Integrals Integration would be a fairly simple matter if we had a list of integral formulas, or a table of integrals, in which we could locate any integral that we ever needed to evaluate. But the diversity of integrals that we encounter in practice, makes it impossible to have such a table. One way to overcome this problem is to have a short table of integrals of elementary functions, and learn the techniques by which the range of applicability of this short table can be extended. Accordingly, we build up a table (Table 1) of standard types of integral formulas by inverting formulas for derivatives, which you have already studied in Block 1. Check the validity of each entry in Table 1, by verifying that the derivative of any integral is the given corresponding function.
Methods of Integration
Table 1 S. No.
Integral
Function
,
I1+1
+
1.
Z
2. 3. 4. 5.
sinx
n+l
1
+C sinx+c tam + c COA + c secx + c cosecx + c
sec2x cosec2x secx tsinx cosec x cobr 1
8.
#
COSX
COSX
6. 7.
c, n
sin'x + c, or  cos' x + c
hT2
cotlx + c 1 1+x2
9.
tan'x + c or I
cot' x + c 1 10.
x
sec'x + c or
J2T
cot' x + c
1
11.
In(x(+c
X
ex+ c (aX/ha). + c coshx + c sinhx+c tanhx + c CON+ c sechx + c cosechx + c
eX
12. 13. 14. 15.
ax sinhx coshx sech2x cosech2x sechx tanhx cosechx cothx
16. 17. 18.
19.
Now let us see how to evaluate some functions which are linear combinations of the functions listed in Table 1. 2.2.2 Algebra of Integrals
You are familiar with the rule for differentiationwhich says
There is a similar rule for integration : R U I ~1
J[af(x) + bg(x)dx = a J f(x)dx + b J g(x)dx
This rule follows from the following two theorems.
Theorem 1 If f is an integrable function, then so is kqx) and
proof Let
J f(x)dx = F(x) + C.
d Thenby definition,  [F(x) dx
. "
+
c] = f ( x )
d [k{F(x) + c)] = kf(x) dx
Again, by definition, we have kf(x)dx = k[F(x)
+ c] = k J f(x)dx
Integral Calculus
Theorem 2 Iff and g are two integrable funcnons, then f+g is integrable, and we have
J [f(x) + g(xlldx = J f(x)dx + J g(x)dx ~ l o o f ~ e ft ( x ) d ~=F(x) + c , j g(x)dx = G ( X ~+ c d Then,  [(F(x) + cl + (G(x) + d l = f(x) + g(x) dx Thus,
J [f(x) + g(x)ldx = [F(x) + e l + [G(x) +el f(x)dx
=
+ J g(x)dx
Rule (1) may be extended to include a finite number of functions, that is, we can write
We can make use of rule (2) to evaluate certain integrals which are not listed in Table 1.
,
Example 1 Let us evaluate We know that
(X
+
1
)'
I (x = X'
X
I
+
;I3 dx
3 1 + 3x + +7 Therefore, . x
X
..........Rule 2 Using integral formulas 1 and 11 from Table 1, we have
Note that c, + 3c2 + 3c3 + c, has been replaced by a single arbitrary constant c.
Example 2 Suppse we want to evaluate
1(2 + 3 sin x
+
4ex )dx
This integral can be written as
k
I
Example 3 To evaluate the defmite integral integral Thus,
(X + 2x'
J (x + 21' 1' dx
J (X + 2 x 2 ) dx =
(x2
+ 4,' + 4 x 4 ) d \
)'
dx. we first find the indefinite
According to our deffition of indefinite integral, this gives an antiderivative of (x + 2 ~for~a given value of c. By using the hdamental Theorem of Calculus we can now evaluate the definite integral.
)
~
Methods of Integration
FTC says that if G(x) is an antiderivative of f(x), then
Ib
f(x) dx = G(x) r
= G@)
Note that for the purpose of evaluating a definte integral, we could take the antiderivative correspondmg to c = 0, that is, 4 s 1 x 3 + X 4 + x , as the constants cancel out. 3 5
See if you can do these exercises now.
E
E 1) Write down the integrals of the following using Table 1 and Rule 2. a) (i) x4 (ii) ymW(iii)4x2 (iv) 3 b) (i) 1 2x +x2
(iii (X 
i)2

(iii) (1 +XY
c) (1") ex + eeX+ 4 (ii) 4cosx 3sinx +ex+x
(iv) 4sech2x + eX
 8x
 ma).
:I
Integral ,Calculus
E
E2) Evaluate the following definite integrals.
b) i)
J
1
4 (X
+
1
d
(ii) Jo
(X
+113 dx
You have seen that with the help of Rule 2 we could evaluate a number of integrals. But still there are certain integrals like
sin 2~ dx which cannot be evaluated by uiing
Rule 2. The method of substitution which we are going to describe in the next section will come in handy in these cases.
2.3
INTEGRATION BY SUBSTITUTION
In this section we shall study the first of the main methods of integration dealt with in this unit: the method of substitution. This is one of the most commonly used techniques of integration. We shall illustrate its application through a number of examples. 2.3.1 Method of Substitution The following theorem will lead us to this method. Theorem 3. If
I f(v)dv
= F(v) + c, then on substituting g(x) for v, we get
Proof We shall make use of the chain rule for derivatives (Unit 3) to prove this theorem. Since
1f(v)dv = F(v) +
C,
dF(v) we can write  f (v). Now if we w t e v as a function of dv
x, say v = g(x), then dFtg(x)l dg(x) d Qg(x) = dx by chain rule
dx
dg(x)
= f[g(x)l.
dg(x)
dx
since v = g(x)
= f[g(x)lg'(x) This shows that F[g(x)] is an antiderivativeof flg(x)g'(x). This means that
The statement of this theorem by itself may not seem very useful to you. But it does simplify our task of evaluating integrals.For example, to evaluate
sin 2x dx, we could take v = g(x) =
2x and get
1 =
2
I
sinv dv by Theorem 3, since g(x) = 2x and gS(x)= 2.
 cos v + C
2
=   cos 2x
+C
2 We make a special mention of the following three cases which follow fiom theorem 3. Case i) If f(v) = v", n #  1 and v = g(x), then by Formula 1 of Table 1.
Case ii) If f(v) = I Iv and v = g(x). then, by formula 11 of Table 1
Case ill) If
Iab
J f(x) dx = F (x) + c, then
flg(x)lgl (x) dx = 'j;(v;
d V, wherev =g(r) [The limits of integration are g(a) and g(b)
Since x = a = > v = g(x) = g(a), and x=b=>v=g(x)=g(b).] We shall be using these three cases very often. Their usefulness is evident from the following examples.
Methods of Integration
Integral Calculus
Example 4 Let us integrate (2x + 1) (x2+ x + 1)' d For this we observe that  (x2 + X + 1) = 2x + I dx Thus, I(2x + I) (x2 + x + dx is of the form [g(xllngf(x) h a n d hence can be evaluated as in i) above. 1 nerefore, ( 2 +~ I) (x'+x+I)' dx = (x' + x+116 +c. 6
1
1
Alternatively, to find
1(2"
+ 1)
(x2 + x + 1)' dx,
can substitute x2 + x + I by u. This
means
1 =  u6 + c by Formula 1 from Table 1.
6
Example 5 Let us evaluate =
J (ax + b)"
dx
=
1 (ax a
J' (ax
+
b)* dx
+ bin dx
Therefore, when n #  1,
and whenn =1
Example 6. Suppose we want to evaluate the definite integral
du We put x2 + 2x + 3 = u. This implies  = 2(x + 1). Further, dx
Example 7 To evaluate
I xe
2x2
do dx. we substitute2x2=u. Since  = 4% we can write, dx
4
On the basis of the rules discussed in this section, yov will be able to solve this exercise.
E E3) Evaluate
Methods of Integration
Integral Calculus
Now we shall use the method of substitution to integrate some trigonometric function. Let's start with sin ax.
Example 8 To evaluate
I sin ax dx, we proceeded in the same manner as we did for
1 sin 2x dx. We make the substitution ax = u du This gives  = a. Thus, dx
du I sinaxdx a1 1sinu..dx dx 1 = 1sinu d u = +c a a =
COS U


E
1  cosax +c
a E4) Proceeding exactly a s in Example 8, fill up the blanks in the table below. S.No.
f(x)
I f(x) dx
1.
sin ax
1 cosax 
2.
cos ax
3.
sec2ax cosec2ax cosec ax cot ax secaXtanax P anx
4.
5. 6. 7. 8.
1 . sin ax a
+c
.............. ............... ............... ............... ............... ...............
Suppse we want to evaluate
Example 9
9 3
+c
a
ii) J t a m dx
Jwtxdx,
and
iii)
1cosec2x dx
We can write Icotxdx, =
cos x
d
dx. Now since  sin x = cos x, this integral falls in the category of sin x dx
case ii) mentioned earlier, and thus, icon: dx = In /sinxI + c
ii)
To evaluate i t a m dx, we write
I tanx dx
=
sec x tan x dx secx .
= Inlsec x(
d + c, as sec x = sec x tan x
dx iii) To integrate cosec 2x we write
J
cosec'2x dx
=
l 2
J
2cosec2x (cosec2x  cot 2x) dx cosec2x  cot 2x
d Here again,  (COSec2x  cot 2x) = 2 cos ec2x(cos ec2x cot 2x) dx This means
J cosec2 dx = 21 In lcosec2x  cot2xl + e
In this example we have used some 'tricks' to put the integrand in some standard form. After you study various examples and try out a number of exercises, you will be able to decide on the particular substitution or the particular trick which will reduce the given integrand to one of the known forms. Let's look at the next example now.
Methods of ~ n t e ~ r a t i o n '
Example LO Let us evaluate r e lfweputs$x=uthe %=fore,
1
E
' sin 2 x dx
du
 = 2 s i n x ~ s x= s h 2 x dx
J esin2' sin2x dx = J.eu du
See if you can solve this exercise now. ES) Evaluate the following integrals
l
a) Isecxdx
b)
JOzI2
sin2 x cos x dx
C) eYnxsec2 X
~ X
Integral Calculus
2.3.2 Integrals using Trigonometric Formulas
In this section, we shall evaluate integrals with the help of the following trigonometric formulas: 1 (1cos2x) sin 2 x = 2 1 (l+cos2x) cos 2 x = 2 3 3 . 1 . sm x =  smx   sm3x 4 4
sinrnx cosnx
=
1 2
 [sin(m + n)x + sin(m  n)x]
In each of these formulas you will find that on the left hand side we have either a power of a trigonometric function or a product of two trigonometric functions. And on the right hand side we have a sum (or difference) oftwo trigonometric functions. You will realise that the functions on the right hand side can be easily integrated by making suitable substitutions. The following examples will illustrate how we make use of the above formulas in evaluating certain integrals. Eumplc 11 To evaluate
I cos ax dx. We write 3
3 1 1(4 cos ax + cos 3ax)dx 4 3 1 c o s a x d x + 1 I c o s 3 a x d x 4 4
cos 3 ax dx = =
 3 sin ax
4a

1 +sin3ax + c
(see E4)
12a
Example 12 Let us evaluate i) Isin3x cos4x and ii)
I sinx iin2x sin3x dx
Here the integrand is the form of a product of trigonometric functions. We shall write it as a sup^ of trigonometric functions so that it can be integrated easily. J'sax c o u l x h =
i) 1 2
 = (sin7x
ii)
1 2
dx   Jsinx
=   1 cos7x
14
I 11 (sm. 7x  sin x) dx
1 2
+  cosx + c
To evaluate
] sin2x cos3x dx, again we express the product sinx sin2x sin3x as a sum of
trigonometric functions. 1 sinx sin2x sin3x =  sin x (cos x  cos 5x) 2 1 2
1 2
=  sinx cosx  sinxcos5x
1
=  sin 2x
4
1
Methods of Integration
.
 (sm6x su14x) 4
Therefore, Isin x sin2x sin3x dx
1   1 cos 2 x  cos4x 16
8
1 +cos6x + 24
c
Try to do some exercises now. You will be able to solve them either by applying the trigonometic formulas mentioned in the beginning of this section or by using the method of substitution. Don't be scared by the number of integrals to be evaluated. The more integrals you evaluate, the more skilled you will become. You have to practise a lot to be able to decide on the best method to be applied for evaluating any given integral.
E 6) Evaluate each of the following integrals. cos X
sin 20 c~~~~~d o
iv)
(I +
9
b)
sin e d e
e ( l + cos4 9)de
i)
6 3
sin2 Ode (15
sec e tan O(1 + sec E I )CIO ~
fii!
4l;J
)
j;1;6s 5 0 ~ 0 s e d e
iv)
ji:s
0 cos 20 cos 46 de
cot 2xcosec22x dx
Integral Calculus
E
E 7) The cost of a transistor radio is Rs. 7001. Its value is depreciating with time according to dv
the formula
500
dl = 3where Rs. v is its value t years after its purchase. What will be its
value 3 years after its purchase? (Don't forget the constant of integration! Think how you can find it which the help of the given information.)
23.3 Trigonometricand Hyperbolic Substitution Various trigonometric and hyperbolic identities like sin% + cos'0 = 1 sinh 0 1 + tan20 = sec20 ,tan h 0 and so on, prove very useful while evaluating certain cosh 0 integrals. In this section we shall see how.
A trigonometric or hyperbolic substitution is generally used to integrate expressions
involving
\/a', Jn, or a' + x 2 .We suggest the following substitutions. ..  . Expression involva
Substitution
J
x = a sine
'../
x = a tan0 or a sinhe
JTZF
x = sece or a coshe
a2 + x2
xatan0 dx
Thus to evaluate
dx
 = a 8.
dB
put x = a s i n 8 Then we h o w that
This means we can write
dx
J
I Jn ' a cos 8 d8
=
=
5J ~ l n Z e 1
a cos 8 de = /d8=8+c acme
Similarly to evaluate
dx I m9
we shall put x = a tan8
dx 2 Since  = a sec 8 de , we get de dx a sec 8d8
Sm=

la'
1  tan' ( x l a )
a
+c dx
We can also evaluate
1 Jm ,by substituting x
dx
a secLOde
=
=a
tan6
1
sec Ode
We can also evaluate this integral by x = asinhe. With this substitution we get, dx = s1nhl (x 1a) + c, and we know that
1Jn
,
sinh'
( x / a ) = In.
x
+
,/n a
(see Unit 5)
Sunilarly,
= In.
x + dx2  a2
+C
Methods of lrrtegretion
1I
Integral Calculus
Let us put these results in the form of a table Table 3 S.No.
I flx)d.
f(x) 1
1.
J.'f
sin' @/a) +c
2.
1 a2 + x 2

a
3.
x
+c
1
+c
J m 
 sec'(xla) a
1
jx'""
JX7
4.
1 tan' (XIa) a
+C
or sinh'(da) + c 1
~n
J77
5.
x
+JTF

a
+c
or cosh'(da) + c Sometimes the integrand does not seem to fall in any of the types mentioned in Table 3, but it is possible to modify or rearrange it so that it conforms to one of these types. We shall illustrate this through some examples. dx Example U Suppose we want to evaluate = J
J12
1
Let us try to rearrange the terms in the integrand
j 2x  2 to suit US. YOUwill see that
dv Ifweputx 1 =v,  =1 and dx dx 1 dv = Note the new limits of integration.
jl2sd
jo
This integrals is finally in the form thit we want and using the fvst formula in Table 3 we get 1
J12
sd
dx
lo
= sin' v
Example 14 The integration in
1
x2
dx does not again
fall into the types mentioned in Table 3. But let's see what we can do. du = 3x 2 ,n u s , ~fwe put x3 = u, dx
1
.:
du, by Theorem 3 ( u = 1 when x = 1 and u =O when x = 0)
Methods of Integration
1
Here the integrand
3can be evaluated using formula 2 in Table 3.
lo 1
n u s , we get
LJ 3
~ dx =A31 tan'~ u 0l+X6
If you have followed this disscussion, you will certainly be able to solve this exercise.
E
E 8) Integrate each of the following with respect to the corresponding variables. I ) 
1
ii)
Jsk
1
LC t2
X
v) 
vi)
JX4_1

G
1
viii)
iii)
u vii). 
h7
1
1 x)
x
xi)  l + x 2 mt:
xZ
 1 2 )1
l+xZ 
1 &+,2
y2,+6y+5
iv)
'1

JGZ
Integral Calculus
2.3.4 Two Properties of Definite Integrals We have already derived some properties of the definite integrals in Unit 11. These are the = c(b
i)
Constant Function Property :
Jab
ii)
Constant Multiple Property
I
 a)
b
iii) Interval Union Property
:
iv) Comparison Property
:
1
f (x) dx =
:
l
b
kf(x)dx = k Jc
f(x) dx
f(x)dx
a
If if c <f(x) < d V then c(b  a) 5
+
Jcb
X E
f ( x ) ~ where x a i c 6 b.
[a,b],
Jab
f(x) dx S d(b a).
Now we shall use the method of substitution to derive two more properties to add to this list. Let's consider them one by one V)
Ji
a12
f(x)dx = Jo
f(x) dx
+
J;IZf(a  x) dx f i r any integrable function f.
We already know that
1;
6
a12
f(x)dx =
f(x) dx
+
Ja;:(x)
dx.
Now if we put x = ay in the second integral on the right hand side, then since
since x is a dummy variable.
The usefulness of this property will be clear to you from the following example.
Example 15 Let us evaluate i) i)
Ji
I'
sin4 x cos5 X
~ and X
ii)
So2'
of Integralion
ms3 x dx
Using proper& v), we can write sin2 xcos xdx =
Jon"
sin4 xcoss xdx
+
sin4(n  x)cos5~n  x)dr (n  x) = sin x, and cos ( x  x) = COS
sin
cos (2n  x)
Our next property greatly simplifies some integrals when the integrands are even or odd functions. vi) Iff is an even function of x, i.e, f (x) = Kx), then faa
f(x) dx = 2
5:
f (x) dx
and iff is an odd function, i.e., f (x) =  f(x), then
This is also obvious from Fig. 1(a) and (b)
= COSX
Inteeral Calculus
We shall prove the result for even functions. The result for odd functions follows easily and is let to you as an exercise. (see E9)c)). So let f be an even function of x in [a,a], that is f(x)
= f(x) Q x E [a,a].
If we put x = y in the first integral on the right hand side, we get
f (x) dx
~h~~
=
2
1;
f (x) dx.
Using this property we can directly say that nl2
I,,,
nl2
sin x dx = 0
cosxdx=2 1%,2
6""
cosxdx=2sinx
I:"
E E9) a) Evaluate %I2
b) show that C) Prove that
sin 2 x In(tan x) dx = 0
dx = 0 iff is an odd h c t i o n of X.
=2
In thls secttoll we have seen how the method of substitution enables us to substantially increase our list of integrable functions. (Here by "integrable function" we mean a function which we can integrate!) We shall discuss.another techmque in the next section.
2.4
INTEGRATION BY PARTS
In this section we shall evolve a method for evaluating integrals of the type u(x)v(x)dx, in which the integrand u(x)v(x) is the product of two functions. In other words, we shall first evolve the integral analogue of
and then use that result to evaluate some standard integrals.
2.4.1 Integrals of a Product of Two Functions We can calculate the denvative of the product of two functions by the formula
Let us rewrite this as
lntegrating both the sides with respect to x, we have
d u ( x ) (v dx
(x)) dx = u ( x ) v ( x ) 
J
V(X)
d
(u(x)) dx
To express this in a more symmetrical form, we replace u(x) by qx), and put d dx
 V(X)= 8(*) This means V(X)=
g(x) dx.
As a result of this substitution, (I) takes the form
This formula may be read as: The integral of the product of two functions = First factor x integral of second factor integral of (derivative of first factor x integral of second factor) It is called the formula for integration by parts. ' h i s formula may appear a little complicated to you. But the success of this method depends upon choosing the first factor in such a way that the second term on the righthand side may be easy to evaluate. It is also essential to choose the second factor such that it can be easily integrated. The following examples will show you the wide variety of integrals which can be evaluated by this technique. You should carefully study our choice of first and second functions in each example. You may also try to evaluate the integrals by reversing the order of functions. This will make you realise why we have chosen these functions the way we have.
Example 16 Let us use the method of integration by parts to evaluate
J xex dx.
In the integrand xeXwe choose x as the first factor and ex as the second factor. Thus, we get d xex'dr = x e x dx  { (x) e x dx) dx dx
5
5
5
Methods of Integration
,
Integral Calculus
Example 17 To evaluate
r I 2
x 2 cos x dx, We shall take x2 as the fxst factor and cos x as the
second. Let us first evaluate the corresponding indefinite integral. d ~ x 2 c o s x d x = x cosxdx{% ( x 2 ) cosxdx) dx
'I
I
I
=x%inx
I
=x2 sinx2
I
2xsinxdx xsinxdx
We shall again use the formula of integration by parts to evaluate
I
Ixsinxdx=x(cosx)
(I)(cosx)dx
]
x sin x dx. Thus
a~(f(x)=x,~(x)=~i~(x)
I cosxdx
=  xcosx+
Hence,
Note that we have written the arbitrary constant as c instead qf 2c 1~12 J0'l2
x2 cosx dx =(x2 sin x +2xcosx 2sin x  7c 4
2
Example 18 Let us now evaluate
I
X
+ c)
1,
In 1x1dx
Here we take In 1x1 as the first factor since it can be differentiated easily, but cannot be integrated that easily. We shall take x to be the second factor. Ixlnlxldx=
I
Inlxlxdx
While choosing In I x I as the first factor, we mentioned that it cannot be integrated easily. The method of integration by parts, in fact, helps us in integrating In x too.
Example 19 We can find
I
lnx dx by taking lnx as the first factor and 1 as the second factor.
Thus,
= xinx  J d x = xinx  x
= x l n x  xlne +c
+c
+ csincelne=l
= xln(x1e)
The trick used in Example 19, that is, considering 1 (unity) as the second factor, helps us to evaluate many integrals which could not be evaluated earlier. You will be able to solve the following exercises by using the method of integration by parts.
E E 10) Evaluate
C)
I I I
d)
J' x 2
a) b)
Methods of Integration
x 2 In x dx
Take f(x) = Inxand g(x) = x2
(I+x)exdx
Take f(x) = 1 + x and g(x)
= ex
( I + x 2 ) c X dx
sin x cosx dx
Take f(x)
= x2 and g(x) =
sinx cos x
E Ell) Evaluate the following integrals by choosing 1 as the second factor. a)
I sin' x dn
1
b)
tan' x dx
c)
cot' x dx
E
E 12)
Integrate
a) xsin'x
Methods of Integration
Evaluations at' eaxsinbx and leal eosbr dx
2.4.2
isaxsin bx dx and I eaxcosbxdx, we use the formula for integration by parts.
To evaluate
eaXsin bx dx = (ea*) ( c1 o s bx) 
I
b
1
I ea\cosbx + a ea\ cosbx dx 
I
b
b
J
1 ( a s a x )(  cor bx) dx b
1 a a . =ea"cnslvt+[em)(s1nbx)b b b 1   b Therefore, you will notlce that the last Integral on the r ~ g hhand t side is the same as the Integral on the left hqnd s ~ d eNow . we transfer the t h ~ r dterm on the right to the left hand s ~ d e , and obta~n,
a' + ;)
(I
b
I e a xslnbx dx
=
eax
a I (2 smbx   cosbx) b
This means,
We can similarly show that 1 eax cosbx dx = eax (acosbx a 2 + b2
1
+ bsinbx) + c
If we put a = rcose, b = rsine, these formulas become
Ie
,
sx sin bx dx =
I
&Ti7
J eax cosbx dx = I
+c
eax sin (bx  0)
1
\la?+b'
eax C O S ( ~X0)
+ C, where0
=
tan'
b
. a
Example 20 Using the foimulas discussed in this ~ubsection,we can eas~lycheck that
and 1 71 l e X c o s , / ~ d d * =  e xc o s ( \ / j ;  ) + c 2 3
ji)
Example21 Toevaluate
[ e2'
,in x c o s ? x d x , weshallfirstwrite
I
sin x cos 2x =  (sin3x  sinx) as in sec. 3. 2 Therefore,
1e2'
x
cog2' dx =
1 2

e 2 " sin 3 x dx 
I 2

J e l x sin x dx
Now the two integrals on the r ~ g h hand t side can be evaluated. We see that
I
,?x
s .~ 3x n dx = L e l x sin (3x  tan'
fi"
1)+ c 2
and
2 1 J'e2x s i n x c o s 2 x dx = e  " [sin(jx
J13
tan'
2) 
1
 sin (x  fan'
2 4 5
)I1 + c 2
53
x 3 sin (a In x) dx
Example 22 Suppose we want to evaluate Let lnx = u. This implies x = eUand duldx = I lx Then,
I x3sin ( a b )dx I x4 sin ( a h ) (llx) dx =
=
1elu sin au du
e4' sin (au JS 

E
1
tan' (a 14)) + c
,/a
x4 sin ( a l n x  tan'
a+ c 4
Why don't you try some exercises now. E 13) Evaluate the following integrals a)
JPCOWIX~X
,
b) JPsin 3x d~ c)
J cos
c)
e4xcoa x OQSZX d~
ax sin bx dx (writecosh ax in tnmr of the exponential function)
2.4.3 Evaluati~nof
I
I/,
d ~ ,
dx, and
Id
In this subsection, we shall see that integrals like and
I4x2
I
n dx, dX,
Shifting the last term on the right hand side to the left we get
Using the formula,
Id
=
dx?
dx can also be evaluated with the help of the formula for integration by pa*
and Table 3.
dx
Methods of Integration
dr
sin' ():
+
c, we obtain
T 7 dx = 12 x Jn + 2 sin 2
);(
X
+c
Similarly, we shall have,
and
=
I xp 7 a' 2
Example 23 Let us evaluate
In x
+
2
JZ dx
Surelv. vou will be able to do these exercises now.
\/X'= a
+C
Integral Calculus
E
B 14) Verify that
in the next subsect~onwe shall consider another type of integrand which occurs quite frequently in mathematics.
2.4.4 Integrals of the Type [ex [fix) + I.(x)] dx We first prove the formula integrating some functions.
I
ex [f(x)
By the formula for integration by parts
= f(x)ex

This implies
J f ' (x) e x dx + c
+ f' (x)] dx = e x f(x) + c
and see how it can be used in
Methods of Integration
Example 24 Let us evaluate the following integrals.
I
We take up (i) f i t ,
Now we shall evaluate ii)
I
=
I
X
X
cos  sin 2 2 2 x 2cos ;
ex12
dx
Now X J' sec c'I2 2
I
i
dx
=
X
(sec 1 (2e"I2) 2
J' (T1 seo
X
X
tan 1 (2e"')
2
dx
Thus,
In this unit we have exposed you to various methods of integration. You have also had a fair
C t
amount of practice in using these methods. We are now giving you some additional exercises. You may like to try your hand at these too. To solve these you will have to first identify'the method which will suit the particular integrand the best. This is the crucial step. The next step where you apply the chosen method to get the answer is relativelyeasy. If you have studied this unit thoroughly, neither of these steps should pose any problem. So good luck!
E E 15) Evaluate the following integrals : x2 + 2
b,
Ix
g)
J sin xecomxdx
dx.
I
c) . sinh (x / 2) cosh (x / 2) dx
~ 1 2sin x
i,
lo
cosx (1 + sinx)' dx.
Integral Calculus
n)
I
ex
(In sin x
+ cot x)
dr
Methods of Integratbn
E E16) Prove that
and use it to evaluate
J
x sin x dx
Integral Calculus
Before we end this unit, here are some general remarks about the existence of integrals. The result Jab
f(x) dx = [~(x)l,b= F(b)  F(a),
where F(x) is the antiderivative of f(x), will make sense only if f(x) exists at every point of the interval. Hence we have to be careful in using this result. Thus,
But l/x is not defined at x = 0, and In 1 x I is also not differentiable at x = 0. As such, at this stage, we should use the result only if the inrerval [a,b] does not include x = 0
J
,
 dx = In
121 = In 2 is not valid.
111
Again, consider 1 Jo
dx = [sin'
71
XI;
= .
2
1 We have used this in Example 3. However does not exist at x = 1, and sin'x is not
dlx'
differentiable at x = 1. L(sin'x)' exists at x = 1, but does not exist when x > 1.
sin'x)' does not exist, since sin'x
itself
However, the above result is true in some sense. This sense will be clear to you in your course on analysis. The antiderivative of every function need not exist, i.e., it need not be any of the functions we are familiar with. For example, there is no function known to us whose derivative is ex2 However, the value of the definite integral
Jab
f(x) dx of every function, where f(x) is
continuous on the interval [a,b], can be found out by numerical methods to any degree of approximation. You can study these methods in detail if you take the course on numerical analysis. You will study two simple numerical methods in Block 4 too. Thus, we cannot find the antiderivative of ex2,but still, we can find the approximate value of Jab
e"' dx, for all real valuer of a and b. In fact, this integral is very important in
probability theory and you will use it very often if you take the course on probability and statistics. That brings us to the end of this unit. Let us summarise what we have studied so far.
3.5
SUMMARY
In this unit we have covered the following points. 1)
If F(x) is an antiderivative of f(x), then the indefinite integral (or simply, integral) of f(x) is
I f(x) d x
= F(X)
+ C, where c is an arbitrary constant.
Methods of Integration
The method of substitution gives :
3)
In particular,
I
"+'
[f (x) [f(x)" fl(x) dx = +c, n
n+ 1
1 a
'
f (x) dx =
Ii2
f (x) dx
+
f
 1 , and
r2
f(a  X) dx
x) dx, if f is even f(x) dx =
4)
Standard formulas :
5)
Integration of a product of two functions (integration by parts):
Thls leads us to
x\i.'+c~n2
Z& i\=fj
J' eax sin bx J' e
x
dx
=
cosbx dx
=
3.6
a
2
JFZF
d
1ex[f(x) + f'(x)]dx
+ $32 +x2 X+J=
."'~j n
2
X
m =
+c
+c a
e a x sin
eax
(bx  tan'
cos (bx

tan'
b
) a
+c
b 1 +c a
exf(x) + c
SOLUTlONS AND ANSWERS 5
X .
El) a) i)  + c 5
ii)2xv2
+c
iii)  4xI
+c
iv) 3x
+c
61
Integral Calculus
L
iii) 4 tanhx + ex  4x2 + c


S. No.
f(x)
1f(x) dx 1
cosax+ c a 1 .  sin ax + c a 1  tan ax + c a
1.
sin ax
2.
cosax
3.
sec2 ax
4.
cosec2 ax
5.
sec ax tan ax
1  sec ax
6.
cosec ax cot ax
a Cosec ax + c
1

a a
1
62
1
cot ax
+c
+c
sec x(sec x + tan x d x = In (secx + tanxl + c sec x + tan x
1

sin x cos xdx =
0
du
2
C) if u = tanx,  = sec x dx
E6) a) i,
V)
2
ii) + c sin2 x
k16 x13
iii)
sin6 x
6+ c

cot 2x cosec2 2x dx =
Ion" O (I + sin
cos4 6) dO =
I
cot 2x (2 cosec2 2x) dx
xl2
sin O dO
0
+
/ox12sin0 oos4 O dO
1 ii) 10 (15tan8)'
(1
iiii )
i)
+ sec814
(i+fi1~2~ 1+12fi 
4
I
4
sin4 0 de =
I
=
I
sin3 9 sine de
,(
3
sin2 O

41 sin O sin 38) dO
4
Methods of Integration
Integral Calculw
= 500 tan' t
+c
700 =  500 tan'0 + c = c ac=700 v(3) = 700  500 tan' 3. E 8) For solution, see P. 104 v(0)
b)
Jo2
=
sin 2 In
t
xx
=
[I4
sin 2x In tan x dx
n
=r4sin2(l
n
,
x) in tan (  x) dx 2
= p d s i n z x in (taoxmtx) dx =
[I:4
sin 2 x in idx =
o
Methods of Integration
~ 1 a) 0
I x2 lnxdr
= lnx
I x 2 d x  I I x2 dx) dx ;(
C) 1 ( 1 + x 2 ) e x dx = ( I + x 2 ) e x
1
 2 1 xex dx = (1 + x 2 ) e x  2[xex  1e x dx]
I
x x2 x sin' x dx =  sin'x 2 x sinL u d x = J  cos u c o s u du put x = sin u in
E12) a)
 1 u  1 sin2u + c = 1 u  1 sinucosu + c 
2 4 2 2 1 =  [sin' x  x cos(sin' x)] + c 2 1 . 1 x2 :. I x s i n  ' x d x =   s2i n  l i   [ n n 4 xxJGFl+e F
= xln (1
1 El3 a)  eZx (2 cos4x 20
+ x 2 )  2[x  tan'
+ 4 sin 4x) + c
XI
+c .
Integral Calculus
[L
= 1 e2' (2cos2r 2 8 cmh ox sinbr dx =
e)
=
(
[em (ssin br 2 (2t b2)
FJ
em
2
t
ZsinZr)
2
s i n b ~&
 b msbx) + srU
1
+ I s Z x ] + c
(asinbx
(a sin bx
 b cm bx)] + c
 bcosbx) dx
x em (a sln br  b cos bx)  a2 +bz 
(a2
+ b212
[aeU (a sin bx
beaX(acosbx + bsinbx)] +c
21
c ) coshr t c
66
,d
 b cos bx) 
Methods of Integration
(1
+
sin X )
x tan' x
[I t 7 dt
1 2
=

=
 (x2 +l)
dx =
1
t 6 dt]
+f
1
4
smh' (x2 + 1) + c
1
9sin9d9,ifx= tan9

=
core +
j
cos0 d0 (inegration by parts)
]' COs (11 + x 2 1 dx X2
1
m) Put x = tan0 in
2

Answer 2[0 tan9 + In ( case I] + c where 8 = tan' x
I
n ) / e x ( l n ~ i n ~ + ~ o t ~ )e d* l~n =s i n x d x +
I
excotxdx
cotx e x dx
J x 3 sinx dx = J x 3 x3 ~
I
excotxdx
2
dx
=  x 3 cosx =
+
o s +i
=  x3 cosx
(rinx) dx
+ 3x2 sinx  6 J'
x sinx dx
3x2 sinx  6 [xmsx +
J
UISX
dx]
+ 3x2 sin x + 6(xcos x  sinx) + c
UNIT 3
REDUCTION FORMULAS
Structure 3.1 32 3.3
3.4
Introduction Reduction Formula Integrals Involving Trigonometric Functions 3.3.1 Reduction Formulas for
I
sinnx dx and
3.3.2 Reduction Formulas for
I
tannx dx and
I 'cosnx dx
I sinnx dx
Integrals Involving Products of Trigonometric Functions 3.4.1 lntegrand of the Type sinmxcosnx 3.4.2 lntegrand o f the Type em sinnx
3.5 3.6 3.7
Integrals Involving Hyperbolic Functions Summary Solutions and Answers
3.1
INTRODUCTION
In the first two units of this block we have introduced the concept of a definite integral and have obtained the values of integrals of some standard forms. We have also studied two irnportmt methods of evaluating integrals, namely, the method of substitution and the method of integration by parts. In the solution of many physical or engineering problems, we have to integrate some integrands involving powers or products of trigonometric hnctions. In this unit we shall devise a quicker method for evaluating these integrals.We shall consider some standard forms of integrands one by one, and derive formulas to integrate them. The integrands which we will discuss here have one thing in common. They depend upon an integer parameter. By using the method of integration by parts we shall try to express such an integral in terms of another similar integral with a lower value of the parameter. You will see that by the repeated use of this technique, we shall be able to evaluate the given integral.
Objectives After reading this unit you should be able to derive and apply the reduction formulas for
[ xnexdx
I
sinnxdx.
I
cosnx d x
I
tannxdx, efs.
J sinmxcosnxdx
3.2
REDUCTION FORMULA
Sometimes the integrand is not only a function of the independent variable, but also depends upon a number n (usually an integer). For example, in on x and n. Similarly, i~
[ sinn x k . the integrand sinnxdepends
[ e x cosmx dx, the integrand excosmx depends on x and m. The
numbers n and m in these two examples are called parameters. We shall discuss only integer parameter here.
On integrating by parts we sometimes obtain the value of the given integral in terms of another similar integral in which the parameter has a smaller value. Thus, after a number of steps we might amve at an integrand which can be readily evaluated. Such a process is called the method of successive reduction, and a formula connecting an integral with parameter n to a similar integral with a lower value of the parameter, is called a reduction formula. Definition 1: A formula of the form
I
f (x, n)dx = g(x)
+
f(x, k)dx.
where k < n, is called a reductio~lformula. Consider the following example as an illustration. Example 1 The integrand in
I
x " e 'dx
depends on x and also on the parameter n which is the
exponent of x, Let
Integrating this by parts, with xn as the first function and ex as the second function gives us
Note that the integrand in the integral on the right hand side is similar to the one we started with. The only difference is that the exponent of x is n1, Or, we can say that the exponent of x is reduced by 1. Thus, we can write Ix =xneXn InI .............................(1) The formula (I) is a reduction formula. Now suppose we want to evaluate I,, that is,
= x 4 e x  4x3eX+ 1212 = x4ex  4x3eX + 12x2eX 2411,using(l)for12 = x4eX 4x3eX + 12x2eX 24 xeX + 241,.
Now lo =
xOe'dx =
exdx = e x +c.
Thus, the method of successive reduction gives us
in five simple steps. You must have noted that we were saved from having to integrate by parts four times. This became possible because of formula (1). In this unit we shall derive many such reduction formulas. These fall into three main categories according as the integrand i) is a power of trigonometric functions. ii) is a product of trigonometric functions, and iii) involves hyperbolic functions. We will take these up in the next three sections.
3.3
INTEGRALS INVOLVING TRIGONOMETRTC FUNCTION
There are many occasions when we have to integrate powers of trigonometric functions. In this section we shall indicate how to proceed in such cases.
Reduction Formulas
I
3.3.1 Reduction Formulas for [sinnx du and eosnx dx ,
In this subsection we will consider integrands which are powers of either sinx or cosx. Let's take n power of sinx fxst. por evaluating sinAxdx=
I
sinn xdx, we write
sinn' x sin x dx, if n > 1.
Taking sip'x as the first function and sin x as the second and integrating by parts, we get
 sinh' xcorx  (n  1) I ninn~'xcorx(corr) dx
1. *
"'
 sin "' x eos x + (n  I) J sin x cos2 x d x =  sinn' xcosx + (nI) [ J sinn' x(lsin2 x) dx . =  sinn' xcosx + (n1) [J ~ i n "  ~dxx  J' sinn x) dx] =
+ (n1) [In2 I n ]
=  sin"' xcosx Hence, I,
+ (n  1 )
In = sinn'x cosx + (n  l)ln2
That is, nIn = sinn' x cos x
I, =
 sinn' x cos x n
+ (n 1)
I n2 Or
+ n n 1 In2
I sin'' xdx (valid for n 2 2).
This is the. reduction formula for
Example 2 We will now use the reduction formula for for n
1 sinn <OSX
=
I
sinnxdx to evaluate the definite
nl2
xdx =
+
I,
I
C
1 sin x
dx
integral,
sin ' x d x
J ~ sinnxdx = ~ ~=
We first observe that
1
17
nl2

xcosx n
+
4
T~US,
sin xdx =  r1'sin3xdx 5 0
4 . 2 1 sin xdx 5 3 n/2 xi2
=
0
8  ( cosx) 15 8
1.
Let us now derive the reduction formula for I,, =
I
c o s " x d x . ~ ~ alet i n us write
J cosn xdx = J cosn' xcosxdx, n > I.
Integrating this integral by parts we get =
I
co."l
x
sin x
= cosn' xsinx

I
+ (nI)
(n  I) cos"'x
(I,,

I,)
( sin r). sin x dx
By rearranging the terms we get I, =
I
COS"
xdx =
+ a n 1 1,2
x sin x
COS"'
n

This formula is valid for a 2 2. What happeas when a 0 or l? You will agree that the integral in each case is easy to evaluate. As we have observed in Example 2,
1
1
Using this formula repreatedly we get
I
5"" sin n1 n  3 n5 3 .........5 dl, n 'n2'114 4 2
n1 n  3 n  5 ........4.2
; Jon'*
sin rdx =
n 'n2'n4
5 3
1
i
x dx, if n is an add number, n L 3.
0
X I ~
o
This means
rXI2
sin "x d x =
I.
n1 n
n3 n2
n1
n3
if n is an even number, nL 2.
4 2 ...... ....  if
3
5
n is o d d , a n d n t 3
',if . . . . . . .. .  4
2
n is e v e n n 2 2
2
We can reverse the order of the factors, and write this as
rni2 s i n " x d x
=
4 ITT
n1 n3 . ......... n2 n
1 3
I
Arguing similarly for
n  1......... n
n ' 2
.
if n is o d d ,
n
2
if n is e v e n , n 2 2
6"
axnx dx we get odd, and
IoXi2 cosnxdx
is even; n We are leaving the proof of this formula to you as an exercise See El)
E El) Prove that
j
Ioni2 cosnxdx
......... n  1 , if n
=
2 4
.........  .. n
2
n is o d d , n 2 3
, if n is e v e n , n 2 2
3
Integral Calculus
E 82)
Evaluate a)
Ci2
E ~ s 6xdx, using the reduction formula
c0s5 X ~ Xb),
derived inEl).
J
3.3.2 Reduction Formulas for tannkdx and
J seenxdx
In this subsection we will take up two other trigonometric functions :tanx and secx. This is, we will derive the reduction formulas for formula for
1tan " xdx and 1sec" xdx. To derive a reduction
I tann xdx, n > 2. we start in a slightly different manner. Instead of writing tannx
=tanxtann'x, as we did in thc case of sinnxand cosnx,.we shall write tank = tann2xtan2x.You will shortly see the reason behind this. So we write
I tann xdx = I tann' x tan2xdx.
In =
=
I tann' r sec
2
rdi 
J tann' idx
......(2)
You must have observed that the second integral on the right hand side is In, Now in the first integral on the right hand side, the integrand is of the form [f(x)lm.f'(x) As we have seen in Unit 11,
nus,
j tann'x sec
2
idx =
tan "I x + c n1
Reduction For

X
Therefore, (2) give 1, =  In2 n1 Thus the reduction formula for tan " x dx is
I
xou must have now realiseo why we wrote tan% bn2x tanW'x

seen xdx (n > 2 ) . we f ~ swrite t seck = S ~ C ~ seei ~ Xx,
To derive the reduction formula for and then integrate by parts. Thus I, =
secn xdx =
secn' x sec xdx
= sec
=
"' x tan x  (n  2) j seen' x tan2 xdx
secnT2x t a n x  ( n  2 ) (I, In,)
After rearranging the terms we get
tan" xdx and
These formulas for integrals
"
S ~ C xdx
are valid for n > 2. For n = 0,1 and 2, the
seen xdx can be easily evaluated. You have come across them in
tan'' xdx and
Units 1 and2.
So
n14
Example3 Let's calculate
sec6xdx =
1)
~ X and
I ' 1 : 1 1"'
sec4 xtanx
4 5
tan5 X
5
4 5
4 +
ii)
sec6x dx
"14
0
sec2 xtanx
 sc4xdx + 23
I"'
o sec2
0
On the basis of our discussion in this section you will be able to solve these exercises.
Reduction Formulas
3.4 INTEGRALS
INVOLVING PRODUCTSOF TRIGONOMETRIC FUNCITONS
In the last section we have seen the reduction formulas for the case where integrands were powers of a single trigonometric function. Here we shall consider some integrands involving products of powers of trigonometric functions. The technique of finding a reduction formula basically involves integration by prqts. Since there can be more than one way of writing the integrand as a product of two functions, you will see that we can have many reduction formulas for the same integral. We start with the first one of the two types of integrands which we shall study in this section.
3.4.1 Integrand of the Q p e sinmxcosn x The function sin"xcos"x depends on two parameters m and n. To find a reduction formula for sinrnx c0sn h,let us fist write
Since we have two parameters here, we shall take a reduction formula tp mean a formula connecting I,,, and Ip,q,where either p < m, or q < n, or both p < m, q < n hold. In 0 t h words, the value of at least one parameter should be reduced. sinm+' x ~ f n  1I,,~ = J  i n m x c o s x d x (lnlsin xl
+ C, when + c, when
m
#
1
m = 1
Hence we assume that n > 1..Now, I
.,
=
J sin
x cosn x dx =
J cosnI x (sin 'x cos x) dx
Integrating by parts we get 1m.n
=
c0sn' x;.sinm+' X m+l
Sinm+l X J (n  1) COS"~ x ( sin X) dx, if m # 1
m+1
Therefore,
+ n  1 Im,n
I m , m +l This gives us,
m+n m+l
 cosn' x sin m+' x
=  1m.n 9
m+l
+ n  11 1. +
.2
cosnl x sin m+l x n1 .......... .(3) Remember +1m.n2 m+ln m+n But, surely this formula will not work if m + n = 0.So, what do we do if m + n = O? Actually we have a simple way out. If m + n = 0,then since, n is positive, we write m = n.
.
Im,n =
Hence I,, =
I ~ i n  ~ x c o sxdx" = I cot
xdx, which is easy to evaluate using the reduc
tion formula derived in Sec. 3 (See E2)). To obtain formula (3) we had started with thk assumption that n > 1. Instead of this, if we assume that m > 1, we can write I
=
1m.n =
sinmxcosnxdx =
 sinm' x cosn+' x n+l
sinm'x (cosn x sin x) dx.Integratiq this by parts we
 {mI)
')dx J ~ i n ~  ~ x (a sc0sn+l x n+1
for n t  I q
we have taken n > 1