NCERT Solutions for Class 12 Maths Chapter 5 continuity and differentiability - Free Download

Chapter 5: Continuity and differentiability.

Exercise 5. Miscellaneous

1.

Differentiate the function w.r.t x

( 3x

2

− 9 x + 5)

9

Solution: Let y = ( 3x 2 − 9 x + 5 )

9

Using chain rule, we obtain 9 dy d = = ( 3x 2 − 9 x + 5) dx dx

= 9 ( 3x 2 − 9 x + 5) .

8

d 3x 2 − 9 x + 5) ( dx

= 9 ( 3x 2 − 9 x + 5) . ( 6 x − 9 ) 8

= 9 ( 3x 2 − 9 x + 5 ) .3 ( 2 x − 3) 8

= 27 ( 3x 2 − 9 x + 5 ) . ( 2 x − 3) 8

2.

Differentiate the function w.r.t x sin3 x + cos6 x

Solution: Let y = sin 3 x + cos6 x 

dy d d = ( sin 3 x ) + ( cos 6 x ) dx dx dx

= 3sin 2 x.

d d ( sin x ) + 6 cos5 x. ( cos x ) dx dx

= 3sin 2 x cos x + 6cos5 x.( − sin x )

= 3sin x cos x ( sin x − 2cos 4 x )

3.

Differentiate the function w.r.t x

( 5x )

3cos2 x

Solution: Let y = ( 5x )

3cos2 x

Taking logarithm on both sides, we obtain log y = 3cos 2 x log5x

Differentiating both sides with respect to x , we obtain 1 dy d d   = 3 log 5 x. ( cos 2 x ) + cos 2 x. ( log 5 x )  y dx dx dx  



dy d 1 d   = 3 y log 5 x ( − sin 2 x ) . ( 2 x ) + cos 2 x. . ( 5 x )  dx dx 5 x dx  



dy cos 2 x   = 3 y  −2sin x log 5 x + dx x  



dy  3cos 2 x  = 3y  − 6sin 2 x log 5 x  dx  x 



4.

dy 3cos 2 x  3cos 2 x  = ( 5x ) − 6sin 2 x log 5 x   dx  x 

Differentiate the function w.r.t x

(

)

sin −1 x x , 0  x  1

Solution:

(

Let y = sin −1 x x

)

Using chain rule, we obtain

(

dy d = sin −1 x x dx dx

1

=

(

1− x x

=

1 − x3



(

d x x dx

)

d  12  x  dx  

=

3 1  .x 2 1 − x3 2

=

3 x

=

5.

1

)

3



)

1

2 1 − x3

3 x 2 1 − x3

Differentiate the function w.r.t x

x 2 , −2  x  2 2x + 7

cos −1

Solution:

x 2 Let y = 2x + 7 cos −1

By quotient rule, we obtain

dy = dx

2x + 7

d  −1 x   −1 x  d  cos  −  cos  dx  2  2  dx

(

2x + 7

)

2

(

2x + 7

)

    −1 d  x    −1 x  1 d  2x + 7  − cos . ( 2x + 7)  2 dx  2    2  2 2 x + 7 dx     1−  x       2 2x + 7 2x + 7 =

−1

x 2  −  cos −1  2  2 2x + 7 4− x  2x + 7 2

− 2x + 7

=

4 − x2 ( 2 x + 7 )

−

x 2 2x + 7 ( 2x + 7) cos−1

(

)

x   cos −1   1 2 = − + 3  2  4 − x 2x + 7 ( 2x + 7)2   

6.

Differentiate the function w.r.t x  cot −1  

(1 + sin x ) + (1 − sin x )   ,0  x  2 (1 + sin x ) − (1 − sin x ) 

Solution:  Let y = cot −1    Then,  

=

=

(

(1 + sin x ) + (1 − sin x )   ............ (1) (1 + sin x ) − (1 − sin x ) 

(1 + sin x ) + (1 − sin x )   1 + sin x − 1 − sin x ( ) ( ) 

(

1 + sin x + 1 − sin x

1 + sin x − 1 − sin x

)

)

2

1 + sin x + 1 − sin x

(1 + sin x ) + (1 − sin x ) + 2 (1 + sin x ) − (1 − sin x ) (1 + sin x ) − (1 − sin x )

2 + 2 1 − sin 2 x = 2sin x =

=

1 + cos x sin x

2 cos 2

x 2

x x 2sin cos 2 2

= cot

x 2

Therefore, equation (1) becomes x  y = cot −1  cot  2  y=



x 2

dy 1 d = ( x) dx 2 dx

dy 1 = dx 2

7.

Differentiate the function w.r.t x

( log x )

log x

, x 1

Solution: Let y = ( log x )

log x

Taking logarithm on both sides, we obtain

log y = log x.log ( log x ) Differentiating both sides with respect to x, we obtain

1 dy d = log x.log ( log x )  y dx dx 



1 dy d d = log ( log x ) . ( log x ) + log x. log ( log x )  y dx dx dx



  dy 1 1 d = y log ( log x ) . + log x. . ( log x )  dx x log x dx  



dy 1 1 = y  log ( log x ) +  dx x x



8.

log ( log x )  dy log x  1 = ( log x )  +  dx x x 

Differentiate the function w.r.t x

cos ( a cos x + b sin x ) , for some constant a and b Solution: Let y = cos ( a cos x + b sin x ) By using chain rule, we obtain dy d = cos ( a cos x + b sin x ) dx dx 

dy d = − sin ( a cos x + b sin x ) . ( a cos x + b sin x ) dx dx

= − sin ( a cos x + b sin x ) . a ( − sin x ) + b cos x 

= ( a cos x + b sin x ) sin ( a cos x + b sin x )

9.

Differentiate the function w.r.t x

( sin x − cos x )(

sin x − cos x )

,

 4

x

Solution: (sin x −cos x )

Let y = ( sin x − cos x )

3 4

Taking logarithm on both sides, we obtain (sin x −cos x ) 

log y = log ( sin x − cos x ) 



 log y = (sin x − cos x ) log (sin x − cos x ) Differentiating both sides with respect to x, we obtain

1 dy d = ( sin x − cos x ) .log ( sin x − cos x ) y dx dx  1 dy d d = log ( sin x − cos x ) . ( sin x − cos x ) + ( sin x − cos x ) . log ( sin x − cos x ) y dx dx dx

1 dy 1 d = log ( sin x − cos x ) . ( cos x + sin x ) + ( sin x − cos x ) . (sin x − cos x ) y dx ( sinx − cosx ) dx 



10.

dy ( sin x − cos x ) = ( sin x − cos x ) ( cos x + sin x ) .log ( sin x − cos x ) + ( cos x + sin x )  dx

dy ( sin x − cos x ) = ( sin x − cos x ) ( cos x + sin x ) 1 + log ( sin x − cos x )  dx

Differentiate the function w.r.t x x x + xa + a x + aa , for such fixed a  0 and x  0

Solution: Let y = x x + x a + a x + a a Also, let x x = u, x a = v, a x = w and a a = s y = u +v+ w+ s

dy du dv dw ds = + + + ................. (1) dx dx dx dx dx

u = xx  log u = log x x

 log u = x log x

Differentiating both side with respect to x we obtain 1 du d d = log x. ( x ) + x. ( log x ) u dx dx dx



du 1  = u log x.1 + x.  dx x 



du = x x  log x + 1 = x x (1 + log x ) .......... ( 2 ) dx

v = xa 

dv d a = (x ) dx dx



dv = ax a −1.......... ( 3) dx

w = ax  log w = log a x

 log w = x log a

Differentiating both sides with respect to x, we obtain 1 dw d . = log a. ( x ) w dx dx 

dw = w log a dx



dw = a x log a.......... ( 4 ) dx

s = aa

Since a is constant, aa is also constant 

ds = 0........... ( 5 ) dx

From 1,2,3,4 and 5, we obtain dy = x x (1 + log x ) + ax a −1 + a x log a + 0 dx

= x x (1 + log x ) + axa−1 + a x log a

11.

Differentiate the function w.r.t x xx

2

−3

+ ( x − 3) , for x  3 x2

Solution: Let y = x x

2

−3

+ ( x − 3)

Also, let u = x x

2

−3

x2

and v = ( x − 3)

x2

y =u+v

Differentiating both sides with respect to x we obtain dv du dv = + ............. (1) dx dx dx

u = xx

2

−3

(

log u = log x x

2

−3

)

log u = ( x2 − 3) logx Differentiating both sides with respect to x, we obtain 1 du d d = log x. ( x 2 − 3) + ( x 2 − 3) . ( log x ) u dx dx dx 

1 du 1 = log x.2 x + ( x 2 − 3) . u dx 3



2  x2 − 3  du = x x −3  + 2  log x  dx  x 

Also, v = ( x − 3)

x2

 log v = log ( x − 3)

x2

 log v = x2 log ( x − 3) Differentiating both sides with respect to x, we obtain 1 dv d d = log ( x − 3) ( x 2 ) + x 2 log ( x − 3)  v dx dx dx  

1 dv 1 d = log ( x − 3) 2 x + x 2 . . ( x − 3) v dx x − 3 dx

 dv x2   = v  2 x log ( x − 3) + .1 dx x − 3   

dv x2 = ( x − 3) dx

 x2   x − 3 + 2 x log ( x − 3)  

Substituting the expression of

du dv and in equation (1), we obtain dx dx

2 2   dy x 2 −3  x − 3 2  x =x  + 2 x log x  + ( x − 3) x  + 2 x log ( x − 3) dx  x   x −3 

12.

Find −

 2

  dy , if y = 12 (1 − cos t ) , x = 10 ( t − sin t ) ,  t  2 2 dx t 

 2

Solution: It is given that y = 12 (1 − cos t ) , x = 10 (t − sin t ) 

dx d d = 10 ( t − sin t )  = 10 ( t − sin t ) = 10 (1 − cos t ) dt dt dt

dy d d = 12 (1 − cos t )  = 12 (1 − cos t ) = 12. 0 − ( − sin t )  = 12sin t dx dx dt

 dy  t t 12.2sin .cos dy  dt  12sin t 2 2 = 6 cot t  = = = t dx  dx  10 (1 − cos t ) 5 2 10.2sin 2   2  dt 

13.

Find

dy , if y = sin −1 x + sin −1 1 − x2 , −1  x  1 dx

Solution: It is given that y = sin −1 x + sin −1 1 − x2 

(

)



dy d d = ( sin −1 x ) + sin −1 1 − x 2 dx dx dx



dy 1 1 d = + .

2 dx 1− x 1 1 − x 2 dx



dy 1 1 1 d = + .

(1 − x2 ) 2 2 dx x dx 1− x 2 1− x



dy 1 1 = + ( −2 x ) dx 1 − x2 2 x 1 − x2



dy 1 1 = − dx 1 − x2 1 − x2



14.

dy d  −1 = sin x + sin −1 1 − x 2   dx dx 

(

)

(

1 − x2

)

dy =0 dx

If x 1 + y + y 1 + x = 0, for −1  x  1, prove that

Solution:

dy 1 =− 2 dx (1 + x )

It is given that, x 1 + y + y 1 + x = 0

x 1+ y = − y 1+ x Squaring both sides, we obtain

x2 (1 + y ) = y 2 (1 + x )  x 2 + x 2 y = y 2 + xy 2  x 2 − y 2 = xy 2 − x 2 y

 x2 − y2 = xy ( y − x )  ( x + y )( x − y ) = xy ( y − x )  x + y = − xy

 (1 + x ) y = − x y=

−x (1 + x )

Differentiating both sides with respect to x, we obtain

y=

−x (1 + x )

dy =− dx

(1 + x )

d d ( x ) − x (1 + x ) (1 + x ) − x 1 dx dx =− =− 2 2 2 (1 + x ) (1 + x ) (1 + x )

Hence, proved 3

15.

If ( x − a ) + ( y − b ) 2

2

  dy 2  2 1 +      dx   2 = c , for some c > 0, prove that  is a constant d2y dx 2

independent of a and b Solution:

It is given that, ( x − a ) + ( y − b ) = c 2 2

2

Differentiating both sides with respect to x, we obtain d  d d 2 2 x − a )  + ( y − b )  = ( c 2 ) (  dx   dx dx   2( x − a)

d d ( x − a) + 2 ( y − b) ( y − b) = 0 dx dx

 2 ( x − a ) .1 + 2 ( y − b )



dy =0 dx

dy − ( x − a ) = ........... (1) dx y −b

d 2 y d  − ( x − a)   2 =   dx dx  y − b  d d   ( y − b ) dx ( x − a ) − ( x − a ) dx ( y − b ) =− 2 ( y − b) dy    ( y − b ) − ( x − a ) dx  = −  2 ( y − b)       − ( x − a)     ( y − b) − ( x − a )  y −b    =− 2   y − b) (      ( y − b )2 + ( x + a )2  = −  2 ( y − b)  

[using (1)]

3

3

 ( x − a ) 2   2  ( y − b )2 + ( x − a )2  2   dy    1 +    2 1 +     ( y − b )2    y − b ( )  dx       2   =  =  2 2 2  d y   ( y − b) + ( x − a)   ( y − b ) + ( x − a )2   dx 2  −  −  3 3   ( y − b) ( y − b)     2

3 2

3 2

 c   2 c2  ( y − b )  3 2 y − b) ( c =− = 3 c2 ( y − b) − 3 ( y − b) 2

= - c, which is constant and is independent of a and b Hence, proved

16.

2 dy cos ( a + y ) If cos y = x cos ( a + y ) with cos a  1, prove that = dx sin a

Solution: It is given that, cos y = x cos ( a + y ) 

d d =  cos y  =  x cos ( a + y )  dx dx 

 − sin y

dy d d = cos ( a + y ) ( x ) + x. cos ( a + y )  dx dx dx

 − sin y =

dy dy cos ( a + y ) + x.  − sin ( a + y )  dx dx

  x sin ( a + y ) − sin y 

dy = cos ( a + y ) dx

Since cos y = x cos ( a + y ) , x =

cos y cos ( a + y )

........... (1)

 cos y  dy Then, equation (1) reduces to  sin ( a + y ) − sin y  = cos ( a + y )  cos ( a + y )  dx  cos y.sin ( a + y ) − sin y.cos ( a + y )   sin ( a + y − y )

dy = cos 2 ( a + y ) dx

dy = cos 2 ( a + b ) dx

2 dy cos ( a + b )  = dx sin a

Hence, proved 17.

If x = a ( cos t + t sin t ) and y = a ( sin t − t cos t ) , find

d2y dx 2

Solution: It is given that, x = a ( cos t + t sin t ) and y = a (sin t − t cos t ) 

dx d = a. ( cot + t sin t ) dt dt

d d   = a  − sin t + sin t. ( t ) + t. ( sin t )  dt dt  

= a − sin t + sin t + t cos t  = at cos t dy d = a. ( sin t − t cos t ) dx dt

 d d   = a cost − cos t. ( t ) + t. ( cos t )  dt dt   

= a cost − cos t − t sin t = a t sin t  dy  dy  dt  a t sin t  = = = tan t dx  dx  a t cos t    dt 

Then,

d 2 y d  dy  d dt =   = ( tan t ) = sec2 t 2 dx dx  dx  dx dx

= sec2 t

=

18.

1 a t cos t

dt 1   dx = at cos t  =  dt dx at cos t 

sec3 t  ,0  t  at 2

If f ( x ) = x , show that f "( x ) exists for all real x, and find it 3

Solution:  x, if x  0 It is known that, x =   − x, if x  0

Therefore, when x  0, f ( x ) = x = x3 3

In this case, f ' ( x ) = 3x2 and hence, f "( x ) = 6x When x  0, f ( x ) = x = ( − x3 ) = x3 3

In this case, f ' ( x ) = 3x2 and hence f "( x ) = 6x Thus,

for

f ( x ) = x , f "( x ) exists 3

for

all

real

x

and

is

given

 6 x, if x  0 f "( x ) =  −6 x, if x  0

19.

Using mathematical induction prove that

d n x ) = nx n −1 for all positive integers n ( dx

Solution: To prove: P ( n ) : For n = 1,

d n ( x ) = nx n−1 for all positive integers n dx

by,

P (1) :

d ( x ) = 1 = 1.x1−1 dx

 p ( n ) is true for n = 1 Let p(k) is true for some positive integer k That is, p ( k ) :

d k x ) = kx k −1 ( dx

It is to be proved that p ( k +1) is also true Consider

xk

d k +1 d x ) = ( x.x k ) ( dx dx

d d ( x ) + x. ( x k ) dx dx

= xk .1 + x.k.xk −1 = xk + kxk

= ( k + 1) xk = ( k + 1) .x( k +1)−1 Thus, P (k+1) is true whenever P(k) is true Therefore, by the principal of mathematical induction, the statement P(n) is true for every positive integer n Hence, proved 20.

Using the fact that sin ( A + B ) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines

Solution:

sin ( A + B ) = sin A cos B + cos A sin B Differentiating both sides with respect to x, we obtain

d d d sin ( A + B )  = ( sin A cos B ) + ( cos A sin B ) dx dx dx  cos ( A + B )  cos ( A + B )

d d d d d ( A + B ) = cos B ( sin A ) + sin A ( cos B ) + sin B ( cos A ) + cos A ( sin B ) dx dx dx dx dx

d d dB dA dB ( A + B ) = cos B.cosA + sin A ( − sin B ) + sin B ( − sin A ) + cos A cos B dx dx dx dx dx

 dA dB   dA dB   cos ( A + B )  + = ( cos A cos B − sin A sin B )  +   dx dx   dx dx 

cos ( A + B ) = cos A cos B − sin Asin B 21.

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer

Solution: Consider f ( x ) = x + x + 1 Since modulus function is everywhere continuous and sum of two continuous function is also continuous Differentiability of f ( x ) : Graph of

f ( x ) shows that f ( x ) is everywhere derivable

except possible at x= 0 and x = 1

At x =0, Left hand derivative

lim−

x →0

f ( x ) − f ( 0) ( x + x − 1 ) − (1) = lim ( − x ) − ( x − 1) − 1 = lim −2 x = −2 = lim− x →0 x → 0− x → 0− x x−0 x x

Right hand derivative =

lim+

x →0

f ( x ) − f ( 0) ( x + x − 1 ) − (1) = lim ( − x ) − ( x − 1) − 1 = lim 0 = 0 = lim+ x →0 x → 0+ x → 0− x x−0 x x

Since L.H .D  R.H .D f ( x ) is not derivable at x = 0 At x = 1 L.H.D

lim−

x →1

f ( x ) − f (1) ( x + x − 1 ) = lim ( x ) − ( x − 1) − 1 = lim 0 = 0 = lim− x →1 x →1− x →1− x − 1 x −1 x −1 x −1

R.H.D

lim+

x →1

f ( x ) − f (1) ( x + x − 1 − 1) = lim ( x ) − ( x − 1) − 1 = lim 2 ( x − 1) = 2 = lim+ x →1 x →1+ x →1− x −1 x −1 x −1 x −1

Since L.H .D  R.H .D f ( x ) is not derivable at x = 1

 f ( x ) is continuous everywhere but not derivable at exactly two points

22.

 f ' ( x ) g ' ( x ) h ' ( x )  f ( x ) g ( x ) h ( x ) dy     = l m n  m n  , prove that If y =  l dx  a  a b c  b c 

Solution:

 f ( x ) g ( x ) h ( x )   y= l m n   a b c 

 y = ( mc − nb) f ( x ) − (lc − na ) g ( x ) + (lb − ma ) h ( x ) dy d d d = ( mc − nb ) f ( x )  − ( lc − na ) g ( x )  + ( lb − ma ) h ( x )  Then, dx dx dx dx

= ( mc − nb) f ' ( x ) − (lc − na ) g ' ( x ) + (lb − ma ) h ' ( x )  f ' ( x ) g ' ( x ) h ' ( x )   = l m n   a b c 

 f ' ( x ) g ' ( x ) h ' ( x ) dy   = l m n  Thus, dx  a b c  23.

If y = ea cos x , −1  x  1, show that (1 − x 2 ) −1

d2y dy − x − a2 y = 0 2 dx dx

Solution: It is given that, y = ea cos

−1

x

Taking logarithm on both sides, we obtain log y = a cos −1 x log e

log y = a cos −1 x

Differentiating both sides with respect to x, we obtain 1 dy 1 = ax y dx 1 − x2 dy −ay = dx 1 − x2

By squaring both the sides, we obtain 2

a2 y 2  dy  =   2  dx  1 − x 2

 dy   (1 − x )   = a 2 y 2  dx  2

2

 2 2 (1 − x )  dy  =a y dx  2

Again, differentiating both sides with respect to x, we obtain 2

dy d 2 y dy  dy     ( −2 x ) + (1 − x2 )  2 = a 2 .2 y 2 dx dx dx  dx 

x

dy d2y + (1 − x 2 ) 2 = a 2 . y dx dx

d2y dy  (1 − x ) 2 − x − a 2 y = 0 dx dx 2

Hence, proved

 dy   dx  0 

Exercise 5.1 1.

Prove that the function f ( x ) = 5x − 3 is continuous at x = 0, x = −3 and at x = 5

Solution: The given function is f ( x ) = 5x − 3 At x = 0, f ( 0) = 5  0 − 3 = 3

lim f ( x ) = lim ( 5 x − 3) = 5  0 − 3 = 3 x →

x →

 lim f ( x ) = f ( 0 ) x →

Therefore, f is continuous at x = 0 At x = −3, f ( −3) = 5x ( −3) − 3 = 18

lim f ( x ) = lim f ( 5 x − 3) = 5 x ( −3) − 3 = −18 x →3

x →3

 lim f ( x ) = f ( −3) x →3

Therefore, f is continuous at x = −3 At x = 5, f ( x ) = f (5) = 5  5 − 3 = 25 − 3 = 22

lim f ( x ) = lim ( 5x − 3) = 5  5 − 3 = 22 x →5

x →5

 lim f ( x ) = f ( 5) x →5

Therefore, is f continuous at x = 5 2.

Examine the continuity of the function at f ( x ) = 2x2 −1 x = 3

Solution: The given function is f ( x ) = 2x2 −1

At x = 3, f ( x ) = f (3) = 2  32 −1 = 17

lim f ( x ) = lim ( 2 x 2 − 1) = 2  32 − 1 = 17 x →3

x →3

 lim f ( x ) = f ( 3) x →3

Thus, f is continuous at x = 3 .

3.

Examine the following functions for continuity a) f ( x ) = x − 5 c) f ( x ) =

b) f ( x ) =

x 2 − 25 , x5 x+5

1 , x5 x−5

d) f ( x ) =| x − 5 |

Solution: a) The given function is f ( x ) = x − 5 It is evident that f is defined at every real number k and its value at k is k − 5 It is also observed that lim f ( x ) = lim f ( x − 5) = k = k − 5 = f ( k ) x →k

x →k

 lim f ( x ) = f ( k ) x →k

Hence, f is continuous at every real number and therefore, it is a continuous function. b) The given function is f ( x ) = lim f ( x ) = lim x→k

x→k

Also, f ( k ) =

1 , x  5 for any real number k  5 , we obtain x−5

1 1 = x−5 k −5

1 k −5

( As k  5 )

lim f ( x ) = f ( k ) x →k

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

c) The given function is f ( x ) =

x 2 − 25 , x5 x+5

For any real number c  −5 , we obtain lim f ( x ) = lim x →c

x →c

Also, f ( c ) =

( x + 5)( x − 5) = lim x − 5 = c − 5 x 2 − 25 = lim ( ) ( ) x →c x →c x+5 x+5

( c + 5)( c − 5) = c c+5

( c − 5)( as c  5)

lim f ( x ) = f ( c ) x →c

Hence f is continuous at every point in the domain of f and therefore. It is continuous function. 5 − x, if x  5 d) The given function is f ( x ) =| x − 5 |=   x − 5, if x  5

This function f is defined at all points of the real line.

Let c be a point on a real time. Then, c  5 or c = 5 or c  5 Case I: c  5 Then, f ( c ) = 5 − c

lim f ( x ) = lim ( 5 − x ) = 5 − c x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all real numbers less than 5.

Case II : c = 5 Then, f ( c ) = f (5) = (5 − 5) = 0

lim f ( x ) = lim ( 5 − x ) = ( 5 − 5) = 0

x →5−

x →5

lim f ( x ) = lim ( x − 5) = 0

x →5+

x →5

 lim+ f ( x ) = lim+ ( x ) = f ( c ) x →c

x →c

Therefore, f is continuous at x = 5 Case III : c  5 Then, f ( c ) = f (5) = c − 5

lim f ( x ) = lim f ( x − 5) = c − 5 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at real numbers greater than 5.

Hence, f is continuous at every real number and therefore, it is a continuous function.

4.

Prove that the function f ( x ) = xn is continuous at is a x = n positive integer.

Solution: The given function is f ( x ) = xn It is evident that f is defined at all positive integers, n , and its value at n is nn .

( )

Then, lim f ( n ) = lim f x n = n n x →n

x →n

 lim f ( x ) = f ( n ) x →n

Therefore, f is continuous at n , where n is a positive integer.

5.

 x, if x  1 Is the function f defined by f ( x ) =  5, if x  1

Continuous at x = 0 ? At x = 1? . At x = 2 ?

Solution:  x, if x  1 The given function f is f ( x ) =  5, if x  1

At x = 0 It is evident that f is defined at 0 and its value of 0 is 0 Then, lim f ( x ) = lim x = 0 x →0

x →0

 lim f ( x ) = f ( 0 ) x →0

Therefore, f is continuous at x = 0 At x = 1 f is defined at 1 and its value at is 1.

The left hand limit of at f i x = 1 s,

lim f ( x ) = lim− x = 1

x →1−

x →1

The right hand limit of at f is x = 1 ,

lim f ( x ) = lim+ f ( 5)

x →1+

x →1

 lim− f ( x )  lim+ f ( x ) x →1

x →1

Therefore, is f is not continuous at x = 1 At x = 2 f is defined at 2 and its value at 2 is 5.

Then, lim f ( x ) = lim f ( 5) = 5 x →2

x →2

 lim f ( x ) = f ( 2 ) x →2

Therefore f is continuous at x = 2 6.

Find all points of discontinuous of f , where f is defined by 2 x + 3, if x  2 f ( x) =  2 x − 3 if x  2

Solution: 2 x + 3, if x  2 The given function f if f ( x ) =  2 x − 3 if x  2

It is evident that the given function f is defined at all the points of the real time.

Let c be a point on the real line. Then, three cases arise.

I.

c2

II.

c2

III.

c=2

Case (i) c  2 Then, f ( x ) = 2x + 3

lim f ( x ) = lim ( 2 x + 3) = 2c + 3 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  2 Case (ii) c  2 Then, f ( c ) = 2c − 3

lim f ( x ) = lim ( 2 x − 3) = 2c − 3 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  2 Case (iii) c = 2 Then, the left hand limit of f at x = 2 is

lim f ( x ) = lim− ( 2 x + 3) = 2  2 + 3 = 7 x →2

x →2

The right hand limit of f at is x = 2

lim f ( x ) = lim+ ( 2 x + 3) = 2  2 − 3 = 1

x → 2+

x →2

It is observed that the left and right hand limit of f at x = 2 do not coincide.

Hence, x = 2 is the only point of discontinuity of f .

7.

Find all points of discontinuity of f , where f is defined by

| x | +3, if x  −3  f ( x ) = −2 x, if − 3  x  3 6 x + 2 if x  3  Solution: The given function f is defined at all the points at the real line.

Let c be a point on the real line.

Case I: If c  −3 , then f ( c ) = −c + 3

lim f ( x ) = lim ( − x + 3) = −c + 3 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  −3 Case II : If , c = −3 then f ( −3) = − ( −3) + 3 = 6

lim f ( x ) = lim− ( − x + 3) = − ( −3) + 3 = 6

x →3−

x →3

 lim+ f ( x ) = lim+ f ( −2 x ) = 2 x ( −3) = 6 x →3

x →3

lim f ( x ) = f ( −3) x →3

Therefore, f is continuous at x = −3

Case III : If , −3  c  3 then f ( c ) = −2c and lim f ( x ) = lim ( −2 x ) = −2c x →c

x →3c

lim f ( x ) = f ( c ) x →c

Therefore, f is continuous in (-3, 3) Case IV: If c = 3 , then the left hand limit of f at x = 3 is

lim f ( x ) = lim− f ( −2 x ) = −2  3 = 6

x →3−

x →3

The right hand limit of f at x = 3 is

lim f ( x ) = lim− f ( 6 x + 2 ) = 6  3 + 2 = 20

x →3+

x →3

It is observed that the left and right hand limit of f at x = 3 do not coincide.

Case V: If c  3 , then f ( c ) = 6c + 2 and lim f ( x ) = lim ( 6 x + 2 ) = 6c + 2 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore f is continuous at all points x , such that x  3 .

Hence, x = 3 is the only point of discontinuity of f .

8.

| x |, if x  0  Find all points of discontinuity of f , where f is defined by f ( x ) =  x if x = 0 0 

Solution: | x |, if x  0  The given function f is f ( x ) =  x if x = 0 0 

It is known that, x  0 | x |= − x and x  0 | x |= x Therefore, the given function can be rewritten as

| x |  x =  f ( x ) 0, | x |  =  x

−x = −1 if x  0 x if x = 0 x if x  0 =1 x

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I: If c  0 , then f ( c ) = 1

lim f ( x ) = lim ( −1) = −1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore f is continuous at all points x  0 Case II: If c = 0 , then the left hand limit of f at x = 0 is,

lim f ( x ) = lim− ( −1) = −1

x →0 −

x →0

The right hand limit of f at x = 0 is

lim f ( x ) = lim+ (1) = 1

x →0+

x →0

It is observed that the left and right hand limit of f at x = 0 do not coincide.

Therefore, f is not continuous at x = 0 .

Case III: If c  0, f ( c ) = 1

lim f ( x ) = lim (1) = 1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  0 Hence, x = 0 is the only point of discontinuity of f .

9.

 x  , if x  0 Find all points of discontinuity of f , where f is defined by f ( x ) =  | x | if x  0 −1 

Solution:

 x  , if x  0 The given function f is f ( x ) =  | x | if x  0 −1  It is known that, x  0 | x |= x Therefore, the given function can be rewritten as

 x  , if x  0  f ( x ) = −1 for all x  R f ( x ) = | x | if x  0 −1  Let c be any real number. Then lim f ( x ) = lim ( −1) = −1 x →c

x →c

Also, f ( c ) = −1 = lim f ( x ) x →c

Therefore, the given function is continuous function.

Hence, the given function has no point of discontinuity.

10.

Find all the points of discontinuity of f , where f is defined by

 x + 1, if x  1 f ( x) =  2 x + 1, if x  1 Solution:

 x + 1, if x  1 The given function f is f ( x ) =  2 x + 1, if x  1 The given function f is defined at all the points of the real lime.

Let c be a point on the real time.

Case I :

(

)

If c  1 , then f ( c ) = c2 + 1 and lim f ( x ) = lim f x 2 + 1 = c 2 + 1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  1 Case II : If c = 1 , then f ( c ) = f (1) = 1 + 1 = 2 The left hand limit of f at x = 1 is

lim− f ( x ) = lim− ( x 2 + 1) = 12 + 1 = 2

x →1

x →1

The right hand limit of f at x = 1 is

lim+ f ( x ) = lim+ ( x 2 + 1) = 12 + 1 = 2

x →1

x →1

 lim f ( x ) = f ( c ) x →1

Therefore, f is continuous at x = 1 Case III: If c  1 , then f ( c ) = c +1

lim f ( x ) = lim ( x + 1) = c + 1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  1 .

Hence, the given function f has no points of discontinuity.

11.

Find all points of discontinuity of f , where f is defined by 3   x − 3, if x  2 f ( x) =  2   x + 1, if x  2

Solution: The given function

3   x − 3, if x  2 f is f ( x ) =  2   x + 1, if x  2

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I :

(

)

If c  2 , then f ( c ) = c3 − 3 and lim f ( x ) = lim x3 − 3 = c 3 − 3 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such at that x  2 Case II : If c = 2 , then f ( c ) = f ( 2) = 23 − 3 = 5

lim+ f ( x ) = lim− ( x3 − 3) = 23 − 3 = 5

x →2

x →2

lim+ f ( x ) = lim+ ( x 2 + 1) = 22 + 1 = 5

x →2

x →2

 lim f ( x ) = f ( 2 ) x →2

Therefore, f is continuous at x = 2 Case III : If c  2 , then f ( c ) = c2 + 1

lim f ( x ) = lim ( x 2 + 1) = c 2 + 1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  2 Thus, the given function f is continuous at every point on the real time.

Hence, f has no point of discontinuity.

12.

10   x − 1, if x  1 Find all points of discontinuity of f , where f is define by f ( x ) =  2 if x 1  x ,

Solution: 10   x − 1, if x  1 The given function f is f ( x ) =  2 if x 1  x ,

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I :

(

)

If c  1 , then f ( c ) = c10 − 1 and lim f ( x ) = lim x10 − 1 = c10 − 1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  1 Case II : If c = 1 , then the left hand limit of f at x = 1 is

lim f ( x ) = lim− ( x10 − 1) = 1010 − 1 = 1 − 1 = 0

x →1−

x →1

The right hand limit of f at x = 1 is ,

lim+ f ( x ) = lim+ ( x 2 ) = 12 = 1

x →1

x →1

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1 Case III : If c  1 , then f ( c ) = c2

lim f ( x ) = lim ( x 2 ) = c 2 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore f is continuous at al points x , such that x  1 Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f .

13.

 x + 5, if x  1 Is the function define by f ( x ) =  a continuous function?  x − 5, if x  1

Solution:  x + 5, if x  1 The given function is f ( x ) =   x − 5, if x  1

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I ; If c  1 , then f ( c ) = c + 5 and lim f ( x ) = lim ( x + 5) = c + 5 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  1 Case II : If c = 1 , then f (1) = 1 + 5 = 6

The left hand limit of f at x = 1 is

lim f ( x ) = lim− ( x + 5) = 1 + 5 = 6

x →1−

x →1

The right hand limit of f at x = 1 is lim− f ( x ) = lim− ( x − 5) = 1 − 5 = −4 x →1

x →1

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore f is not continuous at x = 1 Case III : If c  1 , then f ( c ) = c − 5 and lim f ( x ) = lim ( x − 5) = c − 5 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore f is continuous at all points x , such that x  1 Thus, from the above observations, it can be concluded that x = 1 is the only point of discontinuity of f .

14.

Discuss the continuity of the function f , where f is defined by

3, if 0  x  1  f ( x ) = 4, if 1  x  3 5, if 3  x  10  Solution:

3, if 0  x  1  The given function is f ( x ) = 4, if 1  x  3 5, if 3  x  10  The given function is defined at all the points of the interval 0, 10 .

Let c be a point in the interval 0, 10 Case I ; If , 0  c  1 then f ( c ) = c + 5 f ( c ) = 3 and lim f ( x ) = lim ( 3) = 3 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous in the interval [0,1) Case II : If c = 1 , then f ( 3) = 3 The left hand limit of f at x = 1 is

lim f ( x ) = lim− ( 3) = 3

x →1−

x →1

The left hand limit of f at x = 1 is lim− f ( x ) = lim− ( 3) = 3 x →1

x →1

The right hand limit of f at x = 1 is lim+ f ( x ) = lim+ ( 4 ) = 4 x →1

x →1

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore f is not continuous at x = 1 Case III : If c  1 , then f ( c ) = 4 and lim f ( x ) = lim ( 4 ) = 4 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore f is continuous at all points of the interval (1, 3) Case IV: If c = 3 , then f ( c ) = 5 The left hand limit of f at x = 3 is

lim f ( x ) = lim− ( 4 ) = 4

x →3−

x →3

The right hand limit of f at x = 3 is lim+ f ( x ) = lim+ ( 5) = 5 x →3

x →3

It is observed that the left and right hand limit of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

Case V : If 3  c  10 , then f ( c ) = 5 and lim f ( x ) = lim ( 5) = 5 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points of the interval (3, 10].

Hence, f is not continuous at x = 1 and x = 3 15.

Discuss the continuity of the function f , where f is defined by

2 x, if x  0  f ( x ) = 0, if 0  x  1 4 x. if x  1  Solution:

2 x, if x  0  The given function is f ( x ) = 0, if 0  x  1 4 x. if x  1  The given function is defined at all the points of the real line.

Let c be a point on the real line Case I ; If , c  0 then f ( c ) = 2c , lim f ( x ) = lim ( 2 x ) = 2c x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points , x such that x < 0 Case II : If c = 0 , then f ( c ) = f ( 0) = 3 The left hand limit of f at x = 0 is lim− f ( x ) = lim− ( 2 x ) = 2  0 = 0 x →0

x →0

The right hand limit of f at x = 0 is lim+ f ( x ) = lim+ ( 0 ) = 0 x →0

x →0

lim f ( x ) = f ( 0 ) x →0

Therefore, f is continuous at x = 0 Case III : If 0  c  1 , then f ( x ) = 0 and lim f ( x ) = lim ( 0 ) = 0 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore f is continuous at all points of the interval (0, 1) Case IV: If c = 1 , then f ( c ) = f (1) = 0 The left hand limit of f at x = 1 is lim− f ( x ) = lim− ( 0 ) = 0 x →1

x →1

The right hand limit of f at x = 1 .is lim+ f ( x ) = lim+ ( 4 x ) = 4 1 = 4 x →1

x →1

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1 Case V : If c  1 , then f ( c ) = 4c and lim f ( x ) = lim ( 4 x ) = 4c x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x, such that x > 1 Hence, f is not continuous only at x = 1 16.

Discuss the continuity of the function f , where f is defined by

−2,  f ( x ) =  2 x,  2,  Solution:

if x  −1 if − 1  x  1 if x  1

−2,  The given function f is f ( x ) = 2 x,  2, 

if x  −1 if − 1  x  1 if x  1

The given function is defined at all point of the real time. Let c be a point on the real time.

Case I : If c  −1, then f ( c = −2) and lim f ( x ) = lim ( −2 ) = −2 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  −1 Case –II : If c = −1, then f ( c ) = f ( −1) = −2 The left hand limit of f at x = −1 is, lim f ( x ) = lim− ( −2 ) = −2

x →1−

x →1

The right hand limit of f at x = −1 is , lim f ( x ) = lim+ 2 x ( −1) = −2

x →1+

x →1

lim f ( x ) = f ( −1) x →1

Therefore , f is continuous at x = −1 Case III : If −1  c  1, then f ( c ) = 2c lim f ( x ) = lim ( 2 x ) = 2c x →c

x →c

lim f ( x ) = f ( c ) x →c

Therefore , f is continuous at all points of the interval ( −1,1) .

Case - IV : If c = 1 , then f ( c ) = f (1) = 2 1 = 2 The left hand limit of f at x = 1 is , lim f ( x ) = lim− ( 2 x ) = 2 1 = 2

x →1−

x →1

The right hand limit of f at x = 1 is , lim f ( x ) = lim+ 2 = 2

x →1+

x →1

lim f ( x ) = f ( c ) x →1

Therefore, f is continuous at x = 2 Case –V : If c  1, f ( c ) = 2 and lim f ( x ) = lim ( 2 ) = 2 x →2

x →2

 lim f ( x ) = f ( c ) x →c

Therefore , f is continuous at all points , x , such that x  1 Thus, from the above observations, it can be concluded that f is continuous at all points of the real time.

17.

Find the relationship be a and b so that the function f defined by f (x) = 𝑎𝑥 + 1, 𝑖𝑓 𝑥 ≤ 3 { is continuous at x=3. 𝑏𝑥 + 3, 𝑖𝑓 𝑥 > 3

Solution: The given function f is f (x) = {

𝑎𝑥 + 1, 𝑖𝑓 𝑥 ≤ 3 If f is continuous at x=3, then 𝑏𝑥 + 3, 𝑖𝑓 𝑥 > 3

lim f ( x) = lim- f ( x) = f (3)

x →3-

x →3

Also, lim f ( x) = lim- f (ax + 1) = 3a + 1

x →3−

x →3

lim f ( x) = lim+ f (ax + 1) = 3b + 3

x →3+

x →3

f (3) = 3a+1 therefore, from (1), we obtain 3a+1 = 3b+3 = 3a+1

⇒3a+1 = 3b+3 ⇒3a=3b+2 2

⇒ a=b+ 3 2

Therefore, the required relationship is given by, a=b+ 3

18.

 ( x 2 − 2 x), if x  0 For what value of λ is the function defined by f (x) =   4 x + 1, if x  0 continuous at x = 0 ? What about continuity at x = 1?

 ( x 2 − 2 x), if x  0 Solution: The given function f is f (x) =   4 x + 1, if x  0 If f is continuous at x = 0, then

lim f ( x) lim f ( x) = f (0) x → 0−

x → 0−

 lim  ( x 2 − 2 x) = lim (4 x + 1) =  (02 − 2  0) x → 0−

x → 0+

  (02 − 2  0) = 4  0 + 1 = 0  0 =1 = 0, which is not possible Therefore, there is no value of λ for which f is continuous at x = 0 At x = 1, f (1) = 4x + 1 = 4  1 + 1 = 5

lim (4 x + 1) = 4 1 +1 = 5 x →1

19.

 

Show that the function defined by g ( x ) = x − x is discontinuous at all integral

 

point. Here x denoted the greatest integer less than or equal to x.

Solution:

 

The given function is g ( x ) = x − x

It is evident that g is defined at all integral points.

Let n be a integer Then,

g ( n ) = n −  n = n − n = 0 The left hand limit of f at x = n is

lim g ( x ) = lim−  x −  x = lim− x − lim−  x = n − ( n − 1) = 1

x→n−

x→n

x→n

x→n

The right hand limit of f at x = n is

lim g ( x ) = lim−  x −  x = lim− x − lim−  x = n − n = 0

x→n−

x→n

x→n

x→n

It is observed that the left and right hand limits of f at x = n do not coincide.

Therefore, f is not continuous at x = n Hence, g is discontinuous at all integral points 20.

Is the function defined by f ( x ) = x − sin x + 5 continuous at x =  ?

2

Solution: The given function is f ( x ) = x − sin x + 5 2

It is evident that f is defined at x =  At x =  , f ( x ) = f

( ) =  2 − sin  + 5 =  2 − 0 + 5 =  2 + 5

(

Consider lim f ( x ) = lim f x − sin x + 5 x→

x→

2

)

Put x =  + h If x →  , then it is evident that h → 0

lim f ( x ) = lim ( x2 − sin x ) + 5



x→

x→

= lim ( + h ) − sin ( + h ) + 5  h→0  2

= ( + 0 ) − lim sin  cosh + cos  + sinh  + 5 2

h→0

=  2 − limsin  cosh − limcos  + sinh + 5 h→0

h→0

=  2 − sin  cos0 − cos  sin 0 + 5

=  2 − 0  1 − ( −1)  0 + 5

=2 +5  lim f ( x ) = f ( ) x→ x

Therefore, the given function f is continuous at x =  21.

Discuss the continuity of the following functions. f x = sin x + cos x A) ( ) f x = sin x − cos x B) ( ) f x = sin x  cos x C) ( ) Solution: It is known that if g and h are two continuous functions, then g + h, g − h and g.h are also continuous.

It has to proved first that g ( x ) = sin x and h ( x ) = cos x are continuous functions.

Let g ( x ) = sin x It is evident that g ( x ) = sin x is defined for every real number.

Let c be a real number. Put x = c + h If x → c , then h → 0 g ( c ) = sin c

lim g ( x ) = lim g sin x x →c

x →c

= limsin ( c + h ) h →0

= lim sin c cosh + cos c sinh  h →0

= lim ( sin c cosh ) + lim ( cos c sinh ) h →0

h →0

= sin c cos 0 + cos c sin 0 = sin c + 0 = sin c

 lim g ( x ) = g ( c ) x →c

Therefore, g is a continuous function.

Let h ( x ) = cos x It is evident that h ( x ) = cos x is defined for every real number.

Let c be a real number. x = c + h If x → c, then h → 0

h ( c ) = cos c lim ( x ) = lim cos x x →c

x →c

= lim cos ( c + h ) h →0

= lim cos c cosh − sin c sinh  h →0

= limcos c cosh − limsin c sinh h →0

h →0

= cos c cos 0 − sin c sin 0 = cos c  1 − sin c  0

= cosc  lim ( x ) = h ( c ) h →0

Therefore, h is a continuous function.

Therefore, it can be concluded that a) f ( x ) = g ( x ) + h ( x ) = sin x + cos x is a continuous function b) f ( x ) = g ( x ) − h ( x ) = sin x − cos x is a continuous function c) f ( x ) = g ( x )  h ( x ) = sin x  cos x is a continuous function 22.

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution: It is known that if p and h are two continuous functions, then

i.

h ( x) .g ( x )  0 is continuous g ( x)

ii.

1 .g ( x )  0 is continuous g ( x)

iii.

1 .h ( x )  0 is continuous h ( x) It has to be proved first that g ( x ) − sin x and h ( x ) − cos x are continuous functions.

Let g ( x ) − sin x It has evident that g ( x ) − sin x is defined foc every real number.

Let c be a real number. Put x − c + h If x → C, then h → 0

g ( c ) − sin x

lim g ( c ) − lim sin x x +c

x+c

− limsin ( c + h ) x +c

− lim sin c cosh + cos c sinh  x+c

− lim ( sin c cosh ) + lim ( cos c sinh ) x +c

x +c

− sin c cos 0 + cos c sin 0

− sin c + 0 − sin c  lim g ( x ) − g ( c ) 0+c

Therefore, g is a continuous function.

Let h ( x ) − cos x It is evident that h ( x ) − cos x is defined for every real number.

Let c be a real number. Put x − c + h If x → c, then h → 0 x

h ( c ) − cos c

− lim cos ( c + h ) x →0

− lim  cos c cosh − sin c sinh  x →0

− limcos c cosh − limsin c sinh x →0

x →0

− cos c cos 0 − sin c sin 0 − cos c x1 − sin c x 0

= cosc

 lim h ( x ) − h ( c ) x+c

Therefore, h ( x ) − cos x is continuous function.

It can be conclned that, cos ex −

1 , sin x  0 is continuous sin x

 cos ex x, x  nx ( n  Z ) is continuous Therefore, secant is continuous except at X − np, nIZ sec x =

1 , cos x  0 is continuous cos x

 sec x, x  ( 2n + 1)



2

( n  Z ) is continuous

 Therefore, secant is continuous except at x − ( 2n + 1) ( n  Z ) 2

cos x cot x = , sin x  0 is continuous sin x

 cot x, x  n ( n  Z ) is continuous

23.

Therefore, cotangent is continuous except at x − np, nIZ  sin x , if x  0  Find the points of discontinuity of f , where f ( x ) =  x   x + 1, if x  0

Solution:  sin x , if x  0  The given function f is f ( x ) =  x   x + 1, if x  0 It is evident that f is defined at all points of the real line.

Let c be a real number Case I: sin c  sin x  sin c If c  0 , then f ( c ) = and lim f ( x ) = lim  = x →c x →c c c  x 

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  0 Case II: If c  0 , then f ( c ) = c +1 and lim f ( x ) = lim ( x + 1) = c + 1 x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  0 Case III: If c = 0 , then f ( c ) = f ( 0) = 0 + 1 = 1 The left hand limit of f at x = 0 is, sin x lim f ( x ) = lim− =1 x → 0− x →0 x The right hand limit of f at x = 0 is,

lim f ( x ) = lim+ ( x + 1) = 1

x → 0+

x →0

 lim− f ( x ) = lim+ f ( x ) = f ( 0 ) x →0

x →0

Therefore, f is continuous at x = 0 From the above observations, it can be conducted that f is continuous at all points of the real line.

Thus, f has no point of discontinuity.

24.

1  2  x sin , if  0 Determine if f defined by f ( x ) =  is a continuous function? x  if  0 0,

Solution: 1  2  x sin , if  0 The given function f is f ( x ) =  x  if  0 0, It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I: 1 If c  0 , then f ( c ) = c 2 sin c 1 1 1   lim f ( x ) = lim  x 2 sin  = lim x 2  limsin  = c 2 sin x →c x →c x  x →c x c   x →c

(

)

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all p[oints x  0 Case II: If c = 0 , then f ( 0) = 0 1 1   lim+ f ( x ) = lim+  x 2 sin  = lim  x 2 sin  x →0 x →0  x  x →0  2 1 It is known that, −1  sin  1, x  0 x 1  − x 2  sin  x 2 x 1   lim ( − x 2 )  lim  x 2 sin   lim x 2 x →0 x →0 x  x →0  1   0  lim  x 2 sin   0 x →0 x  1   lim  x 2 sin  = 0 x →0 x   lim f ( x ) = 0 x →0

1 1   Similarly, lim+ f ( x ) = lim+  x 2 sin  = lim  x 2 sin  = 0 x → 0 x →0 x →0  x x   lim− f ( x ) = f ( 0 ) = lim+ f ( x ) = 0 x →0

x →0

Therefore, f is continuous at x = 0 From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus f is a continuous function.

25.

Examine the continuity of f , where f is defined by sin x − cos x, if x  0 f ( x) =  , if x = 0  1

Solution: sin x − cos x, if x  0 The given function f is f ( x ) =  , if x = 0  1 It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I: If c  0 , then f ( c ) = sin c − cos c

lim f ( x ) = lim ( sin x − cos x ) = sin c − cos c x →c

x →c

 lim f ( x ) = f ( c ) x →c

Therefore, f is continuous at all points x , such that x  0 Case II: If c = 0 , then f ( 0) = −1

lim f ( x ) = lim ( sin x − cos x ) = sin 0 − cos 0 = 0 − 1 = −1

x → 0−

x →0

x → 0+

x →0

lim f ( x ) = lim ( sin x − cos x ) = sin 0 − cos 0 = 0 − 1 = −1

 lim− f ( x ) = lim+ f ( x ) = f ( 0 ) x →0

x →0

Therefore, f is continuous at x = 0 From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

26.

Find the value of k so that the function f is continuous at the indicated point.

 k cos x   − 2 x , if   f ( x )  3, if    

 2   x= atx = 2 2

x

Solution:

 2  x= 2

 k cos x   − 2 x , if   if The given function f is f ( x )  3,     The given function f is continuous at x = the f at x =

 2

equals the limit of f at x =

It is evident that f is defined at x = lim x→

 2

f ( x ) = lim

Put x =

x→

 2



2

 2

x

 2

 2

 2

k cos x  − 2x

h→0

  k cos  + h  k cos x 2    lim f ( x ) = lim = lim    − 2x h → 0    x→ x→ 2 2  − 2 + h 2  = k lim x →0

− sinh k sinh k k = lim = .1 = x → 0 −2h 2 h 2 2

   lim f ( x ) = f    x→ 2 2 

k =3 2

k =6 Therefore, the required value of k is 6.

.

  2

and f   = 3

+h

Then, x →

, it is defined at x =

 2

and if the value of

27.

Find the value of k so that the function f is continuous at the indicated point. kx 2 , if x  2 at x = 2 f ( x) =   3, if x  2

Solution: kx 2 , if x  2 The given function is f ( x ) =   3, if x  2 The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of f at x = 2 equals the limit of f at x = 2 2 It is evident that f is defined at x = 2 and f ( 2) = k ( 2) = 4k lim f ( x ) = lim+ f ( x ) = f ( 2 )

x → 2−

x →2

 lim− ( kx ) = lim+ ( 3) = 4k 2

x →2

x →2

 k  2 = 3 = 4k 2

 4k = 3 = 4k  4k = 3 k=

3 4

Therefore, the required value of k is

28.

3 .

4

Find the values of k so that the function f is continuous at the indicated point. kx + 1, if x   f ( x) =  at x =   cos x, if x  

Solution:  kx + 1, if x   The given function is f ( x ) =   cos x, if x   The given function f is continuous at x =  and, if f is defined at x =  and if the value of f at x =  equals the limit of f at x =  It is evident that f is defined at x =  and f ( ) = k + 1

lim f ( x ) = lim+ f ( x ) = f ( )

x → −

x →

 lim− ( kx + 1) = lim+ cos x = k + 1 x →

x →

 k + 1 = cos  = k + 1  k  + 1 = −1 = k  + 1  k  + 1 = −1 = k  + 1 2 k =− 

29.

2 Therefore, the required value of k is − .  Find the values of k so that the function f is continuous at the indicated point.

 kx + 1, if f ( x) =  3x − 5, if

x5 x5

Solution:

 kx + 1, if 3x − 5, if

The given function of f is f ( x ) = 

x5 x5

The given function f is continuous at x = 5 , if f is defined at x = 5 and if the value of f at x = 5 equals the limit of f at x = 5 It is evident that f is defined at x = 5 and f ( 5) = kx + 1 = 5k + 1

lim f ( x ) = lim+ f ( x ) = f ( 5 )

x →5−

x →5

 lim− ( kx + 1) = lim+ ( 3x − 5 ) = 5k + 1 x →5

x →5

 5k + 1 = 15 − 5 = 5k + 1

 5k + 1 = 10  5k = 9 k=

9 5

Therefore, the required value of k is 30.

9 5

Find the vales of a and b such that the function defined

if  5,  f ( x ) = ax + b, if  21 if 

x2 2  x  10 is continuous function. x  10

Solution:

if  5,  The given function f is f ( x ) = ax + b, if  21 if 

x2 2  x  10 x  10

It is evident that the given function f is defined at all points of the real line.

If f is a continuous function, then f is a continuous at all real numbers.

In a particular, f is continuous at x = 2 and x = 10 Since f is continuous at x = 2 , we obtain

lim f ( x ) = lim+ f ( x ) = f ( 2 )

x →2−

x →2

 lim− ( 5 ) = lim+ ( ax + b ) = 5 x →2

x →2

 5 = 2a + b = 5

 2a + b = 5

.......... (1)

Since f is continuous at x = 10 , we obtain

lim f ( x ) = lim+ f ( x ) = f (10 )

x →10−

x →10

 lim− ( ax + b ) = lim+ ( 21) = 21 x →10

x →2

10a + b − 21 = 21 10a + b = 21

.......... (2)

On subtracting equation (1) from equation (2), we obtain

8a = 16 a = 2 By putting a = 2 in equation (1), we obtain

2 2 + b = 5

4+b =5  b =1 Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.

31.

( ) is a continuous function.

Show that the function defined by f ( x ) = cos x

Solution:

( )

The given function is f ( x ) = cos x

2

2

This function f is defined for every real number and f can be written as the composition of two functions as,

f = goh , where g ( x ) = cos x and h ( x ) = x 2

 

( goh )( x ) = g ( h ( x )) = g ( x2 ) = cos ( x2 ) = f ( x )

It has to be first proves that g ( x ) = cos x and

h ( x ) = x 2 are continuous functions.

It is evident that g is defined foe every real number.

Let c be a real number.

Then, g ( c ) = cos c Put x = c + h If x → c , then h → 0

lim g ( x ) = lim cos x x →c

x →c

= lim cos ( c + h ) h →0

= lim  cos c cosh − sin c sinh  h →0

= limcos c cosh − limsin c sinh h→0

h→0

= cos c cos0 − sin c sin0 = cos c 1 − sin c  0

= cosc

 lim g ( x ) = g ( c ) x →c

Therefore, g ( x ) = cos x is a continuous function

h ( x ) = x2 Clearly, h is defined for every real number.

Let k be a real number, then

lim h ( x ) = lim x2 = k 2 x→k

x→k

h(k ) = k 2

lim h ( x ) = h ( k ) x →k

Therefore, h is a continuous function.

It is known that for real valued functions g and h, such that ( goh ) is defined at c, it g is continuous at c and it f is continuous at c.

Therefore, 32.

f ( x ) = ( goh )( x ) = cos ( x 3 ) is a continuous function.

Show that the function defined by f ( x ) = cos x is a continuous function

Solution: The given function is f ( x ) = cos x This function f is defined for every real number and f can be written as the composition of two functions as,

f = goh , where g ( x ) = x and h ( x ) = cos x



( goh )( x ) = g ( h ( x ) ) = g ( cos x ) = cos x = f ( x )

It has to be first proves that g ( x ) = x and h ( x ) = cos x are continuous functions.

g ( x ) = x , can be written as − x if g ( x) =   x if

x0 x0

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I: If c  0 , then g ( c ) = −c and lim g ( x ) = lim ( − x ) = −c x →c

x →c

 lim g ( x ) = g ( c ) x →c

Therefore, g is continuous at all points x, such that x  0 Case II: If c  0 , then g ( c ) = c and lim g ( x ) = lim ( − x ) = c x →c

 lim g ( x ) = g ( c ) x →c

x →c

Therefore, g is continuous at all points x, such that x  0 Case III: If c = 0 , then g ( c ) = g ( 0) = 0 and lim g ( x ) = lim ( − x ) = −c x →c

x →c

lim g ( x ) = lim− ( − x ) = 0

x →0 −

x →0

lim g ( x ) = lim+ ( x ) = 0

x →0 +

x →0

 lim− g ( x ) = lim+ g ( x ) = g ( 0 ) x →c

x →c

Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is continuous at all points.

h ( x ) = cos x It is evident that h ( x ) = cos x is defined for every real number.

Let c be a real number. Put x = c + h If x → c , then h → 0

h ( c ) = cos c lim h ( x ) = limcos x x →c

x →c

= lim cos ( c + h ) h →0

= lim  cos c cosh − sin c sinh  h →0

= limcos c cosh − limsin c sinh h→0

h→0

= cos c cos0 − sin c sin0 = cosc  1 − sin c  0

= cosc

= lim h ( x ) = h ( c ) h →c

Therefore, h ( x ) = cos x is continuous function/

It is known that for real values functions g and h, such that ( goh ) is defined at c, if g is continuous at c and if f is continuous at g ( c ) , then Therefore,

( fog ) is

f ( x ) = ( goh )( x ) = g ( h ( x ) ) = g ( cos x ) = cos x

continuous at c. is a continuous

function.

33.

Examine that sin x is a continuous function

Solution: Let f ( x ) = sin x This function f is defined for every real number and f can be written as the composition of two functions as,

f = goh , where g ( x ) = x and h ( x ) = sin x



( goh )( x ) = g ( h ( x ) ) = g ( sin x ) = sin x = f ( x )

It has to be prove first that g ( x ) = x and h ( x ) = sin x are continuous functions.

g ( x ) = x can be written as − x if g ( x) =   x if

x0 x0

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I: If c  0 , then g ( c ) = −c and lim g ( x ) = lim ( − x ) = −c x →c

x →c

 lim g ( x ) = g ( c ) x →c

Therefore, g is continuous at all points x, that x  0 Case II: If c  0 , then g ( c ) = c and lim g ( x ) = lim x = c x →c

 lim g ( x ) = g ( c ) x →c

x →c

Therefore, g is continuous at all points x, such that x  0 Case III: If c = 0 , then g ( c ) = g ( 0) = 0

lim g ( x ) = lim− ( − x ) = 0

x →0 −

x →0

lim g ( x ) = lim+ ( x ) = 0

x →0 +

x →0

 lim− g ( x ) = lim+ g ( x ) = g ( 0 ) x →c

x →c

Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is continuous at all points.

h ( x ) = sin x It is evident that h ( x ) = sin x is defined for every real number.

Let c be a real number. Put x = c + k If x → c , then k → 0

h ( c ) = sin c lim h ( x ) = limsin x x →c

x →c

= limsin ( c + k ) k →0

= lim sin c cos k + cos c sin k  k →0

= lim ( sin c cos k ) − lim ( cos c sin k ) k →0

k →0

= sin c cos0 + cos c sin0 = sin c + 0

= sinc = lim h ( x ) = g ( c ) x →c

Therefore, h is continuous function.

It is known that for real values functions g and h, such that ( goh ) is defined at c, if g is continuous at c and if f is continuous at g ( c ) , then Therefore,

( foh ) is

continuous at c.

f ( x ) = ( goh )( x ) = g ( h ( x ) ) = g ( sin x ) = sin x

is a continuous

function.

34.

Find all the points of discontinuity of f defined by f ( x ) = x − x + 1 .

Solution: The given function is f ( x ) = x − x + 1 The two functions, g and h, are defined as

g ( x ) = x and h ( x ) = x + 1 Then f = g − h The continuous of g and h is examined first.

g ( x ) = x can be written as − x if g ( x) =   x if

x0 x0

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I: If c  0 , then g ( c ) = g ( 0) = −c and lim g ( x ) = lim ( − x ) = −c x →c

x →c

 lim g ( x ) = g ( c ) x →c

Therefore, g is continuous at all points x, that x  0 Case II: If c  0 , then g ( c ) = c and lim g ( x ) = lim x = c x →c

x →c

 lim g ( x ) = g ( c ) x →c

Therefore, g is continuous at all points x, such that x  0 Case III:

If c = 0 , then g ( c ) = g ( 0) = 0

lim g ( x ) = lim− ( − x ) = 0

x →0 −

x →0

lim g ( x ) = lim+ ( x ) = 0

x →0 +

x →0

 lim− g ( x ) = lim+ g ( x ) = g ( 0 ) x →c

x →c

Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is continuous at all points.

h( x) = x +1

− ( x + 1) , if , x  −1 h( x) =  if , x  −1  x + 1, Clearly, h is defined for every real number.

Let c be a real number Case I: If c  −1 , then h ( c ) = − ( c + 1) and lim h ( x ) = lim − ( x + 1) = − ( c + 1) x→c

x→c

 lim h ( x ) = h ( c ) x →c

Therefore, h is continuous at all points x, such that x  −1 Case II: If c  −1, then h ( c ) = c + 1 and lim h ( x ) = lim x + 1 = ( c + 1) x →c

x →c

 lim h ( x ) = h ( c ) x →c

Therefore, g is continuous at all points x, such that x  −1 Case III: If c = −1, then h ( c ) = h ( −1) = −1 + 1 = 0

lim− h ( x ) = lim− − ( x + 1) = − ( −1 + 1) 0 x→1

x→1

lim+ h ( x ) = lim+ ( x + 1) = ( −1 + 1) = 0 x →1

x →1

 lim− h = lim+ h ( x ) = h ( −1) x →1

x →1

Therefore, h is continuous at x = −1 From the above three observations, it can be concluded that h is continuous at all points of the real line. g and h are continuous functions. Therefore, f = g − h is also a continuous function.

Therefore, f has no point of discontinuity.

Exercise 5.2

1.

(

Differentiate the function with respect to x. sin x 2 + 5

)

Solution:

(

)

Let f ( x ) = sin x2 + 5 , u ( x ) = x 2 + 5, and v ( t ) = sin t

(

)

(

)

Then, ( vou )( x ) = v ( u ( x ) ) = v x 2 + 5 = tan x 2 + 5 = f ( x ) Thus, f is a composite of two functions Put t = u ( x ) = x2 + 5 Then, we obtain dv d = ( sin t ) = cos t = cos ( x 2 + 5 ) dt dt dt d 2 d d = ( x + 5) = ( x 2 ) + ( 5) = 2 x + 0 = 2 x dx dx dx dx

Therefore, by chain rule,

df dv dt = = cos ( x 2 + 5 )  2 x = 2 x cos ( x 2 + 5 ) dx dt dx

Alternate method d d sin ( x 2 + 5 )  = cos ( x 2 + 5 ) ( x 2 + 5 )   dx dx

d d  = cos ( x 2 + 5)  ( x 2 ) + ( 5)  dx   dx

= cos ( x2 + 5)  2 x + 0

= 2 x cos ( x2 + 5)

2.

Differentiate the functions with respect of x. cos ( sin x )

Solution: Let f ( x ) = cos (sin x ) , u ( x ) = sin x, and v (t ) = cos t Then, ( vou )( x ) = v (u ( x )) = v (sin x ) = cos (sin x ) = f ( x ) Thus, f is a composite function of two functions Put t = u ( x ) = sin x 

dv d =  cos t  = − sin t = − sin ( sin x ) dt dt

dt d = ( sin x ) = cos x dx dx

By chain rule,

df dv dt , . = − sin ( sin x ) cos x = − cos x sin ( sin x ) dx dt dx

Alternate method d d cos ( sin x )  = − sin ( sin x ) ( sin x ) = − sin ( sin x ) − cos x = − cos x sin ( sin x ) dx dx

3.

Differentiate the functions with respect of x. sin ( ax + b )

Solution: Let f ( x ) = sin ( ax + b) , u ( x ) = ax + b, and v (t ) = sin t Then, ( vou )( x ) = v (u ( x )) = v ( ax + b) = sin ( ax + b) = f ( x ) Thus, f is a composite function of two functions u and v Put t = u ( x ) = ax + b Therefore,

dv d = ( sin t ) = cos t = cos ( ax + b ) dt dt

dt d d d = ( ax + b ) = ( ax ) + ( b ) = a + 0 = a dx dx dx dx

Hence, by chain rule, we obtain df dv dt = = cos ( ax + b ) a = a cos ( ax + b ) dx dt dx

Alternate method d d sin ( ax + b )  = cos ( ax + b ) ( ax + b ) dx dx

d d  = cos ( ax + b )  ( ax ) + ( b )  dx  dx 

= cos ( ax + b )( a + 0) = a cos ( ax + b) 4.

( ( x ))

Differentiate the functions with respect of x. sec tan

Solution:

( ( x )) , u ( x ) =

Let f ( x ) = sec tan

x , v (t ) = tan t , and w ( s ) = sec s

Then, ( wovou )( x ) = w v (u ( x ) ) = w v 

( x ) = w( tan x ) = sec ( tan x ) = f ( x )

Thus, f is a composite function of three functions, u, v and w Put s = v ( t ) = tan t and t = u ( x ) = x Then,

dw d = ( sec s ) = sec s tan s = sec ( tan t ) tan ( tan t ) ds ds

(

) (

= sec tan x .tan tan x

)

t = x   

ds d = ( tan t ) = sec 2 t = sec 2 x dt dt

dt d = dx dx

( )

x =

d  12  1 12 −1 1  x  = .x = dx   2 2 x

Hence, by chain rule, we obtain

 s = tan t 

dt dw ds dt = dx ds dt dx

(

) (

)

(

) (

= sec tan x .tan tan x  sec2 x 

=

=

1 2 x

sec2 x tan x tan tan x

(

2 x

)

) (

)

(

) (

sec2 x sec tan x tan tan x

1

2 x

Alternate method

(

)

) (

d  d sec tan x  = sec tan x tan tan x tan x  dx  dx

(

) (

)

( x ) dxd ( x )

(

) (

)

( x ). 2 1 x

(

) (

)

( x)

= sec tan x tan tan x sec 2

= sec tan x .tan tan x .sec2

=

5.

)

sec tan x .tan tan x .sec 2 2 x

Differentiate the functions with respect of X.

sin ( ax + b ) cos ( cx + d )

Solution: The given function is f ( x ) =

h ( x ) = cos ( cx + d )

h ( x ) = cos ( cx + d ) f =

g ' h − gh ' h2

sin ( ax + b ) g ( x ) = , where g ( x ) = sin ( ax + b ) and cos ( cx + d ) h ( x )

Consider g ( x ) = sin ( ax + b ) Let u ( x ) = ax + b, v (t ) = sin t Then ( vou )( x ) = v (u ( x )) = v ( ax + b ) = sin ( ax + b ) = g ( x ) g is a composite function of two functions, u and v

Put t = u ( x ) = ax + b dv d = ( sin t ) = cos t = cos ( ax + b ) dt dt dt d d d = ( ax + b ) = ( ax ) + ( b ) = a + 0 = a dx dx dx dx

Therefore, by chain rule, we obtain g'=

dg dv dt = = cos ( ax + b ) .a = a cos ( ax + b ) dx dt dx

Consider h ( x ) = cos ( cx + d ) Let p ( x ) = cx + d , q ( y ) = cos y Then, ( qop )( x ) = q ( p ( x )) = q ( cx + d ) = cos ( cx + d ) = h ( x ) h is a composite function of two functions, p and q

Put y = p ( x ) = cx + d

dq d = ( cos y ) = − sin y = − sin ( cx + d ) dy dy dy d d d = ( cx + d ) = ( cx ) + ( d ) = c dx dx dx dx

Therefore, by chain rule, we obtain

h' =

dh dq dy = = − sin ( cx + d )  c = −c sin ( cx + d ) dx dy dx

 f '=

a cos ( ax + b ) cos ( cx + d ) − sin ( ax + b )−c sin cx + d  cos ( cx + d ) 

2

a cos ( ax + b ) sin ( cx + d ) 1 + c sin ( ax + b )  cos ( cx + d ) cos ( cx + d ) cos ( cx + d )

=

= a cos ( ax + b) sec ( cx + d ) + c sin ( ax + b) tan ( cx + b) sec ( cx + d ) 6.

( )

Differentiate the function with respect to x. cos x3 .sin 2 x5

Solution:

cos x3 .sin 2 ( x5 ) d d d cos x 3 .sin 2 ( x 5 )  = sin 2 ( x 5 )  ( cos x 3 ) + cos3  sin 2 ( x 5 )     dx dx dx  = sin 2 ( x5 )  ( − sin x 3 ) 

d 3 d x ) + cos x 3 + 2sin ( x 5 ) sin x 5  ( dx dx

The given function is = sin x3 sin 2 ( x5 )  3 x 2 + 2sin x 5 cos x 3 .cos x 5 

= 3x2 sin x3 sin3 ( x5 ) + 2sin x5 cos x5 cos x3  5x4 = 10 x 4 sin x5 cos x5 cos x3 − 3x2 sin x3 sin 2 ( x5 ) 7.

Differentiate the functions with respect to x.

2

cot ( x 2 )

Solution: d 2 cot ( x 2 )     dx

1

= 2.

2

cot ( x 2 )



d cot ( x 2 )   dx 

d 5 (x ) dx

sin ( x 2 )

=

cos ( x

2

)

sin ( x 2 )

=

cos ( x

2

)

 − cos ec 2 ( x 2 ) 



1 sin

2

(x ) 2

d 2 (x ) dx

 ( 2x)

−2 x

=

cos x 2 sin x 2 sin x 2 −2 2 x

=

= 8.

2sin x2 cos x2 sin x2

−2 2 x sin x 2 sin 2 x 2

Differentiate the functions with respect to x cos

( x)

Solution: Let f ( x ) = cos

( x)

Also, let u ( x ) = x And, v ( t ) = cos t Then, ( vou )( x ) = v (u ( x )) =v

( x)

= cos x

= f ( x) Clearly, f is a composite function of two functions, u and v, such that t = u ( x ) = x

dt d = Then, dx dx

d  12  x = x  dx  

( )

1 − 12 1 x = 2 2 x

And,

dv d = ( cos t ) = − sin t = − sin x dt dt

By using chain rule, we obtain dt dv dt = dx dt dx

= − sin =−

=−

( x ) 21x

1 2 x sin

sin

( x)

( x)

2 x

Alternate method d  cos dx 

= − sin

( x ) = − sin ( x ) dxd ( x )

( x )  dxd  x

1 2

  

1 −1 = − sin x  x 2 2 =

9.

− sin x 2 x

Prove that the function f given by f ( x ) = x −1 , x  R is not differentiable at x = 1

Solution: The given function is f ( x ) = x −1 , x  R It is known that a function f is differentiable at a point x = c in its domain if both

lim−

k →0

f (c + h) − f (c) f (c + h) − f (c ) are finite and equal and lim+ h →0 k h

To check the differentiability of the given function at x = 1, Consider the left hand limit of f at x = 1

lim−

h →0

f I + h −1 1 −1 f (1 + h ) − f (1) = lim− h →0 h h

lim−

h →0

h −0 −h = lim− h →0 h h

( h  0  h = −h )

=-1 Consider the right hand limit of f at x = 1

f I + h −1 − 1 −1 f (1 + h ) − f (1) = lim+ h →0 h h

lim+

h →0

= lim+ h →0

h −0 h = lim+ h → 0 h h

( h  0  h = h)

=1 Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x =1

10.

Prove that the greatest integer function defined by f ( x ) =  x ,0  x  3 is not differentiable at x = 1 and x = 2

Solution: The given function f is f ( x ) =  x ,0  x  3 It is known that a function f is differentiable at a point x = c in its domain if both

lim−

h →0

f (c + h) − f (c) f (c + h) − f (c ) are finite and equal and lim+ h →0 h h

To check the differentiable of the given function at x = 1, consider the left hand limit of f at x = 1

lim−

h →0

f (1 + h ) − f (1) 1 + h − 1 = lim− h →0 h h

= lim− h →0

0 −1 −1 = lim− = h → 0 h h

Consider the right hand limit of f at x = 1

= lim+ h →0

= lim+ h →0

f (1 + h ) − f (1) 1 + h1 = lim− h →0 h h 1−1 = lim+ 0 = 0 h →0 h

Since the left and right limits of f at x = 1 are not equal, f is not differentiable at x = 1 To check the differentiable of the given function at x=2, consider the left hand limit of f at x = 2

= lim− h →0

= lim− h →0

f ( 2 + h ) − f ( 2)  2 + h  −  2 = lim− h →0 h h 1− 2 −1 = lim− = h →0 h h

Consider the right hand limit of f at x= 1

= lim+ h →0

= lim+ h →0

f ( 2 + h ) − f ( 2)  2 + h  −  2 = lim+ h →0 h h 1− 2 = lim+ 0 = 0 h →0 h

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x=2

Exercise 5.3 11.

Find

dy : 2 x + 3 y = sin x dx

Solution: The given relationship is 2 x + 3 y = sin x Differentiating this relationship with respect to x, we obtain

d d ( 2 x + 3 y ) = ( sin x ) dy dx 

d d ( 2 x ) + ( 3 y ) = cos x dy dx

 2+3

3



12.

dy = cos x dx

dy = cos x − 2 dx

dx cos x − 2 = dy 3

Find

dy : 2 x + 3 y = sin y dx

Solution: The given relationship is 2 x + 3 y = sin y Differentiating this relationship with respect to x, we obtain

d d d ( 2 x ) + ( 3 y ) = ( sin y ) dx dx dx

 2+3

d dy = cos y dx dx

 2 = ( cos y − 3)

dy dx

(by using chain rule)



13.

dy 2 = dx cos y − 3

Find

dy : ax + by 2 = cos y dx

Solution: The given relationship is ax + by 2 = cos y Differentiating this relationship with respect to x, we obtain

d d d ( ax ) + ( by 2 ) = ( cos y ) dx dx dx  a+b

d d by 2 ) = ( cos y ) ( dx dx

Using chain rule, we obtain

And

d 2 d y ) = 2 y ( cos y ) ........ (1) ( dx dx

d d ( cos y ) = sin y ........ ( 2 ) dx dx

From (1) and (2), we obtain

a + bx 2 y

dy dy = − sin y dx dx

 ( 2by + sin y )



14.

dy = −a dx

dy −a = dx 2by + sin y

Find

dy : xy + y 2 = tan x + y dx

Solution: The given relationship is xy + y 2 = tan x + y

Differentiating this relationship with respect to x, we obtain

d d xy + y 2 ) = ( tan x + y ) ( dx dx 

d d d dy ( xy ) + ( y 2 ) = ( tan x ) + dx dx dx dx

d dy dy  d   y. ( x ) + x.  + 2 y = sec2 x + dx  dx dx  dx

y.1 + x

dy dy dy + 2y = sec 2 x + dx dx dx

 ( x + 2 y − 1)



15.

[using product rule and chain rule]

dy = sec 2 x − y dx

dy sec2 x − y = dx ( x + 2 y − 1)

Find

dy 2 : x + xy + y 2 = 100 dx

Solution: The given relationship is x 2 + xy + y 2 = 100 Differentiating this relationship with respect to x, we obtain

d 2 d x + xy + y 2 ) = (100 ) ( dx dx



[Derivative of constant function is 0]

d 2 d d x + ( xy ) + ( y 2 ) = 0 dx dx dx

dy  dy  d  2 x +  y. ( x ) + x.  + 2 y =0 dx  dx  dx

 2 x + y.1 + x.

dy dy + 2y =0 dx dx

 2x + y + ( x + 2 y ).

dy =0 dx

[Using product rule and chain rule]



16.

dy 2x + y =− dx x + 2y

Find

dy 2 : x + x 2 y + xy 2 + y 3 = 81 dx

Solution: The given relationship is x 2 + x 2 y + xy 2 + y 3 = 81 Differentiating this relationship with respect to x, we obtain

d 3 d x + x 2 y + xy 2 + y 3 ) = ( 81) ( dx dx 

d 3 d d d 2 x ) + ( x 2 y ) + ( xy ) + ( y 3 ) = 0 ( dx dx dx dx

dy   d d dy  d   3x 2 +  y ( x 2 ) + x 2  +  y 2 ( x ) + x ( y 2 ) + 3 y 2 =0 dx   dx dx dx  dx   dx   dy  dx  3x 2 +  y.2 x + x 2  +  y 2 .1 + x.2 y  + 3 y 2 =0 dy   dx  dy 

 ( x 2 + 2 xy + 3 y 2 )

dy + ( 3 x 2 + 2 xy + y 2 ) = 0 dx

2 2 dy − ( 3 x + 2 xy + y )  = dx ( x 2 + 2 xy + 3 y 2 )

17.

Find

dx : sin 2 y + cos xy =  dy

Solution: The given relationship is sin 2 y + cos xy =  Differentiating this relationship with respect to x, we obtain

d d sin 2 y + cos xy ) = ( ) ………… (1) ( dx dx



d d sin 2 y ) + ( cos xy ) = 0 ( dx dx

Using chain rule, we obtain

d d dy sin 2 y ) + 2sin y ( sin y ) = 2sin y cos y ( dx dx dx d d dy ( cos xy ) + − sinx y ( xy ) = − sinx y  y ( x ) + x  ......... ( 2 ) dx dx   dx

dy  dy  = − sin xy  y.1 + x  = − y sin xy − x sin xy ………. (3) dx  dx 

From (1), (2) and (3), we obtain

2sin y cos y

dy dy − y sin xy − x sin xy =0 dx dx

 ( 2sin y cos y − x sin xy )

 ( sin 2 y − x sin xy )



18.

dy = y sin xy dx

dx = y sin xy dy

dx y sin xy = dy sin 2 y − x sin xy

Find

dy : sin 2 x + cos 2 y = 1 dx

Solution: The given relationship is sin 2 x + cos 2 y = 1 Differentiating this relationship with respect to x, we obtains

dy d sin 2 x + cos 2 y ) = (1) ( dx dx 

d d sin 2 x ) + cos 2 y = 0 ( dx dx

 2sin x.

d d ( sin x ) + 2 cos y. ( cos y ) = 0 dx dx

 2sin x cos x + 2 cosy ( − siny ) .  sin x − sin 2 y



19.

dy =0 dx

dy =0 dx

dx sin 2 x = dy sin 2 y

Find

dy  2x  : y = sin −1  2  dx  1+ x 

Solution:  2x  The given relationship is y = sin −1  2   1+ x   2x  y = sin −1  2   1+ x 

 sin y =

2x 1 + x2

Differentiating this relationship with respect to x, we obtain d d 2x ( sin y ) =  2  dx dx  1 + x 

 cos y

dy d  2 x  =   ........... (1) dx dx  1 + x 2 

The function

2x u , is of the form of 2 1+ x v

Therefore, by quotient rule, we obtain

d d 1 + x 2 ) ( 2 x ) − 2 x. (1 + x 2 ) d  2x  ( dx dx =  2  2 dx  1 + x  (1 + x )

(1 + x ) .2 − 2 x 0 + 2 x = 2 + 2 x − 4 x = (1 + x ) (1 + x ) 2

2

2 2

Also, sin y =

3

2 2

 2x   cos y = 1 − sin y = 1 −  = 2   1+ x  2

2 2

=

2 2

(1 + x )

2 2

=

((1 + x ) − 4x ) ........(3) 2 2

2

(1 + x )

2 2

1 − x2 1 + x2

From (1) (2) and (3) we obtain 2 1 − x 2 dy 2 (1 − x ) x = 1 + x 2 dx (1 + x 2 )2



20.

dy 2 = dx 1 + x 2

3 dx 1 1 −1  3x − x  : y = tan  ,− x Find 2  dy 3 3  1 − 3x 

Solution:

 3x − x3  The given relationship is y = tan −1  2   1 − 3x   3x − x3  y = tan −1  2   1 − 3x   tan y =

......... ( 2 )

2x 1 + x2 2

(1 − x ) (1 − x )

=

2 (1 + x 2 )

3x − x3 ............(1) 1 − 3x 2

It is known that, tan y =

y y − tan 3 3 3 ............. ( 2 ) y 1 − 3tan 2 3

3tan

Comparing equations (1) and (2), we obtain

x = tan

y 3

Differentiating this relationship with respect to x, we obtain d d y ( x ) =  tan  dx dx  3  1 = sec2

y d  y .   3 dx  3 

 1 = sec 2

y 1 dy .

3 3 dx





21.

dy 3 3 = = y y dx sec 2 1 + tan 2 3 3

dx 3 = dy 1 + x 2

2 dy −1  1 − x  : y cos  ,0  x  1 Find 2  dx  1+ x 

Solution: The given relationship is,

 1 − x2  y cos−1  2   1+ x   cos y =

1 − x2 1 + x2

y 2 2 = 1− x  y 1 + x2 1 + tan 2 2 1 − tan 2

On comparing LHS and RHS of the above relationship, we obtain

tan

y =x 2

Differentiating this relationship with respect to x, we obtain sec2

y d  y d .   = ( x) 2 dx  2  dx

 sec 2



dy 2 = dx sec2 y 2



dy 2 = dx 1 + tan 2 y 2



22.

y 1 d  =1 2 2 dx

dy 1 = dx 1 + x 2

Find

 1 − x2  dy : y = sin −1  ,0  x  1 2  dx  1+ x 

Solution:

 1 − x2  The given relationship is y = sin  2   1+ x  −1

 1 − x2  y = sin −1  2   1+ x  = siny =

1 − x2 1 + x2

Differentiating this relationship with respect to x, we obtain

d d  1 − x2  sin y = ( )   ............. (1) dx dx  1 + x 2  Using chain rule, we obtain

d dy ( sin y ) = cos y. dx dx  1 − x2  cos y = 1 − sin y = 1 −  2   1+ x 

2

2

(1 + x ) − (1 − x ) (1 + x ) 2 2

=



2 2

2 2

=

4x2

(1 + x )

2 2

2

)(1 − x ) − (1 − x )(1 + x ) (1 + x ) 2

2

2 2

(1 + x ) ( −2 x ) − (1 − x ) ( 2 x ) = (1 + x ) 2

2

2 2

=

2x 1 + x2

d 2 x dy ........... ( 2 ) ( sin y ) = dx 1 + x 2 dx

d  1 − x 2  (1 + x  = dx  1 + x 2 

=

=

−2 x − 2 x 3 − 2 x + 2 x 3

(1 + x )

2 2

−4 x

(1 + x )

2 2

.............. ( 3)

From (1), (2) and (3), we obtain 2 x dy −4 x = 2 1 + x dx (1 + x 2 )2



dy −2 = dx 1 + x 2

Alternate method

 1 − x2  y = sin −1  2   1+ x 

2

[using quotient rule]

 sin y =

1 − x2 1 + x2

 (1 + sin y ) x2 = 1 − sin y  x2 =

1 − sin y 1 + sin y 2

y y   cos − sin  2 2  x2 =  2 y y   cos + sin  2 x 

y y − sin 2 2 x= y y cos + sin 2 2 cos

y 2 x= y 1 + tan 2 1 − tan

    x = tan  −  4 2

Differentiating this relationship with respect to x, we obtain d d   y  ( x ) = .  tan  −   dx dx   4 2  

 y  d  y   1 = sec2  −  .  −   4 2  dx  4 2     y   1 dy    1 = 1 + tan 2  −  .  − .    4 2   2 dx   

 1 dy   1 = (1 + x 2 )  −   2 dx 



dx −2 = dy 1 + x 2

23.

Find

dy  2x  : y = cos −1  , −1  x  1 2  dx  1+ x 

Solution:  2x  The given relationship is y = cos −1  2   1+ x   2x  y = cos −1  2   1+ x 

 cos y =

2x 1 + x2

Differentiating this relationship with respect to x, we obtain d d 2x  ( cos y ) =   dx dx  1 + x 2 

d d 1 + x ) ( 2 x ) − 2 x (1 + x ) ( dy dx dx  − sin y. = dx (1 + x ) 2

2

2 2

2 dy (1 + x )  2 − 2 x.2 x  − 1 − cos y = 2 dx (1 + x2 ) 2

2  2 1− x 2   ( )   2 x   dy   1−  = − 2   (1 + x 2 )2    1 + x   dx   



(1 − x ) − 4 x (1 + x )



(1 − x ) (1 + x )

2 2

2 2

2 2 2 2

2

2 dy −2 (1 − x ) = dx (1 + x2 )

2 dy −2 (1 − x ) = 2 dx (1 + x2 )

2 1 − x 2 dy −2 (1 − x )  = 2 1 + x 2 dx (1 + x2 )



24.

dy −2 = dx 1 + x 2

Find

)

(

dy 1 1 : y = sin −1 2 x 1 − x 2 , − x dx 2 2

Solution:

(

Relationship is y = sin −1 2 x 1 − x 2

(

y = sin −1 2 x 1 − x 2

)

)

 sin y = 2x 1 − x2 Differentiating this relationship with respect to x, we obtain cos y =

dy  d = 2 x dx  dx

 1 − sin 2 y

(

)

1 − x2 + 1 − x2

d dx 

 x −2 x  dy = 2 . + 1 − x2  2 dx  2 1− x 

(

 1 − 2 x 1 − x2

)

 − x2 + 1 − x2  dy = 2  2 dx  1− x 

 1 − 2x2  dy  1 − 4 x (1 − x ) = 2  2 dx  1− x  2

(1 − 2 x )



 (1 − 2 x )



25.

2

2

 1 − 2x2  dy = 2  2 dx  1− x 

 1 − 2x2  dy = 2  2 dx  1− x 

dy 2 = dx 1 − x2

Find

dy 1  1  : y = sec−1  2  , 0  x  dx 2  2x −1 

Solution:  1  The given relationship is y = sec−1  2   2x −1   1  y = sec−1  2   2x −1 

 sec y =

1 2x2 −1

 cos y = 2 x 2 − 1  2 x 2 = 1 + cos y

 2 x 2 = 2 cos 2

 x = cos

y 2

y 2

Differentiating this relationship with respect to x, we obtain d d y ( x ) =  cos  dx dx  2

y d  y  1 = sin .   2 dx  2 







−1 1 dy = y 2 dx sin 2 dy −2 −2 = = dx sin y y 1 − cos 2 2 2 dy −2 = dx 1 − x2

Exercise 5.4 1.

Differentiating the following w.r.t x :

Solution:

ex Let y = sin x Differentiating wrt x, we obtain

dy = dx =

sin x

d x d e ) − e x ( sin x ) ( dx dx sin 2 x

sin x. ( e x ) − e x . ( cos x ) sin 2 x

e x ( sin x − cos x ) = , x  n , n  Z sin 2 x

2.

Differentiating the following esin

Solution: Let y = esin

−1

x

Differential wrt x, we obtain

(

dy d sin −1 x = e dx dx 

−1 dy d = esin x . ( sin −1 x ) dx dx

−1

 esin x .



)

esin

−1

x

1 − x2

1 1 − x2

−1

x

ex sin x

−1

dy esin x  = , x  ( −1,1) dx 1 − x2 3.

Differentiating the following wrt x : e x

3

Solution: Let y = ex

3

By using the quotient rule, we obtain

( )

3 3 3 3 dy d d = = e x = e x . ( x 3 ) = e x .3 x 2 = 3x 2 e x dx dx dx

4.

Differentiating the following w.r.t x : sin ( tan −1 e− x )

Solution: Let y = sin ( tan −1 e− x ) By using the chain rule, we obtain

dy d : sin ( tan −1 e − x )  dx dx  = cos ( tan −1 e − x ) . = cos ( tan −1 e− x ) .

=

=

=

cos ( tan −1 e − x ) 1+ e

−2 x

d tan −1 e − x ) ( dx 1+ (e

.e − x .

e − x cos ( tan −1 e − x ) 1 + e −2 x

1

.

d −x (e ) dx

d (−x) dx

( −1)

−e− x cos ( tan −1 e − x ) 1 + e−2 x

)

−x 2

5.

Differentiating the following w.r.t x : log ( cos e x )

Solution: Let y = log ( cos e x ) By using the chain rule, we obtain

dy d = log ( cos e x )  dx dx =

1 d . ( cos e x ) x cos e dx

=

1 d . − sin e x ) ( e x ) x ( cos e dx

− sin e x x = .e cos e x = −e x tan e x , e x  ( 2n + 1) 6.

 2

,n N

Differentiating the following w.r.t x : e x + e x + ........... + e x 2

5

Solution:

(

2 5 d x e + e x + ........... + e x dx

=

( )

)

( )

( )

( )

d x d x2 d x3 d x4 d x5 e )+ e + e + e + e ( dx dx dx dx dx

 2 d 2   x3 d 3   x 4 d 4   x5 d 5  = e x + e x ( x ) + e dx ( x ) + e dx ( x ) + e dx ( x )  dx

(

) (

) (

) (

= e x + e x 2 x + e x 3x 2 + e x 4 x 3 + e x 5 x 4 2

3

= e x + 2 xe x + 3x 2e x + 5 x 4e x 2

3

4

5

5

)

7.

Differentiating the following w.r.t x : e x , x  0

Solution: Let y = e

x

Then, y2 = e

x

By differentiating this relationship with respect to x, we obtain

y2 = e

x

 2y

dy =e dx

 2y

dy =e dx

[by applying the chain rule] x

x



dy e x = dx 4 y x



dy e = dx 4 e

( x)

1 1 2 x

x x

dy e x  = dx 4 xe 8.

d dx

x ,x 0 x

Differentiating the following w.r.t x : log ( log x ) , x  1

Solution: Let y = log ( log x ) By using the chain rule, we obtain

dy d = log ( log x )  dx dx 

=

1 d . ( log x ) log x dx

=

1 1 . log x x

1 , x 1 x log x 9.

Differentiating the following w.r.t x :

cos x ,x 0 log x

Solution: Let y =

cos x log x

By using the quotient rule, we obtain

d d cos x ) log x − cos x ( log x ) dy dx ( dx = 2 dx ( log x ) − sin x log x − cos x

( log x ) =

10.

1 x

2

−  x log x.sin x + cos x  x ( log x )

2

,x 0

Differentiating the following w.r.t x : cos ( log x + e x ) , x  0

Solution: Let y = cos ( log x + e x ) By using the chain rule, we obtain y = cos ( log x + e x )

dy d = − sin log x + e x  ( log x + e x ) dx dx d d  = sin ( log x + e x )  ( log x ) + ( e x )  dx  dx 

1  = − sin ( log x + e x )  + e x  x  1  =  + e x  sin ( log x + e x ) , x  0 x 

Exercise 5.5

1.

Differentiate the following with respect to x. cos x.cos 2 x.cos 3 x

Solution: Let y = cos x.cos 2 x.cos3x Taking logarithm or both the side, we obtain

log y = log ( cos x.cos 2x.cos3x )  log y = log ( cos x ) + log ( cos 2x ) + log ( cos3x ) Differentiating both sides with respect to x, we obtain

1 dy 1 d 1 d 1 d = ( cos x ) + ( cos 2 x ) + ( cos3x ) y dx cos x dx cos 2 x dx cos3x dx 

dy = − cos x.cos 2 x.cos 3 x  tan x + 2 tan 2 x + 3 tan 3 x  dx



2.

dy sin 3x d  sin x sin 2 x d = y − − ( 2x ) − ( 3x ) dx cos x cos 2 x dx cos 3 x dx  

Differentiate the functions with respect to x.

( x − 1)( x − 2 ) ( x − 3)( x − 4 )( x − 5) Solution: Let y =

( x − 1)( x − 2 ) ( x − 3)( x − 4 )( x − 5)

Taking logarithm or both the side, we obtain

log y = log

( x − 1)( x − 2 ) ( x − 3)( x − 4 )( x − 5)

 ( x − 1)( x − 2 )  1  log y = log   2  ( x − 3)( x − 4 )( x − 5 )   log y =

1 log ( x − 1)( x − 2 ) − log ( x − 3 )( x − 4 )( x − 5 ) 2

 log y =

1 log ( x − 1) + log ( x − 2 ) − log ( x − 3) − log ( x − 4 ) − log ( x − 5 )  2

Differentiating both sides with respect to, we obtain 1 dy 1  1 d 1 d 1 d 1 d 1 d =  ( x − 1) + ( x − 2) − ( x − 3) − ( x − 4) − ( x − 5) y dx 2  x − 1 dx x − 2 dx x − 3 dx x − 4 dx x − 5 dx 





3.

dy y  1 1 1 1 1  =  + − − −  dx 2  x − 1 x − 2 x − 3 x − 4 x − 5 

dy 1 = dx 2

( x − 1)( x − 2 )  1 + 1 − 1 − 1 − 1  ( x − 3)( x − 4 )( x − 5)  x − 1 x − 2 x − 3 x − 4 x − 5 

Differentiate the functions with respect to x.

( log x )

cos x

Solution: Let y = ( log x )

cos x

Taking logarithm or both the side, we obtain

log y = cos x.log ( log x ) Differentiating both sides with respect to x, we obtain

1 dy d d = ( cos x )  log ( log x ) + cos x  log ( log x ) y dx dx dx 

1 dy 1 d = − sin x log ( log x ) + cos x  ( log x ) y dx log x dx



 dy cos x 1  = y  − sin x log ( log x ) +  dx log x x  



4.

 dy cos x  cos x = ( log x )  − sin  log ( log x )  dx  x log x 

Differentiate the functions with respect to x. x x − 2sin x

Solution: Let y = x x − 2sin x Also, let x x = u and 2sin x = v y = u −v



dy du dv = − dx dx dx

u = xx

Taking logarithm on both sides, we obtain log u = x log x

Differentiating both sides with respect to x, we obtain 1 du  d d  =  ( x )  log x + x  ( log x )  u dx  dx dx  

du 1  = u 1 log x + x   dx x 



du = x x ( log x + 1) dx



du = x x (1 + log x ) dx

v = 2sin x

Taking logarithm on both the sides with respect to x, we obtain log v = sin x.log 2

Differentiating both sides with respect to x, we obtain

1 dv d = log 2 ( sin x ) v dx dx 

dv = v log 2 cos x dx



dv = 2sin x cos x log 2 dx



5.

dv = x 2 (1 + log x ) − 2sin x cos x log 2 dx

Differentiate the functions with respect to x.

( x + 3) .( x + 4) .( x + 5) 2

3

4

Solution: Let y = ( x + 3) ( x + 4 ) ( x + 5) 2

3

4

Taking logarithm on both sides, we obtain

log y = log ( x + 3) + log ( x + 4 ) + log ( x + 5) 2

3

4

 log y = 2log ( x + 3) + 3log ( x + 4) + 4log ( x + 5) Differentiating both sides with respect to x, we obtain

1 dy 1 d 1 d 1 d = 2. ( x + 3) + 3. ( x + 4) + 4. ( x + 5) y dx x + 3 dx x + 4 dx x + 5 dx 

dy 3 4   2 = y + + dx  x + 3 x + 4 x + 5 



dy 3 4  2 3 4  2 = ( x + 3) ( x + 4 ) ( x + 5 )  + + dx  x + 3 x + 4 x + 5 



dy 2 3 4  2 ( x + 4 )( x + 5 ) + 3 ( x + 3)( x + 5 ) + 4 ( x + 3)( x + 4 )  = ( x + 3) ( x + 4 ) ( x + 5 )   dx ( x + 3)( x + 4 )( x + 5)  



dy 2 3 = ( x + 3)( x + 4 ) ( x + 5 ) .  2 ( x 2 + 9 x + 20 ) + 3 ( x 2 + 9 x + 15 ) + 4 ( x 2 + 7 x + 12 )  dx



6.

dy 2 3 = ( x + 3)( x + 4 ) ( x + 5 ) ( 9 x 2 + 70 x + 133) dx

Differentiate the function with respect to x. x



1

1+  1   x x+  + x x 

Solution: 

x

1

1+  1  Let y =  x +  + x x  x 



x

1

1+  1  Also, let u =  x +  and v = x x  x 

y =u+v



dy du dv = + dx dx dx

1  Then, u =  x +  x 

.......... (1) x

Taking log on both sides

1   log u = log  x +  x 

x

1   log u = x log  x +  x 

Differentiating both sides with respect to x, we obtain 1 du d 1 d   1   = ( x )  log  x +  + x  log  x +   u dx dx x dx   x  



1 du 1 1 d  1  = 1 log  x +  + x  x+   1  dx  u dx x x   x+  x 

   du 1 x 1     = u log  x +  + x  x + 2  1 dx x  x    x+      x 

 1   x −    du  1 1 x    =  x +  log  x +  +  1  dx  x   x   x +   x   x

du  1  =x+  dx  x



x

  1  x2 − 1 log x + +   x  x 2 + 1  

du  1   x2 −1 1   = x+  2 + log  x +   dx  x   x +1 x  

v=x

............. ( 2 )

 1  x+   x

Taking log on both sides, we obtain  1  x+  x

v = x

 1  log v = 1 +  log x  x

Differentiating both sides with respect to x, we obtain 1 dv  d  1    1 d =  1 +    log x + 1 +  log x v dx  dx  x    x  dx



1 dv  1   11 =  − 2  log x + 1 +  v dx  x   xx



1 dv log x 1 1 =− 2 + + 2 v dx x x x



dv  − log x + x + 1 = v  dx x2 



1

 x +   x + 1 − log x  dv  = x x    dx x2  

......... ( 3)

Therefore, from (1), (2) and (3), we obtain   dy  1   x2 −1 1    x + x   x + 1 − log x   =x+   2 + log  x +   + x   dx  x   x +1 x  x2    x

7.

1

Differentiate the functions with respect to x.

( log x )

x

+ xlog x

Solution: Let y = ( log x ) + xlog x x

Also, let u = ( log x ) and v = xlog x x

y =u+v



dy du dv = + dx dx dx

u = ( log x )

......... (1)

x

x  log u = log ( log x )   

 log u = x log ( log x ) Differentiating both sides with respect to x, we obtain 1 du d d = ( x ) x log ( log x ) + x. log ( log x )  u dx dx dx



  du 1 d = u 1x log ( log x ) + x. ( log x ) dx log x dx  



du x 1 x  = ( log x ) log ( log x ) + dx log x x  



du 1  x  = ( log x ) log ( log x ) + dx log x  



du x  log ( log x ) .log x + 1  = ( log x )   dx log x  



du x −1 = ( log x ) 1 + log x.log ( log x )  dx

......... ( 2 )

v = xlog x

 log v = log ( xlog x )  log v = log x log x = ( log x )

2

Differentiating both sides with respect to x, we obtain 1 dv d  2 = ( log x )   v dx dx 

1 dv d = 2 ( log x ) ( log x ) v dx dx



dv log x = 2 x log x dx x



dv = 2 x log x −1 log x dx

.......... ( 3)

Therefore, from (1), (2) and (3), we obtain dy x −1 = ( log x ) 1 + log x.log ( log x )  + 2 x log x −1 log x dx

8.

Differentiate the functions with respect to x

( sin x )

x

+ sin −1 x

Solution: Let y = ( sin x ) + sin −1 x x

Also, let u = ( sin x ) and v = sin −1 x x

y =u+v



dy du dv = − dx dx dx

u = ( sin x )

......... (1)

x

 log u = log ( sin x )

x

 log u = x log (sin x ) Differentiating both sides with respect to x, we obtain 

1 du d d = ( x )  log ( sin x ) + x  log ( sin x )  u dx dx dx



du 1 d  = u 1.log ( sin x ) + x. ( sin x ) dx sin x dx  



du x x   = ( sin x ) log ( sin x ) + .cos x  dx sin x  



du x = ( sin x ) ( x cot x + log sin x ) dx

........... ( 2 )

v = sin −1 x

Differentiating both sides with respect to x, we obtain dv = dx

1 1−

( x)

2

d dx



dv 1 1 = dx 1− x 2 x



dv 1 = dx 2 x − x 2

( x)

......... ( 3)

Therefore, from (1), (2) and (3) we obtain dy 1 2 = ( sin x ) ( x cot x + log sin x ) + dx 2 x − x2

9.

Differentiate the function with respect to x.

xsin x + ( sin x )

cos x

Solution: Let y = xsin x + ( sin x )

cos x

Also u = xsin x and v = ( sin x )

cos x

y =u+v



dy du dv = + dx dx dx

....... (1)

u = xsin x

 log u = log ( xsin x )  log u = sin x log x

Differentiating both sides with respect to x, we obtain 1 du d d = ( sin x ) .log x + sin x ( log x ) u dx dx dx



du 1  = u = cos x log x + sin x.  dx x 



du sin x   = xsin x cos x log x + dx x  

v = ( sin x )

........... ( 2 )

cos x

 log v = log ( sin x )

cos x

 log v = cos x log (sin x ) Differentiating both sides with respect to x, we obtain 1 dv d d = ( cos x )  log ( sin x ) + cos x  log ( sin x )  v dx dx dx



dv 1 d  = v  − sin x.log ( sin x ) + cos x. (sin x ) dx sin x dx  



dv cos x cos x   = ( sin x ) − sin x.log sin x + cos x  dx sin x  



dv cos x = ( sin x )  − sin x log sin x + cot x cos x  dx



dv cos x = ( sin x )  cot x cos x − sin x log sin x  dx

...... ( 3 )

Therefore, from (1), (2) and (3) we obtain dy sin x  cos x  = xsin x  cos x log x +  + ( sin x ) cos x cot x − sin x log sin x  dx x  

10.

Differentiate the function with respect to x.

x x cos x +

x2 + 1 x2 −1

Solution: Let y = x x cos x +

x2 + 1 x2 −1

Also, let u = x x cos x and v =



x2 + 1 x2 −1

dy du dv = + dx dx dx

y =u+v

u = x x cos x

Differentiating both sides with respect to x, we obtain 1 du d d d = ( x ) cos x log x + x ( cos x ) log x + x cos x ( log x ) u dx dx dx dx



du 1  = u 1.cos x.log x + x. ( − sin x ) log x + x cos x.  dx x 



du = x x cos x ( cos x log x − x sin x log x + cos x ) dx



du = x x cos x cos x (1 + log x ) − x sin x log x  dx

v=

x2 + 1 x2 −1

.......... ( 2 )

 log v = log ( x2 + 1) − log ( x2 − 1) Differentiating both sides with respect to x, we obtain 1 dv 2x 2x = = 2 − 2 v dx x + 1 x − 1

 2 x ( x 2 − 1) − 2 x ( x 2 + 1)  dv   = v dx  ( x2 + 1)( x2 −1)   dv x 2 + 1  −4 x   = 2  2 dx x − 1  ( x + 1)( x 2 − 1)    

dv −4 x = dx ( x 2 − 1)2

.......... ( 3)

Therefore, from (1), (2) and (3) we obtain dy 4x = x x cos x cos x (1 + log x ) − x sin x log x  − 2 2 dx ( x − 1)

11.

Differentiate the function with respect to x.

1

( x cos x ) + ( x sin x ) x x

Solution: 1

Let y = ( x cos x ) + ( x sin x ) x x

1

Also, let u = ( x cos x ) and v = ( x sin x ) x x

y =u+v



dy du dv = + dx dx dx

u = ( x cos x )

....... (1)

2

 log u = log ( x cos x )

x

 log u = x log ( x cos x )  log u = x log x + logcos x  log u = x log x + x log cos x

Differentiating both sides with respect to x, we obtain 1 du d d = ( x + log x ) + ( x log cos x ) u dx dx dx



du  d d d d    = u  log x ( x ) + x ( log x )  + log cos x ( x ) + x ( log cos x )  dx dx dx dx dx    



du   1 d x   = ( x cos x )  log x.1 + x.  + log cos x.1 + x. . ( cos x )  dx x  cos x dx  



du   1 d x   = ( x cos x )  log x.1 + x.  + log cos x.1 + x. . ( cos x )  dx x  cos x dx  



du x  x   = ( x cos x ) ( log x + 1) + log cos x + ( − sin x ) dx cos x   



du x = ( x cos x ) (1 + log x ) + ( log cos x − x tan x )  dx



du x = ( x cos x ) 1 − x tan x + ( log x + log cos x )  dx



du x = ( x cos x ) 1 − x tan x + log ( x cos x )  dx 1

v = ( x sin x ) x

1

 log v = log ( x sin x ) x  log v =

1 log ( x sin x ) x

 log v =

1 ( log x + log sin x ) x

.......... ( 2 )

 log v =

1 1 log x + log sin x x x

Differentiating both sides with respect to x, we obtain 1 dv d  1  d 1  =  log x  +  log ( sin x )  v dx dx  x  dx  x 



1 dv  d 1 1 d   d 1 1 d  = log x.   + . ( log x )  + log ( sin x ) .   + . log ( sin x ) v dx  dx  x  x dx dx  x  x dx   



1 dv    1  1 1   1  1 1 d = log x.  − 2  + .  + log ( sin x ) .  − 2  + . . ( sin x )  v dx   x  x x   x  x sin x dx 



 log ( sin x )  1 dv 1 1 = 2 (1 − log x ) + − + .cos x  v dx x x2 x sin x  



1 1 − log x − log ( sin x ) + x cot x  1 dv 1 = 2 ( x sin x ) x +  +  2 v dx x x2  x 



1 1 − log x − log sin x + x cot x  ( ) dv = ( x sin x ) x   dx x2  



1 1 − log x sin x + x cot x  ( ) dv = ( x sin x ) x   2 dx x  

..... ( 3)

Therefore, from (1), (2) and (3), we obtain 1  x cot x + 1 − log x sin x  ( ) dy 2 = ( x cos x ) 1 − x tan x + log ( x cos x ) + ( x sin x ) x   2 dx x  

12.

Find

dy of function. dx

xy + yx = 1

Solution: The given function is x y + y x = 1 Let x y = u and y x = v

Then, the function becomes u + v = 1 

du dv + =0 dx dx

......... (1)

u = xy

 log u = log ( x y )  log u = y log x

Differentiating both sides with respect to x, we obtain  du dy d = log x + y. ( log x ) u dx dx dx



du dy 1  = u log x + y.  dx dx x 



du dy y   = x y  log x +  dx dx x  

........ ( 2 )

v = yx

 log v = log ( y x )  log v = x log y

Differentiating both sides with respect to x, we obtain 1 dv d d = log y ( x ) + x. ( log y ) v dx dx dx



 dv 1 d  = v  log y.1 + x. .  dx y dx  



 dv x dy  = y x  log y +  dx y dx  

............... ( 3 )

Therefore, from (1), (2) and (3) we obtain

dy  x y  log x + + dx 

y x dy  x =0  + y  log y + x y dx  

 ( x 2 + log x + xy y −1 )



13.

dy = − ( yx y −1 + y x log y ) dx

dy yx y −1 + y x log y =− y dx x log x + xy x −1

Find

dy of function y x = x y dx

Solution: The given function is y x = x y Taking logarithm on both sides, we obtain x log y = y log x

Differentiating both sides with respect to x, we obtain log y.

d d d d ( x ) + x. ( log y ) = log x. ( y ) + y. ( log x ) dx dx dx dx

1 dy dy 1  log y.1 + x. . = log x. + y. y dx dx x  log y +

x dy dy y = log x + y dx dx x

x  dy y   − log x  = − log y y  dx x  x − y log x  dy y − x log y  =  y x   dx  x − y log x  dy y − x log y  =  y x   dx 

dy y  y − x log y  =   dx x  x − y log x 

14.

Find

y x dy of friction ( cos x ) = ( cos y ) dx

Solution: The given function is ( cos x ) = ( cos y ) y

x

Taking logarithm on both sides, we obtain y = log cos x = x log cos y

Differentiating both sides with respect to x, we obtain log cos x.

dy d d d + y. ( log cos x ) = log cos y. ( x ) + x. ( log cos y ) dx dx dx dx

 log cos x.

dy 1 d 1 d + y. . ( cos x ) = log cos y.1 + x. ( cos y ) dx cos x dx cos y dx

 log cos x

dy y x dy + ( − sin x ) = log cos y + ( − sin y ) dx cos x cos y dx

 log cos x

dy dy − y tan x = log cos y − x tan y dx dx

 ( log cos x + x tan y )



15.

dy = y tan x + log cos y dx

dy y tan x + log cos y = dx x tan y + log cos x

Find

dy of function xy = e( x − y ) dx

Solution: The given function is xy = e( x − y ) Taking logarithm on both sides, we obtain

log ( xy ) = log ( e x− y )

 log x + log y = ( x − y ) log e

 log x + log y = ( x − y ) 1  log x + log y = x − y

Differentiating both sides with respect to x, we obtain d d d dy ( log x ) + ( log y ) = ( x ) − dx dx dx dx



1 1 dy 1 + = 1− x y dx x

 1  dy x − 1  1 +  = y  dx x 



16.

dy y ( x − 1) = dx x ( y + 1)

(

)(

)(

)

Find the derivative of the function given by f ( x ) = (1 − x ) 1 + x 2 1 + x 4 1 + x8 and hence find f ' (1)

Solution:

(

)(

)(

The given relationship is f ( x ) = (1 − x ) 1 + x 2 1 + x 4 1 + x8

)

Taking logarithm on both sides, we obtain

log f ( x ) = log (1 + x ) + log (1 + x2 ) + log (1 + x4 ) + log (1 + x8 ) Differentiating both sides with respect to x, we obtain

1 d d d d d .  f ( x )  = log (1 + x ) + log (1 + x 2 ) + log (1 + x 4 ) + log (1 + x8 ) f ( x ) dx dx dx dx dx 

1 1 d 1 d 1 d 1 d . f '( x) = . (1 + x ) + . (1 + x 2 ) + . (1 + x 4 ) + . (1 + x8 ) 2 4 8 f ( x) 1 + x dx 1 + x dx 1 + x dx 1 + x dx

1 1 1  1   f '( x) = f ( x)  + .2 x + .4 x3 + .8 x 7  2 4 8 1+ x 1+ x 1 + x 1 + x 

 1 2x 4 x3 8x7   f ' ( x ) = (1 + x ) (1 + x 2 )(1 + x 4 )(1 + x8 )  + + + 2 4 8 1 + x 1 + x 1 + x 1 + x 

 1 2 1 4 13 8 17  + + + Hence, f ' (1) = (1 + 1) (1 + 1 )(1 + 1 )(1 + 1 )  2 4 8  1 + 1 1 + 1 1 + 1 1 + 1  2

4

8

1 2 4 8 = 2 2 2 2  + + +  2 2 2 2  1+ 2 + 4 + 8  = 16    2   = 16 

17.

15 = 120 2

(

)(

)

Differentiate x 2 − 5x + 8 x3 + 7 x + 9 in three easy mentioned below i. By using product rule ii. By expanding the product to obtain a single polynomial iii. By logarithm differentiate Do they all given the same answer?

Solution:

(

)(

Let y = x 2 − 5x + 8 x3 + 7 x + 9

)

(i) Let x = x 2 − 5 x + 8 and u = x3 + 7 x + 9  y = uv



dy du dv = v+u dx dv dx



dy d 2 d = ( x − 5 x + 8) . ( x3 + 7 x + 9 ) + ( x 2 + 5 x + 8) . ( x3 + 7 x + 9 ) dx dx dx



dy = ( 2 x − 5 ) ( x 3 + 7 x + 9 ) + ( x 2 − 5 x + 8 )( 3 x 2 + 7 ) dx



dy = 2 x ( x3 + 7 x + 9 ) − 5 ( x3 + 7 x + 9 ) + x 2 ( 3x 2 + 7 ) − 5 x (3x 2 + 7 ) − 8 (3x 2 + 7 ) dx

(By using product rule)





dy = ( 2 x 4 + 14 x 2 + 18 x ) − 5 x 3 − 35 x − 45 + ( 3 x 2 + 7 x 2 ) − 15 x 3 − 35 x + 24 x 2 + 56 dx

dy = 5 x 4 − 20 x 3 + 45 x 2 − 52 x + 11 dx

(

)(

(ii) y = x 2 − 5x + 8 x3 + 7 x + 9

)

= x 2 ( x3 + 7 x + 9 ) − 5 x ( x3 + 7 x + 9 ) + 8 ( x3 + 7 x + 9 ) = x5 + 7 x3 + 9x2 − 5x4 − 35x2 − 45x + 8x3 + 56x + 72 = x5 − 5x4 + 15x3 − 26x2 + 11x + 72 

dy d 5 = ( x − 5 x 4 + 15 x 3 − 26 x 2 + 11x + 72 ) dx dx



dy d 5 d d d d d = ( x ) − 5 ( x 4 ) + 15 ( x 3 ) − 26 ( x 2 ) + 11 ( x ) + ( 72 ) dx dx dx dx dx dx dx

= 5x4 − 5  4 x3 + 15  3x2 − 26  2 x + 111 + 0 = 5x4 − 20 x3 + 45x2 − 52 x + 11

(iii) Taking logarithm on both sides, we obtain

log y = log ( x2 − 5x + 8) + log ( x3 + 7 x + 9 ) Differentiating both sides with respect to x, we obtain

1 dy d d = log ( x 2 − 5 x + 8) + log ( x3 + 7 x + 9 ) y dx dx dx 

1 dy 1 d 1 d = 2 . ( x 2 − 5 x + 8) + 3 . ( x3 + 7 x + 9 ) y dx x − 5x + 8 dx x + 7 x + 9 dx



dy 1 1   = y 2  ( 2 x − 5) + 3  ( 3x 2 + 7 )  dx x + 7x + 9  x − 5x + 8 



 2x − 5 dy 3x 2 + 7  = ( x 2 − 5 x + 8)( x3 + 7 x + 9 )  3 + 3  dx  x − 5x + 8 x + 7 x + 9 

 2 x − 5 ( x3 + 7 x + 9 ) + ( 3x 2 + 7 )( x 2 − 5 x + 8)  dy 2 3   = ( x − 5 x + 8)( x + 7 x + 9 )  dx   ( x3 − 5 x + 8) + ( x3 + 7 x + 9 ) 

dy = 2 x ( x3 + 7 x + 9 ) − 5 ( x3 + 7 + 9 ) + 3x 2 ( x 2 − 5 x + 8 ) + 7 ( x 2 − 5 x + 8 ) dx



dy = ( 2 x 5 + 14 x 2 + 18 x ) − 5 x 3 − 35 x − 45 + ( 3 x 4 − 15 x 3 + 24 x 2 ) + ( 7 x 2 − 35 x + 56 ) dx



dy = 5 x 2 − 20 x 3 + 45 x 2 − 52 x + 11 dx

From the above three observations, it can be concluded that all the result of

18.

If u, v and w are function of x, then show that

dy are same dx

d du dv dw ( u.v.w ) = v.w + u .w + u .v . dx dx dx dx

In two ways – first by repeated application of product rule, second by logarithmic differentiation.

Solution: Let y = u.v.w = u. ( v.w) By applying product rule, we obtain dy du d = . ( v.w ) + u. ( v.w ) dx dx dx



dy du dv   dv = v.w + u  .w + v.  dx dx dx   dx



dy du dv dw = v.w + u. .w + u.v dx dx dx dx

(Again applying product rule)

By taking logarithm on both sides of the equation y = u.v.w, we obtain log y = log u + log v + log w

Differentiating both sides with respect to x, we obtain

1 dy d d d . = ( log u ) + ( log v ) + ( log w) y dx dx dx dx

1 dy 1 du 1 dv 1 dw  . = + + y dx u dx v dx w dx 

dy  1 du 1 dv 1 dw  = y + +  dx  u dx v dx w dx 



dy  1 du 1 dv 1 dw  = u.v.w  + +  dx  u dx v dx w dx 



dy du dv dw = .v.w + u. .w + u.v dx dx dx dx

Exercise 5.6 1.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy , x = 2at 2 , y = at 4 dx

Solution: The given equation are x = 2at 2 and y = at 4 Then,

dx d d = ( 2at 2 ) = 2a. ( t 2 ) = 2a.2t = 4at dt dt dt dy d d = ( at 4 ) a. ( t 4 ) = a.4.t 3 = 4at 3 dx dt dt  dy  dy  dt  4at 3 2  = = =t dx  dx  4at    dt  2.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

x = a cos , y = b cos Solution: The given equations are x = a cos  and y = b cos Then,

dx d = ( a cos  ) = a ( − sin  ) = −a sin  d d

dy d = ( b cos  ) = b ( − sin  ) = −b sin  d d  dy  dy  d  −b sin  b   = = dx  dx  −a sin  a    d 

3.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

x = sin t , y = cos 2t Solution: The given equations are x = sin t and y = cos 2t Then,

dx d = ( sin t ) = cos t dt dt

dy d d = ( cos 2t ) = − sin 2t. ( 2t ) = −2sin 2t dt dt dt  dy  dy  dx  −2sin 2t −2.2sin t cos t  = = = = −4sin t dx  dx  cos t cos t    dt  4.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

x = 4t , y =

dy dx

4 t

Solution: The equation are x = 4t and y =

4 t

dx d = ( 4t ) = 4 dt dt dy d  4  d 1  −1  −4 =   = 4.   = 4.  2  = 2 dt dt  t  dt  t  t  t

 dy   −4  dy  dt   t 2  −1  = = = 2 dx  dx  4 t    dt 

5.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

x = cos − cos 2 , y = sin  − sin 2 Solution: The given equations are x = cos  − cos 2 and y = sin  − sin 2 Then,

dx d d d = ( cos  − cos 2 ) = ( cos  ) − ( cos 2 ) d d d d

= − sin  ( −2sin 2 ) = 2sin 2 − sin  dy d d d = ( sin  − sin 2 ) = ( sin  ) − ( sin 2 ) d d d d = cos  − 2 cos 

 dy  dy  d  cos  − 2cos   = = dx  dx  2sin 2 − sin     d  6.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

x = a ( − sin  ) , y = a (1 + cos ) Solution: The given equations are x = a ( − sin  ) and y = a (1 + cos ) Then,

dx d d = a  ( ) − ( sin  ) = a (1 − cos  ) d d  d 

dy d d = a  (1) + ( cos  ) = a 0 + ( − sin  ) = −a sin  d d  d 

 dy     −2sin cos − cos dy  d  −a sin  2 2= 2 = − cot   = = =   dx  dx  a (1 − cos  ) 2 2sin 2 sin   2 2  d  7.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

x=

dy dx

sin 3 t cos3 t ,y= cos x 2t cos 2t

Solution: The given equations are x =

sin 3 t cos3 t and y = cos x 2t cos 2t

dx d  sin 3 t  = Then, dt dt  cos 2t 

=

cos 2t −

d d sin 3 t ) − sin 3 t − cos 2t ( dt dt cos 2t

cos 2t .3sin 2 t. =

d 1 d . ( cos 2t ) ( sin t ) − sin 3 tx dt 2 cos 2t dt cos 2t

sin 3 t . ( −2sin 2t ) 2 cos 2t cos 2t cos 2t

3 cos 2t .sin 2 t cos t − =

=

3cos 2t sin 2 t cot t + sin 2 t sin 2t cos 2t cos 2t

dy d  cos3 t  = dt dt  cos 2t 

=

cos 2t .

d d cos3 t ) − cos3 t. ( dt dt cos 2t

(

cos 2t

)

cos 2t .3cos 2 t. =

=

=

d 1 d . ( cos 2t ) ( cos t ) − cos3 t. dt 2 cos 2t dt cos 2t

3 cos 2t cos 2 t. ( − sin t ) − cos3 t.

1 ( −2sin 2t ) cos 2t

cos 2t

−3cos 2t.cos 2 t.sin t + cos3 t sin 2t cos 2t. cos 2t

 dy  dy  dx  −3cos 2t.cos 2 t + cos3 t sin 2t   = dx  dx  3cos 2t sin 2 t cos t + sin 3 t sin 2t    dt 

=

=

3cos 2t.cos 2 t sin t + cos3 t ( 2sin t cos t ) 3cos 2 tsin 2 t.cos t + sin 3 t ( 2sin t cos t ) sin t cos t  −3cos 2t.cos t + 2 cos 3 t  sin t cos t 3cos 2t sin t + 2sin 3 t 

 −3 ( 2cos 2 t − 1) cos t + 2cos3 t   = 3 (1 − 2sin 2 t ) sin t + 2sin 3 t    −4 cos3 t + 3cos t = 3sin t − 4sin 3 t

=

cos 2t = ( 2cos 2 t − 1)   cos 2t = (1 − 2sin 2 t )   

cos 3t = 4 cos3 t − 3cos t    2 sin 3t = 3sin t − 4sin t 

− cos 3t sin 3t

= − cot 3t

8.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

t  x = a  cost + log  , y = a sin t 2 

Solution:

t  The given equations are x = a  cost + log  and y = a sin t 2 

Then,

d d d  t  = a  ( cos t ) +  log tan   dx dt  2   dt

   1 d t  = a  − sin t + . tan   t dt  2   tan  2   t t d  t  = a  − sin t + cot .sec 2 .    2 2 dt  2   

t   cot   1 1 2 = a  − sin t +   t t 2   sin cos 2  2 2      1 = a  − sin t + t t  2sin cos   2 2 1   = a  − sin t +  sin t  

 − sin 2 t + 1  = a   sin t  =a

cos 2 t sin t

dy d = a ( sin t ) = a cos t dt dt  dy  dy  dt  a cos t sin t  = = = = tan t 2 dx  dx   cos t  cos t   a  dt   sin t 

9.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

x = a sec, y = b tan  Solution: The given equations are x = a sec and y = b tan  Then,

dx d = a. ( sec  ) = a sec  tan  d d

dy d = b. ( tan  ) = b sec 2  d d  dy  dy  d  b sec2  b b cos  b 1 b = = = sec cot  = =  = cos ec dx  dx  a sec  tan  a a cos  sin  a sin  a    d  10.

If x and y are connected parametrically by the equation, without eliminating the parameter, find

dy dx

x = a ( cos +  sin  ) , y = a (sin  − cos ) Solution: The given equations are x = a ( cos +  sin  ) and y = a (sin  −  cos ) Then,

dx d d d d = a  cos  + ( sin  ) = a − sin  +  ( sin  ) + sin  ( ) d d d d  d   

= a − sin  +  cos + sin   = a cos dx d  d d  d = a  ( sin  ) − ( cos  ) = a cos  −  ( cos  ) + cos  . ( ) d d d  d   d  

= a cos +  sin  − cos  = a sin 

 dy  dy  d  a sin   = = = tan  dx  dx  a sin     d  11.

If x = asin −1t , y = acos −1t , show that

dy y =− dx x

Solution: The given equation are x = asin −1t and y = acos −1t

x = asin −1t and y = acos −1t 1

 x = ( asin −1t ) and y = ( a cos −1t ) 2 x=a

1 sin −1t 2

1

and y = a 2 1

Consider x = a 2

cos −1t

sin −1t

Taking logarithm on both sides, we obtain

1 log x = sin −1 t log a 2 

1 dx 1 d = log a. ( sin −1 t ) x dt 2 dt



dx x 1 = log a. dt 2 1− t2



dx x log a = dt 2 1 − t 2

Then, consider

y=a

1 cos−1 t 2

Taking logarithm on both sides, we obtain

1 log y = cos −1 t log a 2



1 dy 1 d = log a. ( cos −1 t ) y dx 2 dt



dy y log a  −1  =   dt 2  1− t2 



dy − y log a = dt 2 1 − t 2

 dy   − y log a    dy  dx   2 1 − t 2  y  = = =− x log a dx  dx  x   2  dt  2 1− t

Hence proved

Exercise 5.7 1.

Find the second order derivatives of the function x2 + 3x + 2

Solution: Let y = x 2 + 3x + 2 Then,

dy d 2 d d = ( x ) + ( 3x ) + ( 2 ) = 2 x + 3 + 0 = 2 x + 3 dx dx dx dx

d2y d d d  2 = ( 2 x + 3) = ( 2 x ) + ( 3) = 2 + 0 = 2 dx dx dx dx 2.

Find the second order derivative of the function x 20

Solution: Let y = x 20 Then,

dy d 20 = ( x ) = 20 x19 dx dx

d2y d d  2 = ( 20 x19 ) = 20 ( x19 ) = 20.19.x18 = 380 x18 dx dx dx 3.

Find the second order derivatives of the function x.cos x

Solution: Let y = x.cos x Then,

dy d d d = ( x.cos x ) = cos x. ( x ) + x ( cos x ) = cos x.1 + x ( − sin x ) = cos x − x sin x dx dx dx dx



d2y d d d = cos x − sin x  = ( cos x ) − ( x sin x ) 2 dx dx dx dx

d d   = − sin x − sin x ( x ) + x ( sin x )  dx dx  

= − sin x − (sin x + cos x ) = − ( x cos x + 2sin x ) 4.

Find the second order derivatives of the function log x

Solution: Let y = log x Then,

dy d 1 = ( log x ) = dx dx x



5.

d 2 y d  1  −1 =  = dx 2 dx  x  x 2

Find the second order derivatives of the function x3 log x

Solution: Let y = x3 log x Then,

dy d 3 d d =  x log x  = log x. ( x 3 ) + x 3 ( log x ) dx dx dx dx 1 = log x.3 x 2 + x 3 . = log x.3 x 2 + x 2 x

= x2 (1 + 3log x ) 

d2y d 2 =  x (1 + 3log x ) dx 2 dx 

= (1 + 3log x ) .

d 2 d x ) + x 2 (1 + 3log x ) ( dx dx

= (1 + 3log x ) .2 x + x 3 .

3 x

= 2 x + 6log x + 3x = 5x + 6 x log x = x (5 + 6log x ) 6.

Find the second order derivative of the function ex sin5x

Solution: Let y = e x sin 5 x

dy d x d d = ( e sin 5 x ) = sin 5 x ( e x ) + e x ( sin 5 x ) dx dx dx dx = sin 5 x.e x + e x .cos 5 x.

d ( 5 x ) = e x sin 5 x + e x cos 5 x.5 dx

= ex (sin5x + 5cos5x ) 

d2y d x = e ( sin 5 x + 5cos5 x ) dx 2 dx 

( sin 5 x + 5cos 5 x )

d x d e ) + e x . ( sin 5 x + 5cos 5 x ) ( dx dx

( sin 5x + 5cos 5x ) e x + e x cos 5x. ( 5x ) + 5 ( − sin 5x ) . 

d dx

d (5x ) dx 

= ex (sin5x + 5cos5x ) + ex (5cos5x − 25sin5x ) Thus, ex (10cos5x − 24sin5x ) = 2ex (5cos5x −12sin5x ) 7.

Find the second order derivatives of the function e6 x cos3x

Solution: Let y = e6 x cos 3x

Then,

dy d 6 x d d = ( e cos 3 x ) = cos 3 x. ( e6 x ) + e6 x ( cos 3 x ) dx dx dx dx = cos 3 x.e6 x

d d ( 6 x ) + e6 x ( − sin 3x ) ( 3x ) dx dx

= 6e6 x cos3x − 3e6 x sin3x.........(1) 

d2y d d d = ( 6e6 x cos3x − 3e6 x sin 3x ) = 6. ( e6 x cos3x ) − 3. ( e6 x sin 3x ) 2 dx dx dx dx

d d   = 6. 6e6 x cos 3x − 3e6 x sin 3x  − 3 sin 3x. ( e6 x ) + e6 x . ( sin 3 x )  dx dx  

[using (1)]

= 36e6 x cos3x − 18e6 x sin 3x − 3 sin 3x.e6 x .6 + e6 x .cos3x − 3

= 36e6 x cos3x −18e6 x sin 3x −18e6 x sin 3x − 9e6 x cos3x

= 27e6 x cos3x − 36e6 x sin 3x = 9e6 x (3cos3x − 4sin3x ) 8.

Find the second order derivatives of the function tan −1 x

Solution: Let y = tan −1 x Then,

dy d 1 = ( tan −1 x ) = dx dx 1 + x2 

9.

−1 −2 d d2y d  1  d 1 2x =  = (1 + x 2 ) = ( −1) (1 + x 2 ) 1 + x2 ) −  2x = − ( 2 2 2 2  dx dx  1 + x  dx dx (1 + x 2 ) (1 + x 2 )

Find the second order derivative of the function log ( log x )

Solution: Let y = log ( log x ) Then,

dy d 1 d 1 −1 = log ( log x )  = . ( log x ) = = ( x log x ) dx dx log x dx log x 

10.

d2y d  −1 −2 d = x log x )  = ( −1)( x log x ) ( ( x log x ) 2  dx dx  dx

−1

=

d d   . log x. ( x ) + x. ( log x ) dx dx  ( x log x ) 

=

1  −1(1 + log x )  . log x.1x +  = 2 x ( x log x )  ( x log x )

2

−1

2

Find the second order derivatives of the function sin ( log x )

Solution: Let y = sin ( log x ) Then,

cos ( log x ) dy d d = sin ( log x ) = cos ( log x ) . ( log x ) = dx dx dx x



=

d 2 y d  cos ( log x )  =   dx 2 dx  x  x.

d d cos ( log x )  − cos ( log x ) . ( x ) dx dx x2

d   x  − sin ( log x ) . ( log x ) − cos ( log x ) .1 dx  =  2 x

1 − x sin ( log x ) . − cos ( log x ) x = x2

=

11.

− sin ( log x ) + cos ( log x )  x2

d2y If y = 5cos x − 3sin x , prove that + y=0 dx 2

Solution: It is given that, y = 5cos x − 3sin x Then,

dy d d d d = ( 5cos x ) − ( 3sin x ) = 5 ( cos x ) − 3 ( sin x ) dx dx dx dx dx

= 5 ( − sin x ) − 3cos x = − (5sin x + 3cos x ) 

d2y d = − ( 5sin x + 3cos x )  dx 2 dx 

d  d  = − 5. ( sin x ) + 3. ( cos x )  dx  dx  = 5cos x + 3 ( − sin x )

= − 5cos x − 3sin x = −y



d2y +y=0 dx 2

Hence, proved 12.

If y = cos −1 x , find

Solution:

d2y in terms of y alone dx 2

It is given that, y = cos −1 x Then, −1 dy d −1 = ( cos −1 x ) = = − (1 − x 2 ) 2 dx dx 1 − x2

−1 d2y d  2 2  = − 1 − x )   ( dx 2 dx   −3 d  1 =  −  . (1 − x 2 ) 2 . (1 − x 2 ) dx  2

1

=

(1 − x2 )

3

d2y  2 = dx

 ( −2 x )

−x

(1 − x2 )

......... ( i )

3

y = cos −1 x  x = cos y

Putting x = cos y in equation (1), we obtain

d2y = dx 2

− cos y

(1 − cos y )

d2y  2 = dx

2

3

− cos y

( sin y ) 2

3

− cos y sin 3 y =

− cos y 1  2 sin y sin y



d2y = cot y.cos ec 2 y 2 dx

13.

If y = 3cos ( log x ) + 4sin ( log x ) , show that x 2 y2 + xy1 + y = 0

Solution: It is given that y = 3cos ( log x ) + 4sin ( log x ) and x 2 y2 + xy1 + y = 0 Then,

y1 = 3.

d d cos ( log x )  + 4. sin ( log x )  dx dx

d d     = 3.  − sin ( log x ) . ( log x )  + 4. cos ( log x ) . ( log x ) dx dx    

 y1 =

−3sin ( log x ) 4cos ( log x ) 4cos ( log x ) − 3sin ( log x ) + = x x x

 y2 =

d  4cos ( log x ) − 3sin ( log x )    dx  x 

4cos ( log x ) − −3sin ( log x ) '− 4cos ( log x ) − 3sin ( log x ) .1 =x x2

 −4sin ( log x ) . ( log x ) '− 3cos ( log x )( log x ) ' − 4cos ( log x ) + 3sin ( log x ) =x x2

1 1   −4sin ( logx ) x − 3cos ( logx ) x  − 4cos ( log x ) + 3sin ( log x ) =x x2 =

4sin ( log x ) − 3cos ( log x ) − 4cos ( log x ) + 3sin ( log x ) x2

=

− sin ( log x ) − 7 cos ( log x ) x2

 x 2 y2 + xy1 + y

 − sin ( log x ) − 7 cos ( log x )   4cos ( log x ) − 3sin ( log x )  = x2   + x  + 3cos ( log x ) + 4sin ( log x ) x2 x    

= sin ( log x ) − 7cos ( log x ) + 4cos ( log x ) − 3sin ( log x ) + 3cos ( log x ) + 4sin (log x ) =0 Hence, proved 14.

If y = Aemx + Benx , show that

d2y dy − ( m + n ) + mny = 0 2 dx dx

Solution: It is given that, y = Aemx + Benx Then,

dy d d d d = A. ( e mx ) + B. ( e nx ) = A.e mx . ( mx ) + B.e nx ( nx ) = Ame mx + Bne nx dx dx dx dx dx

d2y d d d = ( Amemx + Bnenx ) = Am. ( emx ) + Bn. ( enx ) 2 dx dx dx dx = Am.e mx .



d d ( mx ) + bn.e nx . ( nx ) = Am 2e mx + Bn 2e nx dx dx

d2y dy − ( m + n ) + mny 2 dx dx

= Am2emx + Bn2enx − ( m + n ) . ( Amemx + Bnenx ) + mn ( Aemx + Bemx )

= Am2emx + Bn2enx − Am2exmx − Bmnemx − Bn2emx + Amnemx + Bmnenx =0 Hence, proved 15.

If y = 500e7 x + 600e −7 x ,show that

d2y = 49 y dx 2

Solution: It is given that, y = 500e7 x + 600e −7 x Then,

dy d d = 500. ( e7 x ) + 600. ( e −7 x ) dx dx dx = 500.e7 x .

d d ( 7 x ) + 600.e−7 x . ( −7 x ) dx dx

= 3500e7 x − 4200e−7 x 

d2y d d = 3500. ( e7 x ) − 4200 ( e−7 x ) 2 dx dx dx

3500.e7 x

d d ( 7 x ) − 4200.e−7 x . ( −7 x ) dx dx

= 7  3500.e7 x + 7  4200e−7 x

= 49  500e7 x + 49  600e−7 x = 49 ( 500e7 x + 600e−7 x )

= 49y Hence, proved 16.

d 2 y  dy  If e ( x + 1) = 1 , show that =  dx 2  dx 

2

y

Solution: The given relationship is e y ( x + 1) = 1

e y ( x + 1) = 1  ey =

1 x +1

Taking logarithm on both sides, we obtain

y = log

1 ( x + 1)

Differentiating this relationship with respect to x we obtain

dy d  1  −1 −1 = ( x + 1)  =  = ( x + 1) 2 dx dx  ( x + 1)  ( x + 1) x + 1

 −1  d2y d  1  1  2 =  =  = −  2 2  dx dx  x + 1   ( x + 1)  ( x + 1)

d 2 y  −1   2 =  dx  x +1  d 2 y  dy   2 =  dx  dx 

2

2

Hence, proved 17.

If y = ( tan −1 x ) , show that ( x 2 + 1) y2 + 2 x ( x 2 + 1) y1 = 2 2

2

Solution: The given relationship is y = ( tan −1 x )

2

Then,

y1 = 2 tan −1 x

d tan −1 x ) ( dx

 (1 + x 2 ) y1 = 2 tan −1 x

Again differentiating with respect to x on both sides, we obtain

(1 + x ) y 2

2

 1  + 2 xy1 = 2  2   1+ x 

 (1 + x2 ) y2 + 2 x (1 + x 2 ) = 2

Hence proved

Exercise 5.8

1.

Differentiate the following with respect to x.

cos x.cos2x.cos3x Solution-1 Let y = cos x.cos 2 x.cos3x Taking logarithm or both the side, we obtain

log y = log ( cos x.cos2 x.cos3x )  log y = log ( cos x ) + log ( cos2 x ) + log ( cos3x )

Differentiating both sides with respect to x, we obtain

1 dy 1 d 1 d 1 d = . ( cos x ) + . ( cos2 x ) + . ( cos3x ) y dx cos x dx cos2 x dx cos3x dx dy sin3x d  sin x sin 2 x d   = y − − . (2 x ) − . ( 3x )  dx cos3 x dx  cos x cos2 x dx  dy  = − cos x.cos 2 x.cos3 x  tan x + 2 tan 2 x + 3tan 3 x  dx

2.

Differentiate the function with respect to x.

( x − 1)( x − 2 ) ( x − 3)( x − 4 )( x − 5)

Solution-2 Let y =

( x − 1)( x − 2 ) ( x − 3)( x − 4 )( x − 5)

Taking logarithm or both the side, we obtain

log y = log

( x−1)( x−2) x − ( 3)( x−4)( x−5)

 1  log y = log  2 

( x − 1)( x − 2 )   ( x − 3)( x − 4 )( x − 5) 

1  log y = log ( x − 1)( x − 2 ) − log ( x − 3)( x − 4 )( x − 5 ) 2

1  log y = log ( x − 1) + log ( x − 2 ) − log ( x − 3) − log ( x − 4 ) − log ( x − 5 )  2 Differentiating both sides with respect to, we obtain

1 d 1 d  1 d  . x − 1 + . x − 2 − . x − 3 ( ) ( ) ( )  1 dy 1  x − 1 dx x − 2 dx x − 3 dx =   1 d 1 d y dx 2   − . ( x − 4) − . ( x − 5)   1 − 4 dx x − 5 dx dy y  1 1 1 1 1   =  + − − −  dx 2  x − 1 x − 2 x − 3 1 − 4 x − 5 



3.

( x − 1)( x − 2 )  1 + 1 − 1 − 1 − 1  ( x − 3)( x − 4 )( x − 5)  x − 1 x − 2 x − 3 1 − 4 x − 5 

dy 1 = dx 2

Differentiate the function with respect to x.

( log x )

cos x

Solution-3 Let y = ( log x )

cos x

Taking logarithm or both the side, we obtain

log y = cos x.log ( log x )

Differentiating both sides with respect to x, we obtain

1 dy d d . = ( cos x )  log ( log x ) + cos x  log ( log x ) y dx dx dx 1 dy 1 d  . = − sin x log ( log x ) + cos x  . ( log x ) y dx log x dx  dy cos x 1   = y − sin x log ( log x ) +   dx log x x 



 dy cos x  cos x = ( log x )  − sin x log ( log x ) dx  log x 

.

4.

Differentiate the function with respect to x. x − 2 x

Solution-4 Let y = x − 2 x

sin x

Also, let x = u and 2 x

sin x

=v

sin x

y = u − v dy du dv  = − dx dx dx u = xx Taking logarithm on both sides, we obtain

1 du  d d  =  ( x )  log x + x  ( log x ) u dx  dx dx  du 1   = u 1 log x + x   dx x  du  = x x ( log x + 1) dx du  = x x (1 + log x ) dx v = 2sin x Taking logarithm on both the sides with respect to x, we obtain

log v = sin x.log 2

Differentiating both sides with respect to x, we obtain

1 dv d . = log 2. ( sin x ) v dx dx dv  = v log 2cos x dx dv  = 2sin x cos x log 2 dx dy  = x 2 (1 + log x ) − 2sin x cos x log 2 dx 5.

Differentiate the function with respect to x.

( x + 3) .( x + 4 ) . ( x + 5 ) 2

3

4

Solution-5 Let y = ( x + 3) .( x + 4 ) .( x + 5 ) 2

3

4

Taking logarithm on both sides, we obtain.

log y = log ( x + 3) + log ( x + 4 ) + log ( x + 5 ) 2

3

4

 log y = 2log ( x + 3) + 3log ( x + 4) + 4log ( x + 5) Differentiating both sides with respect to x, we obtain



dy 2 3 = ( x + 3)( x + 4 ) ( x + 5 ) .  2 ( x 2 + 9 x + 20 ) + 3 ( x 2 + 9 x + 15 ) + 4 ( x 2 + 7 x + 12 )  dx

dy 3 4   2 = y + + dx  x + 3 x + 4 x + 5  dy 3 4  2 3 4 2  = ( x + 3) ( x + 4 ) ( x + 5 )  + + dx  x + 3 x + 4 x + 5 





dy 3 4  2 3 4  2 ( x + 4 )( x + 5 ) + 3 ( x + 3)( x + 5 ) + 4 ( x + 3)( x + 4 ) = ( x + 3) ( x + 4 ) ( x + 5 ) .  + +  dx x + 4 x + 5 ( x + 3)( x + 4 )( x + 5) 



dy 2 3 = ( x + 3)( x + 4 ) ( x + 5 ) .  2 ( x 2 + 9 x + 20 ) + 3 ( x 2 + 9 x + 15 ) + 4 ( x 2 + 7 x + 12 )  dx



dy 2 3 = ( x + 3)( x + 4 ) ( x + 5 ) ( 9 x 2 + 70 x + 133) dx x

6.



1

1+  1  Differentiate the function with respect to x.  x +  + x x  x 

Solutoin-6 x



1

1+  1  Let y =  x +  + x x  x  

x

1

1+  1  x Also, let u =  x +  and v = x   x 

y =u+v

dy du dv = + dx dx dx x 1  Then, u =  x +  x  

................ (1)

Taking log on both sides x

1   log u = log  x +  x  1   log u = x log  x +  x  Differentiating both sides with respect to x, we obtain

1 du d 1 d   1   = ( x )  ( log )  x +  + x  log  x +   u dx dx x dx   x   1 du 1 1 d  1   = 1 ( log )  x +  + x  . x+  1  dx  u dx x x   x+  x     du 1 x 1     = u ( log )  x +  + x x + 2  1   dx x  x    x+     x  x du  1  1  x2 − 1   =  x +  ( log )  x +  + 2  dx  x  x  x + 1 

du  1   x2 − 1 1    =  x +  2 + ( log )  x +   dx  x   x +1 x   v=x

............(2)

 1 1+   x

Taking log on both sides, we obtain

log v = log x

 1 1+   x

 1  log v = 1 +  log x  x Differentiating both sides with respect to x, we obtain

1 dv  d  1    1 d =  1 +    log x + 1 +  . log x v dx  dx  x  x  dx  1 dv  1  1 1   =  − 2  log x +  x +  . v dx  x  x x 

1 dv log x 1 1 =− 2 + + 2 v dx x x x dv  − log x + x + 1  = v  dx x2  



1

1+   x + 1 − log x  dv  = x x    dx x2 

.......... (3)

Therefore from (1) , (2) and (3), we obtain 

1

dy  1   x 2 − 1 1    x+   x + 1 − log x   = 1 +   2 + log x  x +   + x x    dx  x   x x  x2    x

7.

Differentiating both sides with respect to x.

( log x )

x

+ x log x

Solution-7 Let y = ( log x ) + x x

log x

Also, let u = ( log x ) and v = x x

log x

y = u + v

dy du dv = + dx dx dx x u = ( log x )



............(1)

 log u = log ( log x )     log u = x log ( log x ) x

Differentiating both sides with respect to x, we obtain

1 du d d = ( x ) x log ( log x ) + x. log ( log x )  u dx dx dx   du 1 d  = u 1x log ( log x ) + x. . ( log x ) dx log x dx  



du x 1 x = ( log x ) log ( log x ) + . dx log x x  



du 1  x = ( log x ) log ( log x ) + dx log x  



du x  log ( log x ) .log x + 1  = ( log x )   dx log x  

du x −1 = ( log x ) 1 + log x.log ( log x )  dx v = x log x  log v = log x log x

(

.......(2)

)

 log v = log x log x = ( log x )

2

Differentiating both sides with respect to x, we obtain

1 dv d  2 . = log x )  (  v dx dx  1 dv d  . = 2 ( log x ) . ( log x ) v dx dx



dv = 2 x log x −1.log x dx

........(3)

Therefore, from (1), (2) and (3), we obtain

dv x −1 = ( log x ) 1 + log x.log ( log x )  + 2 x log x−1.log x dx 8.

Differentiating both sides with respect to x.

( sin x )

x

+ sin −1 x

Solution-8 Let y = ( sin x ) + sin

−1

x

x

Also, let u = ( sin x ) and v = sin −1 x x

y =u+v

dy du dv = − dx dx dx x u = ( sin x ) 

............ (1)

 log u = log ( sin x )

x

 log u = x log ( sin x )

Differentiating both sides with respect to x, we obtain

1 du d d = ( x ) x log ( sin x ) + x  log ( sin x )  u dx dx dx du 1 d    = u 1.log ( sin x ) + x. . ( sin x ) dx sin x dx   du x x   = ( sin x ) log ( sin x ) + .cos x  dx sin x   du x  = ( sin x ) ( x cot x + logsin x ) ..........(2) dx v = sin −1 x 

Differentiating both sides with respect to x, we obtain

dv = dx

1 1−

( x)

2

.

d dx

dv 1 1 = . dx 1− x 2 x dv 1  = dx 2 x − x 2

( x)



............(3)

Therefore, from (1), (2) and (3), we obtain

dy 1 2 = ( sin x ) ( x cot x + logsin x ) + dx 2 x − x2 9.

Differentiate the function with respect to x. x

sin x

+ ( sin x )

cos x

Solution-9 Let y = x

sin x

Also u = x

+ ( sin x )

sin x

y = u + v

cos x

and v = ( sin x )

dy du dv = + dx dx dx u = xsin x  log u = log xsin x 

(

cos x

.............(1)

)

 log u = sin x log x Differentiating both sides with respect to x, we obtain

1 du d d = ( sin x ) .log x + sin x. ( log x ) u dx dx dx du 1   = u = cos x log x + sin x.  dx x  du sin x    = xsin x = cos x log x + dx x   cos x v = ( sin x )

 log v = log ( sin x )

cos x

 log v = cos x log ( sin x )

Differentiating both sided with respect to x, we obtain

1 dv d d = ( cos x )  log ( sin x ) + cos x  log ( sin x )  v dx dx dx dv 1 d    = v − sin x.log ( sin x ) + cos x. . ( sin x ) dx sin x dx   dv cos x cos x    = ( sin x )  − sin x logsin x + cos x  dx sin x   dv cos x  = ( sin x )  − sin x logsin x + cot x cos x  dx dv cos x  = ( sin x )  cot x cos x − sin x logsin x  dx

Therefore, from (1), (2) and (3), we obtain

dv sin x  cos x  = xsin x  cos x log x +  + ( sin x ) cos x cot x − sin x logsin x  dx x   10.

Differentiate the function with respect to x. x

cos x

x2 + 1 + 2 x −1

Solution-10 Let y = x

x cos x

Also, let u = x



x2 + 1 + 2 x −1 x cos x

x2 + 1 and v = 2 x −1

dy du dy = + dx dx dx

y = u + v u = x x cos x

Differentiating both sides with respect to x, we obtain

1 du d d d = ( x ) .cos x log x + x. ( cos x ) .log x + x cos x. ( log x ) u dx dx dx dx du 1   = u 1.cos x log x + x.( − sin x ).log x + x cos x.  dx x  du  = x x cos x ( cos x log x − x sin x.log x + cos x ) dx du  = x x cos x cos x (1 + log x ) − x sin x log x  .......(2) dx x2 + 1 v= 2 x −1  log v = log x 2 + 1 − log x 2 − 1

(

)

(

)

Differentiating both sides with respect to x, we obtain

1 dv 2x 2x = = 2 − 2 v dx x + 1 x − 1  2 x ( x 2 − 2 ) − 2 x ( x 2 + 1)  dv  = v  dx ( x2 + 1)( x2 − 1)    dv x 2 + 1  −4 x  = 2  2  dx x − 1  ( x + 1)( x 2 − 1)   



dv −4 x = dx ( x 2 − 1)2

..............(3)

Therefore, from (1), (2) and (3), we obtain

dy 4x = x x cos x cos x (1 + log x ) − x sin x log x  − 2 dx ( x2 − 1) 11.

1

( x cos x ) + ( x sin x ) x x

Differentiate the function with respect to x

Solution-11 1

Let y = ( x cos x ) + ( x sin x ) x x

Also, let u = ( x cos x ) and v = ( x sin x ) x

1 x

y = u + v

dy du dy = + dx dx dx 2 u = ( x cos x ) 

 log u = log ( x cos x )

..............(1)

x

 log u = x log ( x cos x )

x

 log u = x log x + logcos x  log u = x log x + x log cos x

Differentiating both sides with respect to x , we obtain

1 du d d = ( x + log x ) + ( x log cos x ) u dx dx dx 

du  d d d d    = u  log x. ( x ) + x. ( log x )  + log cos x. ( x ) + x. ( log cos x )  dx dx dx dx dx     du 1  1 d x    = ( x cos x )  log x.1 + x.  + log cos x.1 + x. . ( cos x )  dx x  cos x dx   du x x   = ( x cos x ) ( log x + 1) + log cos x + .( − sin x )  dx cos x    du x  = ( x cos x ) (1 + log x ) + (1 + log cos x − x tan x )  dx du x  = ( x cos x ) 1 − x tan x + ( log x + log cos x )  dx





du x = ( x cos x ) 1 − x tan x + + log ( x cos x )  dx

..............(2)

1

v = ( x sin x ) x

1 x

 log v = log ( x sin x ) 1  log v = log ( x sin x ) x 1  log v = ( log x + logsin x ) x 1 1  log v = log x + logsin x x x Differentiating both sides with respect to x , we obtain

1 dv d  1  d 1  =  log x  +  log ( sin x )  v dx dx  x  dx  x  

1 dv  d 1 1 d   d 1 1 d  = log x.   + . ( log x )  + log ( sin x ).   + . log ( sin x ) v dx  dx  x  x dx dx  x  x dx    1 dv    1  1 1   1  1 1 d  = log x. − 2  + .  + log ( sin x ) . − 2  + . ( sin x ) v dx   x  x x   x  x sin x dx   log ( sin x )  1 dv 1 1  = 2 (1 − log x ) + − + .cos x  2 v dx x x x sin x   1 1 − log x − log ( sin x ) + x cot x  1 dv 1 = 2 ( x sin x ) x +  +  2 v dx x x x2   1 1 − log x − log sin x + x cot x  dv ( )  = ( x sin x ) x   dx x2  



1 1 − log x sin x + x cot x  dv ( )  = ( x sin x ) x   dx x2  

..............(3)

Therefore, from (1), (2) and (3), we obtain 1  x cot x + 1 − log x sin x  dv ( ) 2 = ( x cos x ) 1 − x tan x + log ( x cos x ) + ( x sin x ) x   2 dx x  

12.

Find

dy y x of function. x + y = 1 dx

Solution-12 The given function is x + y = 1 y

x

Let x = u and y = v Then, the function becomes u + v = 1 y

x

du dv + =0 dx dx u = xy  log u = log x y



.........(1)

( )

 log u = y log x Differentiating both sides with respect to x , we obtain

1 du dy d = log x + y. ( log x ) u dx dx dx du dy 1   = u log x + y.  dx dx x  du dy y    = x y  log x +  dx dx x   v = yx

...........(2)

 log v = log ( y x )

 log v = x log y Differentiating both sides with respect to x , we obtain

1 dv d d = log y ( x ) + x. ( log y ) v dx dx dx  dv 1 dy   = v log y.1 + x. .  dx y dx  



 dv x dy  = y x  log y + dx y dx  

...........(3)

Therefore, from (1), (2) and (3), we obtain

 dy y  x dy   x y  log x +  + y x  log y + =0 dx x  y dx    dy  ( x 2 + log x + xy y −1 ) = − ( yx y −1 + y x log y ) dx y −1 x dy yx + y log y  =− y dx x log x + xy x−1

13.

Find

dy x y of function y = x dx

Solution:13 The given function y = x Taking logarithm on both sides, we obtain x

y

x log y = y log x

Differentiating both sides with respect to x, we obtain

d d d d ( x ) + x. ( log y ) = log x. ( y ). ( log x ) dx dx dx dx 1 dy dy 1  log y.1 + x. . = log x. + y. y dx dx x x dy dy y  log y + . = log x. + y dx dx x x  dy y   − log x  = − log y y  dx x log y.

 x − y log x  dy y − log y   dx = y x   dy y  y − x log y  .  =  dx x  x − y log x  14.

Find

dy y x of function ( cos x ) = ( cos y ) dx

Solution:14 The given function ( cos x ) = ( cos y ) y

x

Taking logarithm on both sides, we obtain

y = log cos x = x log cos y

Differentiating both sides with respect to x, we obtain

d d d d + y. ( log cos x ) = log cos y. ( x ). + x ( log cos y ) dx dx dx dx dy 1 d 1 d  logcos x. + y . ( cos x ) = logcos y.1 + x. . ( cos y ) dx cos x dx cos y dx log cos x.

 logcos x.

dy y x dy + .( − sin x ) = logcos y + ( − sin y ) dx cos x cos y dx

dy dy − y tan x = log cos y − x tan y dx dx dy  ( log cos x + x tan y ) = y tan x + log cos y dx dy y tan x + logcos y  = dx x tan y + logcos x  log cos x.

15.

Find

dy ( x− y ) of function xy = e dx

Solution-15 ( x− y )

The given function is xy = e Taking logarithm on both sides, we obtain.

log ( xy ) = log ( e x − y )

 log x + log y = ( x − y ) log e  log x + log y = ( x − y ) 1  log x + log y = x − y Differentiating both sides with respect to x , we obtain

d d d dy ( log x ) + ( log y ) = ( x ) − dx dx dx dx 1 1 dy 1  + =1− x y dx x  1  dy x − 1  1 +  = y  dx x  

16.

dy y ( x − 1) = dx x ( y − 1)

Find the derivative of the function given by

f ( x ) = (1 − x ) (1 + x 2 )(1 + x 4 )(1 + x8 ) and hence find f ' (1) Solution-16

(

The given relationship is f ( x ) = (1 − x ) 1 + x

2

Taking logarithm on both sides, we obtain

)(1 + x )(1 + x ) 4

8

log f ( x ) = log (1 − x ) + log (1 + x 2 ) + log (1 + x 4 ) + log (1 + x8 ) Differentiating both sides with respect to x, we obtain.

1 d d d d d .  f ( x )  = log (1 − x ) + log (1 + x 2 ) + log (1 + x 4 ) + log (1 + x8 ) f ( x ) dx dx dx dx dx 

1 1 d 1 d 1 d 1 d . f '( x ) = . (1 − x ) + . (1 + x 2 ) + . log (1 + x 4 ) + . log (1 + x8 2 4 8 f ( x) 1 + x dx 1 + x dx 1 + x dx 1 + x dx

1 1 1  1   f '( x ) = f ( x )  + .2 x + + .8x 7  2 4 8 1+ x 1+ x 1 + x 1 + x   1 2x 4 x3 8x7   f ' ( x ) = (1 + x ) (1 + x 2 )(1 + x 4 )(1 + x8 )  + + + 2 4 8 1 + x 1 + x 1 + x 1 + x   1 2  1 4  13 8  17  Hence, f ' (1) = (1 + 1) (1 + 1 )(1 + 1 )(1 + 1 )  + + + 2 4 1 + 1 1 + 1 1 + x 1 + x8   2

4

8

1 2 4 8 = 2 2 2 2 + + +  2 2 2 2 1 + 2 + 4 + 8  = 16   2  15 = 16  = 100 2 17.

(

Differentiate x − 5 x + 8 i.

2

)( x

3

+ 7 x + 9 ) in three ways mentioned below

By using product rule ii.

By expanding the product to obtain a single polynomial iii.

By iii.

By logarithm Differentiate Do they all given the same answer?

Solution-17

(

Let y = x − 5 x + 8 2

(i)



dy dx dy  dx



)( x

3

+ 7 x + 9)

Let x = x − 5 x + 8 and u = x + 7 x + 9 2

3

y = uv du dv = v+u (By using product rule) dv dx d d = ( x 2 − 5 x + 8 ) .( x 3 + 7 x + 9 ) + ( x 2 − 5 x + 8 ) . ( x 3 + 7 x + 9 ) dx dx



dy = ( 2 x − 5 ) ( x3 + 7 x + 9 ) + ( x 2 − 5 x + 8 )( 3 x 2 + 7 ) dx



dy = 2 x ( x3 + 7 x + 9 ) − 5 ( x3 + 7 x + 9 ) + x 2 ( 3x 2 + 7 ) − 5 x ( 3x 2 + 7 ) − 8 ( 3x 2 + 7 ) dx



dy = ( 2 x 4 + 14 x 2 + 18 x ) − 5 x3 − 35 x − 45 + ( 3x 4 + 7 x 2 ) − 15 x 3 − 35 x + 24 x 2 + 56 dx dy  = 5 x 4 − 20 x3 + 45 x 2 − 52 x + 11 dx 2 3 (ii) y = x − 5 x + 8 x + 7 x + 9

(

)(

)

= x 2 ( x3 + 7 x + 9 ) − 5 x ( x3 + 7 x + 9 ) + 8 ( x3 + 7 x + 9 ) = x5 + 7 x3 + 9 x 2 − 5 x 4 − 35 x 2 − 45 x + 8 x3 + 56 x + 72 = x5 − 5 x 4 + 15x3 − 26 x 2 + 11x + 72 dy d  = = ( x5 − 5 x 4 + 15 x 3 − 26 x 2 + 11x + 72 ) dx dx d d d d d d = = ( x5 ) − 5 ( x 4 ) + 15 ( x3 ) − 26 ( x 2 ) + 11 ( x ) + ( 72 ) dx dx dx dx dx dx 4 3 2 = 5 x − 5  4 x + 15  3x − 26  2 x + 11 1 + 0 = 5 x 4 − 20 x3 + 45 x 2 − 52 x + 11 (iii) Taking logarithm on both sides, we obtain.

log y = log ( x 2 + 5 x + 8 ) + log ( x3 + 7 x + 9 )

Differentiating both sides with respect to x, we obtain

1 dy d d = log ( x 2 − 5x + 8) + log ( x3 + 7 x + 9 ) y dx dx dx Differentiating both sides with respect to x, we obtain



1 dy d d = log ( x 2 − 5x + 8). log ( x3 + 7 x + 9 ) y dx dx dx

1 dy 1 d 1 d = 2 . ( x 2 − 5 x + 8) + 3 . log ( x3 + 7 x + 9 ) y dx x − 5x + 8 dx x + 7 x + 9 dx dy 1 1    = y 2  ( 2 x − 5) + 3  ( 3x 2 + 7 )  dx x + 7x + 9  x − 5x + 8   2x − 5 dy 3x 2 + 7   = ( x 2 − 5x + 8)( x3 + 7 x + 9 )  2 + 3  dx  x − 5x + 8 x + 7 x + 9  

 ( 2 x − 5) ( x3 + 7 x + 9 ) + ( 3x 2 + 7 )( x 2 − 5 x + 8 )  dy 2 3  = ( x − 5 x + 8)( x + 7 x + 9 )   dx ( x 2 − 5 x + 8) + ( x3 + 7 x + 9 )   

dy = 2 x ( x3 + 7 x + 9 ) − 5 ( x3 + 7 x + 9 ) + 3x 2 ( x 2 − 5 x + 8 ) + 7 ( x 2 − 5 x + 8 ) dx



dy = ( 2 x 4 + 14 x 2 + 18 x ) − 5 x3 − 35 x − 45 + ( 3x 4 − 15 x3 + 24 x 2 ) + ( 7 x 2 − 35 x + 56 ) dx dy  = 5 x 4 − 20 x3 + 45 x 2 − 52 x + 11 dx dy From the above three observations, it can be concluded that all the result of are dx

same 18.

If u , v and w are functions of x, then show that

d du dv dw ( u.v.w) = v.w + u .w + u.v dx dx dx dx In two ways-first by repeated application of product rule, second by logarithmic differentiation.

Solution – 18 Let y = u.v.w = u.( v.w) By applying product rule, we obtain

dy du d = .( v.w ) + u. .( v.w ) dx dx dx dy du dv   dv  = v.w + u  .w + v.  dx dx dx   dx dy du dv dw  = v.w + u. .w + u.v dx dx dx dx

(Again applying product rule)

By taking logarithm on both sides of the equation y = u.v.w , we obtain

log y = log u + log v + log w Differentiating both sides with respect to x, we obtain

1 dy d d d . = ( log u ) + ( log v ) + ( log w) y dx dx dx dx 1 dy 1 d 1 d 1 d  . = + + y dx u dx v dx w dx dy  1 du 1 dv 1 dw   = y + +  dx  u dx v dx w dx 

dy  1 du 1 dv 1 dw  = u.v.w  + +  dx  u dx v dx w dx  dy du dv dw  = u.w + u w + u.v dx dx dx dx 