Page 1

Ÿ Differentiate with respect to x y HxL = I2 x3 + 5M I7 x4 + 3M

1.

Tick to show solution

By using the formula for the derivative of a product we obtain

âu

uHxL = 2 x3 +5,

= 6 x2

âx âv

Out[16]=

vHxL = 7x4 +3,

= 28 x3

âx ây âx

= 28 H2 x3 + 5L x3 + 6 H7 x4 + 3L x2

ây or

âx

= 2 x2 H49 x4 + 70 x + 9L

2 ãx yHxL =

2.

5

x

Tick to show solution

By using the quotient rule we obtain

âu

uHxL = 2ã x ,

vHxL =

Out[48]=

= 2 ãx

âx âv

5

=

x,

âx ây =

âx

2 ãx 5 x4•5

ây =

or

âx 3.

+ 2 ãx

5

1 5 x4•5

x

2 ã x H5 x+1L 5 x4•5

yHxL = lnHxL I12 x3 - 4M


2 | Numerical Integration

Tick to show solution

By using the product rule we obtain

âu uHxL = ln x,

=

âx âv

Out[17]=

vHxL =12x3 -4,

1 x

= 36 x2

âx ây =

12 x3 -4 x

âx

+ 36 x2 logHxL

f HxL = Ix2 + 3M

2

4.

Tick to show solution

By using the chain rule we get

âu

uHxL = x2 + 3,

= 2x

âx

Out[33]=

f HuL = u2 ,

âf âx

âf = 2u

âu

= 4 x Hx2 + 3L

6 x2 - 4 x + 9

5.

yHxL =

ln HxL

Tick to show solution

By using the quotient rule we obtain

âu

uHxL = 6x2 -4x+9,

= 12 x - 4

âx

Out[13]=

âv vHxL = lnHxL,

=

âx ây

1

=

âx 6.

HlnHxLL2

1 x

H12 x - 4L ln HxL - I6 x2 - 4 x + 9M x-1 =

y HxL = H2 x + 4L4

12 x-4 logHxL

-

6 x2 -4 x+9 x log2 HxL


Numerical Integration | 3

Tick to show solution

By using the chain rule we get

âu uHxL = 2x + 4,

=2

âx

Out[32]=

f HuL = u4 ,

âf âx

âf

= 4 u3

âu

= 8 H2 x + 4L3

y HxL = H7 -6 xL2

7.

Tick to show solution

By using the chain rule we get

âu uHxL =7 - 6x, f HuL = u2 ,

âf âx 8.

= -6

âx âf

Out[31]=

= 2u

âu

= -12 H7 - 6 xL 3

yHxL = 7 ãx

x

Tick to show solution

By using the product rule we get

âu

uHxL = 7ã x , Out[19]=

vHxL =

= 7 ãx

âx âv

3

=

x,

âx ây =

âx ây

=

or

âx

9.

7 ãx 3 x2•3

+ 7 ãx

3

1 3 x2•3

x

7 ã x H3 x+1L 3 x2•3

y HxL =

2 x5 x+1


4 | Numerical Integration

Tick to show solution

By using the formula for the derivative of a quotient we obtain

âu

uHxL = 2x5 ,

= 10 x4

âx âv

Out[8]=

vHxL = x+1,

=1

âx ây =

10 x4 Hx + 1L - 2 x5 Hx + 1L

=

2

âx

10 x4 x+1

-

2 x5

Hx+1L2

yHxL = Ix2 + 3 x + 4M

10.

5

Tick to show solution

By using the chain rule we get

âu

uHxL = x2 + 3x + 4,

= 2x+3

âx

Out[29]=

f HuL = u5 ,

âf âx

âf

= 5 u4

âu

= 5 H2 x + 3L Hx2 + 3 x + 4L4

y HxL =

11.

x3 - 2 x2 + 1

Tick to show solution

By using the quotient rule we obtain

âu

uHxL = x3 -2,

= 3 x2

âx âv

Out[7]=

vHxL = x2 +1,

= 2x

âx ây =

âx

12.

3 x2 Ix2 + 1M - Ix3 - 2M 2 x Ix2 + 1M

2

y HxL =

2 x2 - 5 3

x +8

=

3 x2 x2 +1

-

2 x Ix3 -2M Ix2 +1M

2


Numerical Integration | 5

Tick to show solution

By using the quotient rule we obtain

âu

uHxL = 2 x2 -5,

= 4x

âx âv vHxL = 3 x +8,

Out[9]=

2

âx ây

4x

=

âx ây

2 J3

âx

13.

3 I2 x2 -5M

2 J3

2

x +8N

x

x

64 x3•2 +18 x2 +15

=

or

-

x +8

3

3

=

yHxL =

2

x +8N

x

6x+5

Tick to show solution

By using the chain rule we get

âu uHxL = 6x+5,

=6

âx

Out[25]=

f HuL = u1•2 ,

âf

âf

1

=

âu

2

u

3

=

6 x+5

âx

lnHxL 14.

yHxL =

5 x3 - 4

Tick to show solution

By using the quotient rule we obtain

âu uHxL = ln x, Out[14]=

vHxL = 5x3 -4,

=

âx âv

1 x

= 15 x2

âx ây =

âx

15.

x-1 I5 x3 - 4M - ln HxL 15 x2 I5 x3 - 4M

2

yHxL = H3 x + 4L-2

=

1

x I5 x3 -4M

-

15 x2 logHxL I5 x3 -4M

2


6 | Numerical Integration

Tick to show solution

By using the chain rule we get

âu uHxL = 3x + 4,

=3

âx

Out[28]=

f HuL = u-2 ,

âf =-

âu

âf = -

âx

2 u3

6

H3 x+4L3

3 yHxL =

16.

3

2

x -5

x

Tick to show solution

By using the quotient rule we obtain

âu uHxL = 3,

=0

âx âv

3

vHxL = 2 x -5 x ,

=

âx

2 3 x2•3

5

2

x

Out[49]= 2

3

ây = -

âx

K2

3 x2•3 3

ây

x

x O

2

2 K2-5

âx

2

x -5

15

=

or

5

-

6

x -4 6

x O x4•3 2

If we rewrite yHxL = 3 J2

3

x -5 x N

yHxL = H1 - 8 xL-3

17.

Tick to show solution

By using the chain rule we get

âu uHxL = 1 - 8x,

= -8

âx

Out[27]=

f HuL = u-3 ,

âf =

âx

24

âf

H1-8 xL4

=-

âu

3 u4

-1

then the chain rule can be used.


Numerical Integration | 7

18.

ã x I4 x - 2 x3 + 3M

yHxL =

Tick to show solution

By using the product and chain rules we obtain

âu

ãx ,

uHxL =

ãx

=

2

âx

âv

Out[37]=

vHxL = 4x - 2x3 + 3,

= 4 - 6 x2

âx ây

1

=

2

âx ây

=

or

âx

ã x H-2 x3 + 4 x + 3L +

ã x H-2 x3 - 12 x2 + 4 x + 11L

1 2

1

19.

ã x H4 - 6 x2 L

yHxL = lnHsinHxLL 2

cos2 H4 xL

Tick to show solution

By using the linearity and chain rules we obtain for the first term

âu uHxL = sinHxL,

= cosHxL

âx âf

f HuL = ln u,

=

âu

1 u

for the second term Out[43]=

âu uHxL = cosH4 xL,

= -4 sinH4 xL

âx f HuL =

1

u2 ,

âf =u

âu

2

Finally we get

ây = cotHxL + 4 sinH4 xL cosH4 xL

âx ây = 2 sinH8 xL + cotHxL

or

âx

20.

1 yHxL = 4

tan4 HxL -

1 2

tan2 HxL - lnHcosHxLL


8 | Numerical Integration

Tick to show solution

By using the linearity and chain rules we obtain Out[47]=

창y 창x

= tanHxL + tan3 HxL sec2 HxL - tanHxL sec2 HxL

창y or

창x

= tan5 HxL

Differentiation rules  

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