ü Differentiate with respect to x

1. y HxL = I2 x3

+ 5M I7 x4

+ 3M

Out[16]=

Tick to show solution

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

uHxL = 2x3+5,

„ u

„ x

= 6 x2

vHxL = 7x4+3,

„ v

„ x

= 28 x3

„ y

„ x

= 28 H2 x3 + 5L x

3 + 6 H7 x4 + 3L x

2

or

„ y

„ x

= 2 x2 H49 x

4 + 70 x + 9L

2. yHxL =2 ‰x

x5

Out[48]=

Tick to show solution

By using the quotient rule we obtain

uHxL = 2‰x,

„ u

„ x

= 2 ‰x

vHxL = x5

,

„ v

„ x

=1

5 x4ê5

„ y

„ x

=2 ‰x

5 x4ê5 + 2 ‰x

x5

or

„ y

„ x

=2 ‰x H5 x+1L

5 x4ê5

3. yHxL = lnHxL I12 x3 - 4M

Out[17]=

Tick to show solution

By using the product rule we obtain

uHxL = ln x,

„ u

„ x

=1

x

vHxL =12x3-4,

„ v

„ x

= 36 x2

„ y

„ x

=12 x

3-4

x+ 36 x

2logHxL

4. f HxL = Ix2+ 3M2

Out[33]=

Tick to show solution

By using the chain rule we get

uHxL = x2+ 3,

„ u

„ x

= 2 x

f HuL = u2,

„ f

„ u

= 2 u

„ f

„ x

= 4 x Hx2 + 3L

5. yHxL =6 x

2 - 4 x + 9

ln HxL

Out[13]=

Tick to show solution

By using the quotient rule we obtain

uHxL = 6x2-4x+9,

„ u

„ x

= 12 x - 4

vHxL = lnHxL,„ v

„ x

=1

x

„ y

„ x

=

1

HlnHxLL2H12 x - 4L ln HxL - I6 x

2- 4 x + 9M x

-1=

12 x-4

logHxL -6 x

2-4 x+9

x log2HxL

6. y HxL = H2 x + 4L4

2 | Numerical Integration

Out[32]=

Tick to show solution

By using the chain rule we get

uHxL = 2x + 4,

„ u

„ x

= 2

f HuL = u4,

„ f

„ u

= 4 u3

„ f

„ x

= 8 H2 x + 4L3

7. y HxL = H7 -6 xL2

Out[31]=

Tick to show solution

By using the chain rule we get

uHxL =7 - 6x,

„ u

„ x

= -6

f HuL = u2,

„ f

„ u

= 2 u

„ f

„ x

= -12 H7 - 6 xL

8. yHxL = 7 ‰xx

3

Out[19]=

Tick to show solution

By using the product rule we get

uHxL = 7‰x,

„ u

„ x

= 7 ‰x

vHxL = x3

,

„ v

„ x

=1

3 x2ê3

„ y

„ x

=7 ‰x

3 x2ê3 + 7 ‰x

x3

or

„ y

„ x

=7 ‰x H3 x+1L

3 x2ê3

9. y HxL =2 x

5

x + 1

Numerical Integration | 3

Out[8]=

Tick to show solution

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

uHxL = 2x5,

„ u

„ x

= 10 x4

vHxL = x+1,

„ v

„ x

= 1

„ y

„ x

=

10 x4 Hx + 1L - 2 x

5

Hx + 1L2=

10 x4

x+1-

2 x5

Hx+1L2

10. yHxL = Ix2+ 3 x + 4M5

Out[29]=

Tick to show solution

By using the chain rule we get

uHxL = x2+ 3x + 4,

„ u

„ x

= 2 x + 3

f HuL = u5,

„ f

„ u

= 5 u4

„ f

„ x

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

11. y HxL =x

3 - 2

x2 + 1

Out[7]=

Tick to show solution

By using the quotient rule we obtain

uHxL = x3-2,

„ u

„ x

= 3 x2

vHxL = x2+1,

„ v

„ x

= 2 x

„ y

„ x

=

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

Ix2 + 1M2=

3 x2

x2+1

-2 x Ix3-2M

Ix2+1M2

12. y HxL =2 x

2 - 5

3 x + 8

4 | Numerical Integration

Out[9]=

Tick to show solution

By using the quotient rule we obtain

uHxL = 2x2-5,

„ u

„ x

= 4 x

vHxL = 3 x +8,

„ v

„ x

=3

2 x

„ y

„ x

=4 x

3 x +8

-3 I2 x

2-5M

2 J3 x +8N2 x

or

„ y

„ x

=64 x

3ê2+18 x2+15

2 J3 x +8N2 x

13. yHxL = 6 x + 5

Out[25]=

Tick to show solution

By using the chain rule we get

uHxL = 6x+5,

„ u

„ x

= 6

f HuL = u1ê2

,

„ f

„ u

=1

2 u

„ f

„ x

=3

6 x+5

14. yHxL =lnHxL

5 x3 - 4

Out[14]=

Tick to show solution

By using the quotient rule we obtain

uHxL = ln x,

„ u

„ x

=1

x

vHxL = 5x3-4,

„ v

„ x

= 15 x2

„ y

„ x

=

x-1 I5 x

3 - 4M - ln HxL 15 x2

I5 x3 - 4M2

=1

x I5 x3-4M -

15 x2

logHxLI5 x

3-4M2

15. yHxL = H3 x + 4L-2

Numerical Integration | 5

Out[28]=

Tick to show solution

By using the chain rule we get

uHxL = 3x + 4,

„ u

„ x

= 3

f HuL = u-2

,

„ f

„ u

= -2

u3

„ f

„ x

= -6

H3 x+4L3

16. yHxL =3

2 x3

- 5 x

Out[49]=

Tick to show solution

By using the quotient rule we obtain

uHxL = 3,

„ u

„ x

= 0

vHxL = 2 x3

-5 x ,

„ v

„ x

=2

3 x2ê3 -

5

2 x

„ y

„ x

= -

32

3 x2ê3

-5

2 x

K2 x3

-5 x O2

or

„ y

„ x

=15 x

6

-4

2 K2-5 x6 O

2

x4ê3

If we rewrite yHxL = 3 J2 x3

- 5 x N-1

then the chain rule can be used.

17. yHxL = H1 - 8 xL-3

Out[27]=

Tick to show solution

By using the chain rule we get

uHxL = 1 - 8x,

„ u

„ x

= -8

f HuL = u-3

,

„ f

„ u

= -3

u4

„ f

„ x

=24

H1-8 xL4

6 | Numerical Integration

18. yHxL = ‰x I4 x - 2 x

3+ 3M

Out[37]=

Tick to show solution

By using the product and chain rules we obtain

uHxL = ‰x,

„ u

„ x

=‰x

2

vHxL = 4x - 2x3 + 3,

„ v

„ x

= 4 - 6 x2

„ y

„ x

=1

2‰x H-2 x

3 + 4 x + 3L + ‰x H4 - 6 x2L

or

„ y

„ x

=1

2‰x H-2 x

3 - 12 x2 + 4 x + 11L

19. yHxL = lnHsinHxLL -1

2

cos2H4 xL

Out[43]=

Tick to show solution

By using the linearity and chain rules we obtain

for the first term

uHxL = sinHxL,„ u

„ x

= cosHxL

f HuL = ln u,

„ f

„ u

=1

u

for the second term

uHxL = cosH4 xL,„ u

„ x

= -4 sinH4 xL

f HuL =

1

2

u2,

„ f

„ u

= u

Finally we get

„ y

„ x

= cotHxL + 4 sinH4 xL cosH4 xL

or

„ y

„ x

= 2 sinH8 xL + cotHxL

20. yHxL =1

4

tan4HxL -

1

2

tan2HxL - lnHcosHxLL

Numerical Integration | 7

Out[47]=

Tick to show solution

By using the linearity and chain rules we obtain

„ y

„ x

= tanHxL + tan3HxL sec

2HxL - tanHxL sec2HxL

or

„ y

„ x

= tan5HxL

8 | Numerical Integration