﻿ Numerical Derivatives and Integrals Worksheet | Problems & Solutions

# Numerical Derivatives and Integrals Worksheet

Numerical Derivatives and Integrals Worksheet
• Page 1
1.
Find $f$ '($x$), if $f$ ($x$) = 8$x$2 + 10$x$ + 19.
 a. 10 b. 16$x$ - 10 c. 16$x$ d. 16$x$ + 10

#### Solution:

f(x) = 8x2 + 10x + 19

f ' (x) = limh0f(x + h) - f(x)h
[Use the definition of f'(x).]

= limh0 [8(x + h)2 +10(x + h) + 19] - [8x2 +10x + 19] h

= limh0[8x2 + 8h2 + 16xh + 10x + 10h + 19 - 8x2 - 10x - 19]h

= limh08h2 + 16xh + 10hh = limh0h[8h + 16x + 10]h

= limh0 8h + 16x + 10
[As h ≠ 0, hh = 1.]

= 16x + 10

2.
Find $f$ ′($x$), if $f$ ($x$) = 8$x$3.
 a. 8 $x$2 b. 24$x$2 c. 24$x$3 d. - 24$x$2

#### Solution:

f(x) = 8x3

f '(x) = limh0f(x + h) - f(x)h
[Use the definition of f ′(x).]

= limh08(x + h)3 - 8(x)3h

= limh08x3+24x2h+24xh2+8h3-8x3h
[Use (a+b)3 =a3+3a2b+3ab2+b3.]

= limh0[24x2+24hx+8h2]
[Simplify.]

= 24x2

3.
Find $g$ ′(8), if $g$($x$) = $\frac{1}{{x}^{2}}$.
 a. - $\frac{1}{4}$ b. - 16 c. - $\frac{1}{256}$ d. $\frac{1}{64}$

#### Solution:

g(x) = 1x2

g ′(8) = limh0g(8 + h) - g(8)h
[Use the definition of g ′(8).]

= limh0 1h[1(8 + h)2 -182]

= limh0 1h[64 -(8 + h)264(8 + h)2]

= limh0 1h[- (h2 + 16h)64(8 + h)2]

= limh0 [- (h + 16)64(8 + h)2]
[As h ≠ 0, hh = 1.]

= - 1 / 256

4.
Find $g$ ′(4), if $g$ ($x$) = $x$3 + $x$2.
 a. - 56 b. 56 c. 69 d. 13

#### Solution:

g(x) = x3 + x2

g ′(4) = limh0g(4 + h) - g(4)h
[Use the definition of g ′(x).]

= limh0 1h[(4 + h)3 +(4 + h)2 - 64 -16]

= limh0 1h[h3+12h2+48h+h2+8h]

= limh0 1h[h3 + 13h2 + 56h]
[As h ≠ 0, hh = 1.]

= limh0[h2 + 13h + 56]

= 56

5.
Find $f$ ′(4), if $f$($x$) = 2$x$ - $x$2.
 a. - 8 b. - 6 c. 6

#### Solution:

f(x) = 2x - x2

f '(4) = limh0f(4 + h) - f(4)h
[Use the definition of f ′(x).]

= limh0 1h[2(4 + h)-(4 + h)2 - 8 + 16]

= limh0 1h[8 + 2h - 16 -h2 - 8h - 8 + 16]

= limh0[- 6 - h]
[As h ≠ 0, hh = 1.]

= - 6

6.
Find $f$ '($\pi$), if $f$($x$) = sin 18$x$.
 a. - 18 b. 18 c. 1

#### Solution:

f(x) = sin 18x

f '(π) = limh0f(π + h) - f(π)h
[Use the definition of f '(π).]

= limh0 1h[sin 18(π + h) - sin 18π]

= limh0 1 / h[sin (18π + 18h) - sin 18π]

= limh0 sin 18hh
[Use sin(x + h) = sin xcos h + cos xsin h.]

= 18
[Use limθ0  sin kθθ = k.]

7.
Find $f$ ′($\frac{\pi }{2}$), if $f$ ($x$) = 42cos $x$.
 a. - 42 b. Does not exist c. 42

#### Solution:

f(x) = 42 cos x

f ′(π2) = limh0f(π2 + h) - f(π2)h
[Use the definiton of f ′(π2).]

= limh0 1h[42cos (π2 + h) - 42cos(π2)]

= limh0 (- 42) sin hh

= (- 42) limh0 sin hh

= - 42.
[Use limθ0  sin θθ = 1.]

8.
Find NDER $f$(3), if $f$ ($x$) = 1 + 3$x$2, by calculating the symmetric difference quotient with $h$ = 0.001.
 a. 14 b. 18 c. 19 d. 16

#### Solution:

f(x) = 1 + 3x2

NDER f(3) = f(3 + 0.001) - f(3 - 0.001)0.002
[Use NDER f(a) = f(a + h) - f(a - h)2h.]

= f(3.001) - f(2.999)0.002

= [1 + 3(3.001)2] - [1 + 3(2.999)2]0.002

= 18.
[Use a calculator and simplify.]

9.
Find NDER $f$(- 1), if $f$ ($x$) = 4 + $x$3 by calculating the symmetric difference quotient with $h$ = 0.001.

#### Solution:

f(x) = 4 + x3

NDER f(-1) = f(- 1 + 0.001) - f(- 1 - 0.001)0.002
[Use NDER f(a) = f(a + h) - f(a - h)2h.]

= f(-0.999) - f(- 1.001)0.002

= [4 +(-0.999)3] - [4 +(- 1.001)3]0.002

= 3.000001.
[Use a calculator and simplify.]

10.
Find NDER $f$($\frac{\pi }{4}$), if $f$($x$) = sin $x$ by calculating the symmetric difference quotient with $h$ = 0.001.
 a. 0.707 b. 1 c. -0.7 d. -1

#### Solution:

f(x) = sin x

NDER f(π4)= f(π4 + 0.001) -  f(π4 - 0.001)0.002
[Use NDER f (a) = f(a + h) - f(a - h)2h.]

= sin(π4 + 0.001) -  sin(π4 - 0.001)0.002

= 0.707.
[Use a calculator and simplify.]