# The value h ′ ( 1 ) . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.4, Problem 53E

a)

To determine

## To find: The value h′(1).

Expert Solution

The value h(1) is h(1)=30.

### Explanation of Solution

Given:

The function is h(x)=f(g(x)).

Result used: Chain Rule:

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product

F(x)=g(h(x))h(x) (1)

Calculation:

Obtain the derivative of h(x)=f(g(x)).

h(x)=ddx(h(x))=ddx(f(g(x)))

Apply the chain rule as shown in equation (1),

h(x)=f(g(x))g(x) (2)

Substitute x=1 in equation (2),

h(1)=f(g(1))g(1)

Consider the values from given table,g(1)=2 and g(1)=6,

h(1)=f(2)6.

Consider the values from given table f(2)=5,

h(1)=56=30

Therefore, the derivative of h(1)=f(g(1)) is h(1)=30_.

(b)

To determine

### To find: The value H′(1).

Expert Solution

The value H(1) is H(1)=36_.

### Explanation of Solution

Given:

The function is H(x)=g(f(x)).

Calculation:

Obtain the derivative of H(x)=g(f(x)).

H(x)=ddx(H(x))=ddx(g(f(x)))

Apply the chain rule as shown in equation (1),

H(x)=g(f(x))f(x) (3)

Substitute x=1 in equation (3),

H(1)=g(f(1))f(1)

Consider the values from given table f(1)=3 and f(1)=4,

H(1)=g(3)4

Consider the values from given table g(3)=9,

H(1)=94=36

Therefore, the value H(1) is H(1)=36_.

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