# The value F ′ ( 2 ) .

### 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 54E

(a)

To determine

## To find: The value F′(2).

Expert Solution

The value F(2) is F(2)=20_.

### Explanation of Solution

Given:

The function is F(x)=f(f(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 F(x)=f(f(x)).

F(x)=ddx(F(x))=ddx(f(f(x)))

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

F(x)=f(f(x))f(x) (2)

Substitute x=2 in equation (2),

F(2)=f(f(2))f(2)

Consider the values from given table in exercise (63), f(2)=1 and f(2)=5.

F(2)=f(1)5

Consider the values from given table in exercise (63), f(1)=4.

F(2)=45=20

Therefore, the value F(2) is F(2)=20_.

(b)

To determine

### To find: The value G′(3).

Expert Solution

The value G(3) is G(3)=63_.

### Explanation of Solution

Given:

The function is G(x)=g(g(x)).

Calculation:

Obtain the derivative of G(x)=g(g(x)).

G(x)=ddx(G(x))=ddx(g(g(x)))

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

G(x)=g(g(x))g(x) (3)

Substitute x=3 in equation (2),

G(3)=g(g(3))g(3)

Consider the values from given table in exercise (63), g(3)=2 and g(3)=9 .

G(3)=g(2)9

Consider the values from given table in exercise (63), g(2)=7.

G(3)=79=63

Therefore, the value G(3) is G(3)=63_.

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