# The value F ′ ( 0 ) .

### 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 63E
To determine

## To find: The value F′(0).

Expert Solution

The value F(0) is 96 .

### Explanation of Solution

Given:

The function is F(x)=f(3f(4f(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(3f(4f(x))).

F(x)=ddx(F(x))=ddx(f(3f(4f(x))))

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

F(x)=f(3f(4f(x)))ddx(3f(4f(x)))

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

F(x)=f(3f(4f(x)))ddx(3f(4f(x)))=f(3f(4f(x)))[3ddx(f(4f(x)))]=f(3f(4f(x)))[3f(4f(x))4ddx(f(x))]=f(3f(4f(x)))[12f(4f(x))f(x)]

Substitute x=0 in the above equation,

F(0)=f(3f(4f(0)))[12f(4f(0))f(0)]

Substitute the given values f(0)=0 and f(0)=2,

F(0)=f(3f(4(0)))[12f(4(0))(2)]=f(3f(0))[24f(0)]=f(0)[24f(0)]=2(242)=96

Therefore, the value F(0) is 96.

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