# The value of F ′ ( 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 64E
To determine

## To find: The value of F′(1).

Expert Solution

The value of F(1) is 96.

### Explanation of Solution

Given:

The function is F(x)=f(xf(xf(x))),

Formula used:

The 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)

Product Rule:

If f(x).and g(x) are both differentiable function, then

ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)] (2)

Calculation:

Obtain the derivative of F(x)=f(xf(xf(x))).

Apply the chain rule as shown in equation (1)

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

Apply the product rule as shown in equation (2)

ddx(F(x))=f(xf(xf(x)))(xddx(f(xf(x)))+f(xf(x))ddx(x))F(x)=f(xf(xf(x)))(xddx(f(xf(x)))+f(xf(x))(1x11))=f(xf(xf(x)))(xddx(f(xf(x)))+f(xf(x)))

Apply the chain rule as shown in equation (1)

F(x)=f(xf(xf(x)))(x(f(xf(x))ddx(xf(x)))+f(xf(x)))

Apply the product rule as shown in equation (2)

F(x)=f(xf(xf(x)))(x(f(xf(x))[xddx(f(x))+f(x)ddx(x)])+f(xf(x)))=f(xf(xf(x)))(x(f(xf(x))[x(f(x))+f(x)(1x11)])+f(xf(x)))=f(xf(xf(x)))(x(f(xf(x))[x(f(x))+f(x)])+f(xf(x)))

Substitute x=1

F(1)=f(1f(1f(1)))(1(f(1f(1))[1(f(1))+f(1)])+f(1f(1)))=f(f(f(1)))((f(f(1))[(f(1))+f(1)])+f(f(1)))

Given f(1)=2 and f(1)=4

F(1)=f(f(2))((f(2)[(4)+2])+f(2))=f(f(2))((f(2))+f(2))

Given, f(2)=3 and f(2)=5

F(1)=f(3)((5(6))+3)=f(3)((30)+3)=33f(3)

Given, f(3)=6

F(1)=33(6)=198

Therefore, the value of F(1) is 198.

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