# The value r ′ ( 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 61E
To determine

## To find: The value r′(1).

Expert Solution

The value r(1) is 120.s

### Explanation of Solution

Given:

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

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

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

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

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

r(x)=f(g(h(x)))[g(h(x))h(x)]

Substitute x=1 in the above equation,

r(1)=f(g(h(1)))[g(h(1))h(1)]

Substitute the given values h(1)=2 and h(1)=4,

r(1)=f(g(2))[g(2)(4)]

Substitute the given values g(2)=3 and g(2)=5,

r(1)=f(3)[(5)(4)]=20f(3)

Substitute f(3)=6,

r(1)=20(6)=120

Therefore, the value r(1) is 120.

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