# 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 52E
To determine

## To find: The value h′(1).

Expert Solution

The value h(1) is h(1)=65_.

### Explanation of Solution

Given:

The function is h(x)=4+3f(x), where f(1)=7,f(1)=4.

Result used: The Power Rule combined with the Chain Rule

If n is any real number and g(x) is differentiable function, then

ddx[g(x)]n=n[g(x)]n1g(x) (1)

Calculation:

Obtain the derivative of h(x)=4+3f(x).

h(x)=ddx(h(x))=ddx(4+3f(x))=ddx[4+3f(x)]12

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

h(x)=12[4+3f(x)]121ddx(4+3f(x))=12[4+3f(x)]122(ddx(4)+ddx(3f(x)))=12[4+3f(x)]12(0+(3f(x)))

h(x)=3f(x)24+3f(x) (2)

Substitute x=1 in equation (2),

h(1)=3f(1)24+3f(1)

Substitute f(1)=7 and f(1)=4 in the above equation,

h(1)=3(4)24+3(7)=1224+21=12225=1225

Simplify further and obtain the values of h(1).

h(1)=1210=65

Therefore, the derivative of h(1)=4+3f(1) is h(1)=65_.

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