# The value 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 3.5, Problem 17E
To determine

## To find: The value f′(1).

Expert Solution

The value of f(1)=1613.

### Explanation of Solution

Given:

The equation is f(x)+x2[f(x)]3=10 and f(1)=2.

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x)  are both differentiable function, then

dydx=dydududx.

(2) Product Rule: If f1(x) and f2(x) are both differentiable, then

ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x)).

Calculation:

Obtain the value of f(1).

f(x)+x2[f(x)]3=10

Differentiate with respect to x on both sides,

ddx(f(x)+x2[f(x)]3)=ddx(10)f(x)+ddx(x2[f(x)]3)=0

Apply the product rule (2) and simplify the terms,

f(x)+[x2ddx([f(x)]3)+[f(x)]3ddx(x2)]=0f(x)+[x2ddx([f(x)]3)+2x[f(x)]3]=0

Let u=f(x) and apply the chain rule,

f(x)+[x2ddx(u3)+2x[f(x)]3]=0f(x)+[x2[ddu(u3)dudx]+2x[f(x)]3]=0f(x)+[3x2u2dudx+2x[f(x)]3]=0

Substitute u=f(x) in the above equation,

f(x)+[3x2(f(x))2ddx(f(x))+2x[f(x)]3]=0f(x)+[3x2(f(x))2f(x)+2x[f(x)]3]=0

Substitute 1 for x in f(x),

f(1)+[3(1)2(f(1))2f(1)+2(1)[f(1)]3]=0f(1)+3(f(1))2f(1)+2(1)[f(1)]3=0

Substitute the value f(1)=2,

f(1)+3(2)2f(1)+(2(1)(2)3)=0f(1)+12f(1)+16=013f(1)+16=0f(1)=1613

Therefore, the value of f(1)=1613.

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