# The value g ′ ( 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 3.5, Problem 18E
To determine

## To find: The value g′(0).

Expert Solution

The value of g(0)=0.

### Explanation of Solution

Given:

The equation is g(x)+xsing(x)=x2.

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 g(0).

g(x)+xsing(x)=x2

Differentiate with respect to x on both sides,

ddx(g(x)+xsing(x))=ddx(x2)ddx(g(x))+ddx(xsing(x))=2xg(x)+ddx(xsing(x))=2x

Apply the product rule and simplify the terms,

g(x)+[xddx(sing(x))+sing(x)ddx(x)]=2xg(x)+xddx(sing(x))+sing(x)=2x

Let u=g(x) and apply the chain rule (1),

g(x)+xddx(sinu)+sinu=2xg(x)+x[ddu(sinu)dudx]+sinu=2xg(x)+xcosududx+sinu=2x

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

g(x)+x(cosg(x))ddx(g(x))+sing(x)=2xg(x)+x(cosg(x))g(x)+sing(x)=2x

Substitute 0 for x in g(x),

g(0)+x(cos0)g(0)+sin0=2(0)g(0)+01g(0)+0=0g(0)=0

Therefore, the value of g(0)=0.

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