# Find the derivative of the given function.

### 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 24E
To determine

Expert Solution

## Answer to Problem 24E

The derivative of the given function is G'(y)=5y9(y+2)(y+1)6 .

### Explanation of Solution

Given:

The given function is G(y)=(y2y+1)5 .

Calculation:

G(y)=(y2y+1)5

Apply chain rule.

Let f=a5,a=y2y+1

ddyG(y)=dda(a5)ddy(y2y+1)

Use quotient rule.

(fg)'=gf'fg'g2

G'(y)=dda(a5)(y+1)ddy(y2)(y2)ddy(y+1)(y+1)2

Use derivative rule.

ddx(xn)=nxn1 .

G'(y)=5a4(y+1)2yy2(y+1)2G'(y)=5a4(2y2+2yy2)(y+1)2G'(y)=5a4(y2+2y)(y+1)2

Substitute the value of a=y2y+1 .

G'(y)=5(y2y+1)4y(y+2)(y+1)2G'(y)=5y9(y+2)(y+1)6

Hence the derivativeof the given function is G'(y)=5y9(y+2)(y+1)6 .

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