# 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 19E
To determine

Expert Solution

## Answer to Problem 19E

The derivative of the given function is dydt=8(2x5)3(8x25)348x(2x5)4(8x25)4 .

### Explanation of Solution

Given:

The given function is y=(2x5)4(8x25)3 .

Calculation:

y=(2x5)4(8x25)3

Apply product rule.

(fg)=f'g+fg'

dydt=ddt(2x5)4(8x25)3+(2x5)4ddt(8x25)3

Apply chain rule.

Let f=a4,a=2x5 and f=a3,a=8x25

dydt=[dda(a4)ddx(2x5)](8x25)3+(2x5)4[dda(a3)ddx(8x25)]

Apply difference rule.

dydt=[dda(a4){ddx(2x)ddx(5)}](8x25)3+(2x5)4[dda(a3){ddx(8x2)ddx(5)}] Use derivative rule.

ddx(xn)=nxn1 .

dydt=[4a3{20}](8x25)3+(2x5)4[(3a4){16x0}]dydt=8a3(8x25)3+(2x5)4(48xa4)

Substitute the value of a=2x5 and a=8x25 .

dydt=8(2x5)3(8x25)348x(2x5)4(8x25)4

Hence the derivativeof the given function is dydt=8(2x5)3(8x25)348x(2x5)4(8x25)4 .

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