# 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 20E
To determine

## Find the derivative of the given function.

Expert Solution

The derivative of the given function is h'(t)=12t2(t41)2(t3+1)3[2t4+t1] .

### Explanation of Solution

Given:

The given function is h(t)=(t41)3(t3+1)4 .

Calculation:

h(t)=(t41)3(t3+1)4

Apply product rule.

(fg)=f'g+fg'

ddth(t)=ddt(t41)3(t3+1)4+(t41)3ddt(t3+1)4

Apply chain rule.

Let f=a3,a=t41 and f=a4,a=t3+1

h'(t)=[dda(a3)ddt(t41)](t3+1)4+(t41)3[dda(a4)ddt(t3+1)]

Apply sum/difference rule.

f±g=f'±g'

h'(t)=[dda(a3){ddt(t4)ddt(1)}](t3+1)4+(t41)3[dda(a4){ddt(t3)+ddt(1)}] Use derivative rule.

ddx(xn)=nxn1 .

h'(t)=[3a2{4t30}](t3+1)4+(t41)3[4a3{3t2+0}]h'(t)=[12a2t3](t3+1)4+(t41)3[12a3t2]

Substitute the value of a=t41 and a=t3+1 .

h'(t)=[12(t41)2t3](t3+1)4+(t41)3[12(t3+1)3t2]h'(t)=12t3(t41)2(t3+1)4+12t2(t41)3(t3+1)3h'(t)=12t2(t41)2(t3+1)3[t(t3+1)+(t41)]h'(t)=12t2(t41)2(t3+1)3[t4+t+t41]h'(t)=12t2(t41)2(t3+1)3[2t4+t1]

Hence the derivativeof the given function is h'(t)=12t2(t41)2(t3+1)3[2t4+t1] .

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!