# 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 30E
To determine

## Find the derivative of the given function.

Expert Solution

The derivative of the given function is f'(t)=4t22t(t2+4)32 .

### Explanation of Solution

Given:

The given function is f(t)=tt2+4 .

Calculation:

f(t)=tt2+4=(tt2+4)12

Apply chain rule.

Let f=a12,a=tt2+4

ddtf(t)=dda(a12)ddt(tt2+4)

Use quotient rule.

(fg)'=gf'fg'g2

f'(t)=dda(a12)(t2+4)ddt(t)tddt(t2+4)(t2+4)2

Use derivative rule ddx(xn)=nxn1 .

f'(t)=12a(t2+4)t(2t)(t2+4)2f'(t)=12a(t2+42t2)(t2+4)2f'(t)=12a(4t2)(t2+4)2

Substitute the value of a=tt2+4 .

f'(t)=12tt2+4(4t2)(t2+4)2f'(t)=4t22t(t2+4)32

Hence the derivativeof the given function is f'(t)=4t22t(t2+4)32 .

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