# use part 1 of the fundamental theorem of calculus to find the derivative of the function.

BuyFind

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 5.4, Problem 14E
To determine

Expert Solution

## Answer to Problem 14E

The derivative of the function is h(x)=2x·1+x6

### Explanation of Solution

Given information: The function is h(x)=0x21+r3 dr

Let’s remind of the fundamental theorem of calculus part 1:

The fundamental theorem of calculus part 1: If f is continuous on [a,b] then the function of g defined by

g(x)=axf(t) dt where axb is continuous on [a,b] and differentiable on (a,b) and g(x)=f(x)

We have, h(x)=0x21+r3 dr

Since the upper limit of integration is not x , we apply the chain rule

Let , u=x2, then u=2x Consider the new function

f(u)

=0u1+r3 dr

Differentiation with respect to u,

dduf(x)=ddu[0u1+r3 dr]

f(u)=1+u3

So, f(u)=h(x)

f(x2)=h(x)

h(x)=[f(x2)]

h(x)=f(x2)·(x2)

h(x)=1+(x3)2·2x

h(x)=2x·1+x6

Thus, h(x)=2x·1+x6

Hence, the derivative of the function is h(x)=2x·1+x6

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