# use part 1 of the fundamental theorem of calculus to find the derivative of the 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 5.4, Problem 17E
To determine

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

Expert Solution

The derivative of the function is g(x)=8x2-24x2+1+18x2-29x2+1

### Explanation of Solution

Given information: The function is g(x)=2x3xu2-1u2+1du

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, g(x)=2x3xu2-1u2+1du

First, we will use the properties of definite integral to match the form in fundamental theorem of calculus part 1

2x3xu2-1u2+1du

=2x0u2-1u2+1du+03xu2-1u2+1duwhere 2x3xf(u) du=2x0f(u)du+03xf(u) du

Now again, g(x)

=2x0u2-1u2+1du

we will use the properties of definite integral to match the form in fundamental theorem of calculus part 1

2x0u2-1u2+1du

=02xu2-1u2+1du where f(x)=-f(x)

g(x)=02xu2-1u2+1du

so, g(x)=02xu2-1u2+1du

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

Let, v=2x, then v=2

h(v)=0vu2-1u2+1du

Differentiation with respect to v

ddvh(v)=ddv[0vu2-1u2+1du]

h(v)=v2-1v2+1

Thus, h(v)=g(x)

g(x)=[h(2x)]

g(x)=h(2x)·(2x)

g(x)=(2x)2-1(2x)2+1·2

g(x)=8x2-24x2+1

Again, g(x)=03xu2-1u2+1du

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

Let, w=3x, then v=3

h(w)=0wu2-1u2+1du

Differentiation with respect to w

ddwh(w)=ddw[0vu2-1u2+1du]

h(w)=w2-1w2+1

Thus, h(w)=g(x)

g(x)=[h(3x)]

g(x)=h(3x)·(3x)

g(x)=(3x)2-1(3x)2+1·3

g(x)=18x2-29x2+1

Hence , The derivative of the function is g(x)=8x2-24x2+1+18x2-29x2+1

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