# 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 16E
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

### Explanation of Solution

Given information: The function is

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) .

First , we will use the properties of the definite integral to make the integral match the form in the fundamental theorem, so we have, ex0sin3 t dt=0exsin3 t dt [ where f(x)=-f(x) ]

y=0exsin3 t dt

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

Let, y=f(u), ex=u

Consider the new function

f(u)

=0usin3 t dt

Differentiation with respect to u,

dduf(u)=ddu[0usin3 t dt]

f(u)=-sin3u

So y=f(u)

y=[f(ex)]

y=f(ex)·(ex)

Hence, The derivative of the function is y=f(ex)·(ex) .

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