# 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 15E
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 tanx+tanx·sec2x .

### Explanation of Solution

Given information: The function is tanx+tanx·sec2x

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, y=0tan xt+t dt

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

Let , y=f(u) , u=tanx

Then, u=tanx , u=sec2x

Consider the new function

f(u)=0ut+t dt

Differentiation with respect to u,

dduf(u)=ddu[0ut+t dt]

dduf(u)

=u+u

f(u)=u+u

Now y=f(u)

y=[f(tanx)u]

y=f(tanx)·u

y=tanx+tanx·sec2x

Hence, the derivative is tanx+tanx·sec2x .

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