# 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 13E
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 u(x)=(1x2)·arctan(1x)

### Explanation of Solution

Given information: The function is h(x)=21xarctan t dt

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)=21xarctan t dt .

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

Let u=1x , u=1x2

Consider the new function

f(u)=2uarctan t dt

Differentiation with respect to u,

df(u)du=ddu[2uarctan t dt]f(u)=arc tan uf(u)=u(x)u(x)=f(1x).u(x)=[f(1x)]u(x)=f(1x)(1x)u(x)=arctan(1x)·(1x)u(x)=arctan(1x)·(1x2)u(x)=(1x2)·arctan(1x)

Thus u(x)=(1x2)·arctan(1x)

Hence, The derivative of the function is u(x)=(1x2)·arctan(1x)

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