Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises: 14

Answer

$h'(x) = \frac{x}{2\sqrt x(x^2+1)}$

Work Step by Step

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $h(x) = \int ^{\sqrt{x}}_{1}\frac{z^2}{z^4+1}dz$ Since the upper bound is a function we need to use the chain rule: $u = \sqrt{x}$ $u' = \frac{1}{2\sqrt x}$ Since $h(x) = \int^{b}_{a}f(t)dt$ $h'(x) = f(x)$ $h'(u) = \frac{u^2}{u^4+1}u'$ $h'(x) = \frac{\sqrt x^2}{\sqrt x^4+1}\times \frac{1}{2\sqrt x}$ $h'(x) = \frac{x}{x^2+1}\times \frac{1}{2\sqrt x}$ $h'(x) = \frac{x}{2\sqrt x(x^2+1)}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.