Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises - Page 198: 68


$f'(x) = g'(tan(\sqrt x)) \frac {sec^{2} (\sqrt x)} {2\sqrt x}$

Work Step by Step

$f(x) = g(tan(\sqrt x))$ Use chain rule here $f'(x) = \frac{d}{dx} g(tan(\sqrt x))$ $f'(x) = g'(tan(\sqrt x)) \frac{d}{dx} tan(\sqrt x))$ $f'(x) = g'(tan(\sqrt x)) sec^{2} (\sqrt x)\frac{d}{dx} \sqrt x$ $f'(x) = g'(tan(\sqrt x)) sec^{2} (\sqrt x)\frac{1}{2\sqrt x}$ $f'(x) = g'(tan(\sqrt x)) \frac {sec^{2} (\sqrt x)} {2\sqrt x}$
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.