Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 17


$$y' = \frac{-tan\sqrt x}{2}$$

Work Step by Step

$$y = \int\limits^{\frac{\pi}{4}}_{\sqrt x} {\theta tan\theta}d\theta$$ Switch the bounds, making the integral negative. $$y = -\int\limits_{\frac{\pi}{4}}^{\sqrt x} {\theta tan\theta}d\theta$$ Substitute in the upper bound for $\theta$ and multiply by the derivative of the upper bound. $$y' = - (\sqrt{x}tan(\sqrt x))(\sqrt x)'$$ Simplify. $$y' = - (\sqrt{x}tan(\sqrt x))(\frac{1}{2\sqrt x})'$$ $$y' = \frac{-tan\sqrt x}{2}$$
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.