Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 46

Answer

$(sec\sqrt{\theta})[\frac{tan \sqrt{\theta}tan(\frac{1}{\theta})}{2\sqrt{\theta}}-\frac{sec^2(\frac{1}{\theta})}{\theta^2}]$

Work Step by Step

$r=(sec\sqrt{\theta}tan(\frac{1}{\theta}))$ Apply the chain rule: $\frac{dr}{d\theta}=(sec \sqrt{\theta})(sec^2 \frac{1}{\theta})(\frac{-1}{\theta^2})+tan(\frac{1}{\theta}(sec\sqrt{\theta})tan \sqrt{\theta})(\frac{1}{2\sqrt{\theta}})$ =$\frac{-1}{\theta^2}sec\sqrt{\theta}sec^2(\frac{1}{\theta})+\frac{1}{2\sqrt{\theta}}tan(\frac{1}{\theta})sec \sqrt{\theta} tan \sqrt{\theta}$ =$(sec\sqrt{\theta})[\frac{tan \sqrt{\theta}tan(\frac{1}{\theta})}{2\sqrt{\theta}}-\frac{sec^2(\frac{1}{\theta})}{\theta^2}]$
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.