Calculus 8th Edition

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

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 27


$$y'=\tan^{-1} \left(4x\right)+\frac{4x}{16x^2+1}$$

Work Step by Step

Given $$y=x\tan^{-1}\left(4x\right)$$ Then \begin{align*} y'&=\frac{d}{dx}\left(x\right)\tan^{-1} \left(4x\right)+\frac{d}{dx}\left(\tan^{-1} \left(4x\right)\right)x\\ &=\tan^{-1} \left(4x\right)+\frac{4x}{16x^2+1} \end{align*}
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.