Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 464: 36


$y'=x^{cosx}(\frac{cosx}{x}-lnx sinx)$

Work Step by Step

Given: $y =x^{cos x}$ Taking logarithmic on both sides of the function $y =x^{cos x}$ $lny=cosx .lnx$ Take implicit differentiation with respect to $x$. Apply product rule of differentiation. $\frac{d}{dx}(lny)=\frac{d}{dx}(cosx. lnx)$ $\frac{1}{y}\frac{d}{dx}(y)=cosx\frac{d}{dx}(lnx)+lnx\frac{d}{dx}(cosx)$ $\frac{d}{dx}(y)=y[cosx\times(\frac{1}{x})+lnx(-sinx)]$ Hence, $y'=x^{cosx}(\frac{cosx}{x}-lnx sinx)$
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.