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: 33

Answer

$y'=\frac{x(cotx)}{ln4}+log_{4}(sinx)$

Work Step by Step

Differentiate y with respect to x. $y'=\frac{1}{ln4}\frac{d}{dx}(xlog(sinx))$ $=\frac{1}{ln4}[x\frac{d}{dx}(logsinx)]+log_{4}(sinx)\frac{d}{dx}(x)$ $=\frac{1}{ln4}[\frac{x}{sinx}(cosx)]+log_{4}(sinx)$ Hence, $y'=\frac{x(cotx)}{ln4}+log_{4}(sinx)$
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