Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 50

Answer

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

Work Step by Step

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