Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 48

Answer

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

Work Step by Step

Given: $y =x^{cos x}$ Taking logarithmic on both sides of the function$y =x^{cos x}$. Use logarithmic property $ln(x^{y})=yln(cos x)$ $lny=cosx .lnx$ 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, $y'=x^{cosx}(\frac{cosx}{x}-sinx.lnx)$
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