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: 48

Answer

$y'=(sinx)^{lnx}[lnx cotx+\frac{ln(sinx)}{x}]$

Work Step by Step

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