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 - Page 223: 45


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

Work Step by Step

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