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: 43



Work Step by Step

Given: $y =x^{x}$ Taking logarithmic on both sides of the function$y =x^{x}$. Use logarithmic property $ln(x^{y})=ylnx$ $lny=xlnx$ Take implicit differentiation with respect to $x$. Apply product rule of differentiation. $\frac{d}{dx}(lny)=\frac{d}{dx}(xlnx)$ $\frac{1}{y}\frac{d}{dx}(y)=x\frac{d}{dx}(lnx)+lnx\frac{d}{dx}(x)$ $\frac{d}{dx}(y)=y[x\times(\frac{1}{x})+lnx\times1]$ Hence, $y'=x^{x}(1+lnx)$
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