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: 51

Answer

$y'=(cos)^{x}[-xtanx+ln(cosx)]$

Work Step by Step

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