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 - Page 437: 53



Work Step by Step

Given: $y =(tanx)^{\frac{1}{x}}$ Taking logarithmic on both sides of the function $y =(tanx)^{\frac{1}{x}}$ $lny=\frac{1}{x} ln(tanx)$ Take implicit differentiation with respect to $x$. Apply product rule of differentiation. $\frac{d}{dx}(lny)=\frac{d}{dx}[\frac{1}{x} ln(tanx)]$ $\frac{1}{y}\frac{d}{dx}(y)=\frac{1}{x}\frac{d}{dx}[ln(tanx)]+ln(tanx)\frac{d}{dx}(\frac{1}{x})$ $\frac{d}{dx}(y)=y[\frac{1}{x}\frac{d}{dx}[ln(tanx)]+ln(tanx) (-\frac{1}{x^{2}})]$ To make the tedious calculations easier, we will solve term$\frac{d}{dx}[ln(tanx)]$ separately such as: $\frac{d}{dx}[ln(tanx)]=\frac{1}{tanx}\frac{d}{dx}(tanx)=\frac{1}{tanx}\frac{d}{dx}(sec^{2}x)$ Thus, $y'=(tanx)^{\frac{1}{x}}[\frac{1}{x}\frac{1}{tanx}\frac{d}{dx}(sec^{2}x)-\frac{1}{x^{2}}ln(tanx)]$ Hence, $y'=(tanx)^{\frac{1}{x}}[\frac{sec^{2}x}{xtanx}-\frac{1}{x^{2}}ln(tanx)]$
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