## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 96

#### Answer

$$y'=\frac{(\ln(\ln x)+1)(\ln x)^{\ln x}}{x}$$

#### Work Step by Step

$$y=(\ln x)^{\ln x}$$ We can use logarithmic differentiation to solve these problems. Take the natural logarithm of both sides: $$\ln y =\ln(\ln x)^{\ln x}$$ $$\ln y =\ln x\ln(\ln x)$$ Next, use implicit differentiation to solve for $y'$ with respect to $x$: $$\frac{y'}{y}=\Big((\ln x)'\ln(\ln x)\Big)+\Big(\ln x(\ln(\ln x))'\Big)$$ $$\frac{y'}{y}=\frac{\ln(\ln x)}{x}+(\ln x)\times\frac{1}{\ln x}\times(\ln x)'$$ $$\frac{y'}{y}=\frac{\ln(\ln x)}{x}+1\times\frac{1}{x}$$ $$\frac{y'}{y}=\frac{\ln(\ln x)}{x}+\frac{1}{x}=\frac{\ln(\ln x)+1}{x}$$ Now solve for $y'$: $$y'=\frac{(\ln(\ln x)+1)y}{x}$$ $$y'=\frac{(\ln(\ln x)+1)(\ln x)^{\ln x}}{x}$$

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