University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 29

Answer

$$y'=\frac{1}{x\ln x}$$

Work Step by Step

$$y=\ln(\ln x)$$ Recall the following Derivative Rules: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ We have $$y'=\Big(\ln(\ln x)\Big)'=\frac{1}{\ln x}(\ln x)'$$ $$y'=\frac{1}{\ln x}\times\frac{1}{x}=\frac{1}{x\ln x}$$
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