Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 23


$$\frac{{dy}}{{dx}} = \frac{1}{{x\ln x}}$$

Work Step by Step

$$\eqalign{ & y = \ln \left( {\ln x} \right) \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {\ln x} \right)} \right] \cr & {\text{use the rule }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr & {\text{then}}{\text{,}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln x}}\frac{d}{{dx}}\left[ {\ln x} \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln x}}\left( {\frac{1}{x}} \right) \cr & {\text{simplify}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{x\ln x}} \cr} $$
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