Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 425: 20


$$y' = \frac{1}{{x\left( {\ln x} \right)\ln \left( {\ln x} \right)}}$$

Work Step by Step

$$\eqalign{ & y = \ln \left( {\ln \left( {\ln x} \right)} \right) \cr & {\text{differentiate}} \cr & y' = \left[ {\ln \left( {\ln \left( {\ln x} \right)} \right)} \right]' \cr & {\text{chain rule}}{\text{, recall that }}\left( {\ln u} \right)' = \frac{{u'}}{u} \cr & y' = \frac{1}{{\ln \left( {\ln x} \right)}}\left( {\ln \left( {\ln x} \right)} \right)' \cr & y' = \frac{1}{{\ln \left( {\ln x} \right)}}\frac{1}{{\ln x}}\left( {\ln x} \right)' \cr & y' = \frac{1}{{\ln \left( {\ln x} \right)}}\frac{1}{{\ln x}}\left( {\frac{1}{x}} \right) \cr & {\text{simplify}} \cr & y' = \frac{1}{{x\left( {\ln x} \right)\ln \left( {\ln x} \right)}} \cr} $$
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