Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 30


$$ g'(x)= \frac{1}{x(1+(\ln x)^2)}.$$

Work Step by Step

Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$ Recall that $(\ln x)'=\dfrac{1}{x}$ Since $ g(x)=\tan^{-1}\ln x $, then the derivative, using the chain rule, is given by $$ g'(x)= \frac{1}{1+(\ln x)^2}(\ln x)'= \frac{1/x}{1+(\ln x)^2}= \frac{1}{x(1+(\ln x)^2)}.$$
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