Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 34

Answer

$\frac{1}{x[1+(\ln x)^2]}$

Work Step by Step

$y= tan^{-1}(\ln x)$ Derive using the chain rule: $\frac{dy}{dx}=\frac{(\frac{1}{x})}{1+(\ln x)^2}$ =$\frac{1}{x[1+(\ln x)^2]}$
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