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: 42

Answer

$-tan^{-1}(\frac{x}{2})$

Work Step by Step

$y=\ln (x^2+4)-x \tan^{-1}(\frac{x}{2})$ Derive using the chain rule: $\frac{dy}{dx}=\frac{2x}{x^2+4}-tan^{-1}(\frac{x}{2})-x[\frac{\frac{1}{2}}{1+(\frac{x}{2})^2}]$ $=\frac{2x}{x^2+4}-tan^{-1}(\frac{x}{2})-\frac{2x}{4+x^2}$ $=-tan^{-1}(\frac{x}{2})$
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