Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 37


$$ y'=\tan^{-1} x+\frac{x}{1+x^2}.$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$ Since $ y=x\tan^{-1} x $, the derivative $ y'$ by using the product rule, is given by $$ y'=\tan^{-1} x+\frac{x}{1+x^2}.$$
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