Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises - Page 221: 13



Work Step by Step

Recall that $tan^{-1} x= \frac{1}{1+x^{2}}$ Then, using chain rule, we get $tan^{-1}(10x)= \frac{1}{(10x)^{2}+1}\cdot\frac{d(10x)}{dx}=\frac{10}{100x^{2}+1}$
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