University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 33


$\frac{1}{tan^{-1}{x}}$ $\frac{1}{{1+x^2}}$

Work Step by Step

Given that $y$ = $\ln tan^{-1}{x}$, on substituting $tan^{-1}{x}$ = $u$ and applying the chain rule derivative, we get: $\frac{dy}{dx}$ = $\frac{{d\ln{u}}}{du}$ $\times$ $\frac{du}{dx}$ or $\frac{dy}{dx}$ = $\frac{1}{u}$ $\times$ $\frac{du}{dx}$ so $\frac{dy}{dx}$ = $\frac{1}{tan^{-1}{x}}$ $\times$ $\frac{dtan^{-1}{x}}{dx}$ or $\frac{dy}{dx}$ = $\frac{1}{tan^{-1}{x}}$ $\times$ $\frac{1}{{1+x^2}}$ The final answer is: $\frac{1}{tan^{-1}{x}}$ $\frac{1}{{1+x^2}}$
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