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: 34


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

Work Step by Step

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