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: 39



Work Step by Step

Given that $y$ = $\tan^{-1}{\sqrt{({x^2-1})}}+\csc^{-1}{x}$, on applying the product rule of the derivative, we get: $\frac{dy}{dx}$ = ${\frac{1}{{({\sqrt{({x^2-1})})}^2+1}}\frac{d(\sqrt{x^2-1})}{dx}-\frac{1}{|x|\sqrt{x^2-1}}}$ or $\frac{dy}{dx}$ = ${ \frac {2x}{2x^2\sqrt{x^2-1}}-{\frac{1}{|x|\sqrt{(x^2-1)}}}}$ On simplifying: $\frac{dy}{dx}$ = ${ \frac {(1)}{x\sqrt{x^2-1}}}$ - ${ \frac {(1)}{x\sqrt{x^2-1}}}$ =0 The final answer is: = 0
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