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: 40



Work Step by Step

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