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: 52


${\frac{d{cot^{-1}{u}}}{du}}$ = -${\frac{1}{1+u^2}}\frac{du}{dx}$

Work Step by Step

We know that: ${cot^{-1}{u}}$ = ${\frac{\pi}{2}}-{tan^{-1}{u}}$ differentiating both sides: ${\frac{d{cot^{-1}{u}}}{du}}={\frac{d\pi}{2du}}-{\frac{d{tan^{-1}{u}}}{du}}$ differentiation of constant is zero and ${\frac{d{\tan^{-1}{u}}}{du}={\frac{1}{1+u^2}}}$ thus: ${\frac{d{cot^{-1}{u}}}{du}}$ = -${\frac{1}{1+u^2}}\frac{du}{dx}$
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