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: 49


${\frac{d{csc^{-1}{u}}}{du}}$ = -${\frac{1}{|u|{\sqrt{({(u^2-1)})}}}}$

Work Step by Step

${csc^{-1}{u}}$ value determined by the given equation: ${csc^{-1}{u}}$ = ${\frac{\pi}{2}}-{sec^{-1}{u}}$ on differentiating both sides: ${\frac{d{csc^{-1}{u}}}{du}}={\frac{d\pi}{2du}}-{\frac{d{sec^{-1}{u}}}{du}}$ differentiation of constant is zero: and ${\frac{d{\sec^{-1}{u}}}{du}={\frac{1}{|u|{\sqrt{({(u^2-1)})}}}}}$ so ${\frac{d{csc^{-1}{u}}}{du}}$ = -${\frac{1}{|u|{\sqrt{({(u^2-1)})}}}}$ thus the final answer is: ${\frac{d{csc^{-1}{u}}}{du}}$ = -${\frac{1}{|u|{\sqrt{({(u^2-1)})}}}}$
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