Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises - Page 221: 4


$\frac{d}{dx}\sin^{-1}x = -\left(\frac{d}{dx}\cos^{-1}x\right)$

Work Step by Step

$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt {1-x^2}}$ $\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt {1-x^2}}$ Thus, $\frac{d}{dx}\sin^{-1}x = -\left(\frac{d}{dx}\cos^{-1}x\right)$
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