## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

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

#### Answer

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

#### 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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