## Calculus: Early Transcendentals (2nd Edition)

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