## Calculus 8th Edition

$\frac{dy}{dx} = \frac{-\sin y -y\cos x}{x\cos y + \sin x}$
$x\sin y + y\sin x = 1$ Differentiate both sides with respect to x $\sin y + \frac{dy}{dx}x\cos y + \frac{dy}{dx}\sin x + y \cos x = 0$ Isolate $\frac{dy}{dx}$ $\frac{dy}{dx}x\cos y+ \frac{dy}{dx}\sin x = -\sin y - y\cos x$ $\frac{dy}{dx} (x\cos y + \sin x) = (-\sin y - y\cos x)$ $\frac{dy}{dx} = \frac{-\sin y -y\cos x}{x\cos y + \sin x}$