Implicit differentiation of sin(xy) = x + y

Differentiate both sides with respect to $x$. The left side requires the chain rule: $\frac{d}{dx}[\sin(xy)] = \cos(xy)\cdot \frac{d}{dx}[xy]$, and since $xy$ is a product, $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$ by the product rule. The right side gives $\frac{d}{dx}[x+y] = 1 + \frac{dy}{dx}$. So the equation becomes $\cos(xy)\left(y + x\frac{dy}{dx}\right) = 1 + \frac{dy}{dx}.$

Distribute $\cos(xy)$ and collect all terms with $\frac{dy}{dx}$ on one side: $y\cos(xy) + x\cos(xy)\frac{dy}{dx} - \frac{dy}{dx} = 1.$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}\left(x\cos(xy) - 1\right) = 1 - y\cos(xy),$ so dividing gives $\frac{dy}{dx} = \frac{1 - y\cos(xy)}{x\cos(xy) - 1}.$