Implicit differentiation of y = sin(xy)

Differentiate both sides with respect to $x$. The left side gives $\dfrac{dy}{dx}$. For the right side, apply the chain rule to $\sin(xy)$: $\dfrac{d}{dx}[\sin(xy)] = \cos(xy)\cdot \dfrac{d}{dx}[xy]$. Now use the product rule on $xy$: $\dfrac{d}{dx}[xy] = y + x\dfrac{dy}{dx}$. Putting this together gives $\dfrac{dy}{dx} = \cos(xy)\left(y + x\dfrac{dy}{dx}\right).$

Distribute $\cos(xy)$ on the right side to get $\dfrac{dy}{dx} = y\cos(xy) + x\cos(xy)\dfrac{dy}{dx}$. Move all terms with $dy/dx$ to the left side: $\dfrac{dy}{dx} - x\cos(xy)\dfrac{dy}{dx} = y\cos(xy)$. Factor out $dy/dx$ and solve: $\dfrac{dy}{dx} = \dfrac{y\cos(xy)}{1 - x\cos(xy)}.$

The product rule is used to differentiate xy, since it is a product of two functions of x. Applying it gives y plus x times dy/dx, which then gets multiplied by cosine of xy from the chain rule step. This is distinct from the chain rule, which was applied to the outer sine function.