## Calculus (3rd Edition)

$$\frac{d}{dx}(x^2y^3)= 2xy^3+3x^2y^2\frac{dy}{dx}.$$
Usin the product rule: $(uv)'=uv'+u'v$ and using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, we get $$\frac{d}{dx}(x^2y^3)=y^3\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(y^3)\\=2xy^3+3x^2y^2\frac{dy}{dx}.$$