1. **State the problem:** Find the derivative of the function $$y = x^2 \sin x$$. 2. **Formula used:** We will use the product rule for derivatives, which states: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $u(x) = x^2$ and $v(x) = \sin x$. 3. **Find derivatives of each part:** - Derivative of $u(x) = x^2$ is $u'(x) = 2x$. - Derivative of $v(x) = \sin x$ is $v'(x) = \cos x$. 4. **Apply the product rule:** $$y' = u'(x)v(x) + u(x)v'(x) = 2x \sin x + x^2 \cos x$$ 5. **Final answer:** $$\boxed{y' = 2x \sin x + x^2 \cos x}$$