1. **State the problem:** Find the derivative $y'$ of the function $y = x^2 \cos x$. 2. **Recall the formula:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $u(x) = x^2$ and $v(x) = \cos x$. 3. **Differentiate each part:** - $u'(x) = \frac{d}{dx} x^2 = 2x$ - $v'(x) = \frac{d}{dx} \cos x = -\sin x$ 4. **Apply the product rule:** $$y' = u'(x)v(x) + u(x)v'(x) = 2x \cos x + x^2 (-\sin x)$$ 5. **Simplify the expression:** $$y' = 2x \cos x - x^2 \sin x$$ 6. **Final answer:** $$\boxed{y' = 2x \cos x - x^2 \sin x}$$ This derivative tells us the rate of change of $y$ with respect to $x$ for the given function.