To find the derivative of the function $y = x^2 \cos x$, we will apply the product rule of differentiation, which states:

$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$

Here:

• $u(x) = x^2$,
• $v(x) = \cos x$.

Steps:

1. Differentiate $u(x) = x^2$: $u'(x) = 2x.$

2. Differentiate $v(x) = \cos x$: $v'(x) = -\sin x.$

3. Apply the product rule: $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x).$

Substitute the values: $\frac{dy}{dx} = (2x)(\cos x) + (x^2)(-\sin x).$

Simplify: $\frac{dy}{dx} = 2x \cos x - x^2 \sin x.$

Final Answer:

$\frac{dy}{dx} = 2x \cos x - x^2 \sin x.$

Would you like a detailed explanation of any step or clarification?

Here are 5 related questions to extend your understanding:

1. What is the second derivative of $y = x^2 \cos x$?
2. How would the derivative change if $y = x^3 \cos x$?
3. Can you find critical points of $y = x^2 \cos x$ by setting $\frac{dy}{dx} = 0$?
4. How does the product rule compare to the chain rule for this function?
5. What happens to $\frac{dy}{dx}$ as $x \to 0$ or $x \to \infty$?

Tip: Always simplify derivatives as much as possible for easier application in subsequent problems!