To compute the derivative of $x \cos^2(x)$, we'll apply the product rule and the chain rule. Here's the step-by-step process:

Given:

$f(x) = x \cos^2(x)$

Step 1: Recall the product rule

$\frac{d}{dx}[u \cdot v] = u'v + uv'$ Here, $u = x$ and $v = \cos^2(x)$.

Step 2: Differentiate each part

1. Derivative of $u = x$: $u' = 1$

2. Derivative of $v = \cos^2(x)$: Using the chain rule: $\frac{d}{dx}[\cos^2(x)] = 2\cos(x) \cdot (-\sin(x)) = -2\cos(x)\sin(x)$

Step 3: Apply the product rule

$\frac{d}{dx}[x \cos^2(x)] = (1)(\cos^2(x)) + (x)(-2\cos(x)\sin(x))$

Simplify: $\frac{d}{dx}[x \cos^2(x)] = \cos^2(x) - 2x\cos(x)\sin(x)$

Final Answer:

$\boxed{\frac{d}{dx}[x \cos^2(x)] = \cos^2(x) - 2x\cos(x)\sin(x)}$

Alternative Form (Optional):

Using the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$, the derivative can also be expressed as: $\boxed{\frac{d}{dx}[x \cos^2(x)] = \cos^2(x) - x\sin(2x)}$

Let me know if you'd like further clarification! 😊

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Questions for Expansion:

1. What is the derivative of $x\sin^2(x)$?
2. How would the derivative change if it were $x^2\cos^2(x)$?
3. Can you simplify $-2x\cos(x)\sin(x)$ further using trigonometric identities?
4. What happens if you integrate $x \cos^2(x)$?
5. How does the double-angle identity simplify trigonometric derivatives in general?

Tip:

Always check for trigonometric identities to simplify derivatives when $\sin(x)$ and $\cos(x)$ appear together!