Question

Differentiate (x^2)(cos^2x)

Answer 1

To differentiate the function \( y = (x^2)(\cos^2 x) \), we will use the product rule. The product rule states that if you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), then the derivative \( y' \) is given by:

\[
y' = u'v + uv'
\]

In this case, let:
- \( u = x^2 \)
- \( v = \cos^2 x \)

First, we find the derivatives \( u' \) and \( v' \).

1. **Differentiate \( u \):**
\[
u' = \frac{d}{dx}(x^2) = 2x
\]

2. **Differentiate \( v = \cos^2 x \):**
We can apply the chain rule here. Let \( w = \cos x \), then \( v = w^2 \) and:
\[
v' = 2w \frac{dw}{dx} = 2\cos x (-\sin x) = -2\cos x \sin x = -\sin(2x) \quad \text{(using the double angle identity)}
\]

Now, we can apply the product rule:
\[
y' = u'v + uv'
\]

Substituting the values we found:
- \( u' = 2x \)
- \( v = \cos^2 x \)
- \( u = x^2 \)
- \( v' = -\sin(2x) \)

Thus,
\[
y' = (2x)(\cos^2 x) + (x^2)(-\sin(2x))
\]

This simplifies to:
\[
y' = 2x \cos^2 x - x^2 \sin(2x)
\]

So, the derivative of \( y = (x^2)(\cos^2 x) \) is:
\[
\boxed{2x \cos^2 x - x^2 \sin(2x)}
\]