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To differentiate the given function \( y = x^2 e^{-ax} \) with respect to \( x \), we will use the product rule. The product rule states:

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

For the function \( y = x^2 e^{-ax} \), we can identify:
- \( u(x) = x^2 \)
- \( v(x) = e^{-ax} \)

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

### Step 2: Differentiate \( v(x) = e^{-ax} \)
Using the chain rule, where \( v(x) = e^{f(x)} \) and \( f(x) = -ax \):
\[
v'(x) = \frac{d}{dx} (e^{-ax}) = e^{-ax} \cdot \frac{d}{dx} (-ax) = -ae^{-ax}
\]

### Step 3: Apply the product rule
\[
\frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Substitute the derivatives:
\[
\frac{dy}{dx} = (2x) \cdot e^{-ax} + (x^2) \cdot (-ae^{-ax})
\]

### Step 4: Simplify the expression
\[
\frac{dy}{dx} = 2x e^{-ax} - ax^2 e^{-ax}
\]

Therefore, the derivative of \( y = x^2 e^{-ax} \) with respect to \( x \) is:
\[
\frac{dy}{dx} = e^{-ax}(2x - ax^2)
\]

Would you like more details on this process or have any questions?

### Related Questions:
1. What is the derivative of \( y = x^n e^{-ax} \) for a general \( n \)?
2. How do you find the second derivative of \( y = x^2 e^{-ax} \)?
3. Can you explain how the chain rule was applied to \( v(x) = e^{-ax} \)?
4. What are some real-world applications of functions like \( y = x^2 e^{-ax} \)?
5. How does the parameter \( a \) affect the behavior of the function and its derivative?

### Tip:
When differentiating products of functions, always apply the product rule first, and remember to simplify the expression afterward for easier interpretation or further analysis.

Learn how to differentiate the function y = x^2 e^{-ax} with respect to x using the product rule and chain rule. This detailed step-by-step solution covers key differentiation techniques and their applications in calculus.

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