To differentiate the function \( f(x) = x \cdot e^{-2x} \), we need to apply the **product rule**, since it's a product of two functions: \( x \) and \( e^{-2x} \).

The product rule states that:
\[
(f \cdot g)' = f' \cdot g + f \cdot g'
\]
where \( f(x) = x \) and \( g(x) = e^{-2x} \).

1. Differentiate \( f(x) = x \):
   \[
   f'(x) = 1
   \]

2. Differentiate \( g(x) = e^{-2x} \). Using the chain rule:
   \[
   g'(x) = e^{-2x} \cdot (-2) = -2e^{-2x}
   \]

Now, applying the product rule:
\[
\frac{d}{dx} \left( x \cdot e^{-2x} \right) = 1 \cdot e^{-2x} + x \cdot (-2e^{-2x})
\]
Simplifying:
\[
\frac{d}{dx} \left( x \cdot e^{-2x} \right) = e^{-2x} - 2x \cdot e^{-2x}
\]
Thus, the derivative is:
\[
f'(x) = e^{-2x}(1 - 2x)
\]

Would you like more details or explanations on any part of the solution?

Here are 5 questions related to this topic:
1. How is the chain rule applied in differentiation?
2. What is the product rule, and when do we use it?
3. How do you simplify expressions involving exponentials after applying the product rule?
4. Can you differentiate more complex functions like \(x^2 e^{-2x}\)?
5. How do you compute second derivatives using the product and chain rules?

**Tip:** Always simplify the expression after applying the product or chain rule to make the derivative clearer and easier to interpret.

This page explains how to differentiate the function x * e^(-2x) using the product and chain rules. Step-by-step instructions cover key calculus concepts, such as the product rule for differentiating two functions and the chain rule for exponentials. Suitable for Grades 11-12 and early university calculus courses.

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