To find $f'(2)$, we first need to determine the derivative of the function $f(x) = 3x e^{-x}$.

Step 1: Differentiate $f(x) = 3x e^{-x}$

To find $f'(x)$, we’ll use the product rule for differentiation. Recall that if $u(x)$ and $v(x)$ are functions of $x$, then the derivative of their product is given by: $(uv)' = u'v + uv'$

In this case, let:

• $u = 3x$ and $u' = 3$
• $v = e^{-x}$ and $v' = -e^{-x}$ (since the derivative of $e^{-x}$ is $-e^{-x}$)

Now, applying the product rule: $f'(x) = u'v + uv' = (3)(e^{-x}) + (3x)(-e^{-x})$ Simplifying this, we get: $f'(x) = 3e^{-x} - 3x e^{-x}$ or equivalently, $f'(x) = e^{-x}(3 - 3x)$

Step 2: Substitute $x = 2$ into $f'(x)$

Now we need to find $f'(2)$: $f'(2) = e^{-2}(3 - 3 \cdot 2)$ $= e^{-2}(3 - 6)$ $= e^{-2}(-3)$ $= -3e^{-2}$

Step 3: Simplify the answer with no negative exponents

Since we are asked to express the answer without negative exponents: $f'(2) = -\frac{3}{e^2}$

Final Answer:

$f'(2) = -\frac{3}{e^2}$

Would you like more details on any part of this solution?

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Here are 5 related questions that might interest you:

1. How is the product rule applied in differentiation?
2. How do you find the derivative of functions involving $e^x$ terms?
3. What would be $f'(x)$ if $f(x) = 5x e^{2x}$?
4. How would you find the second derivative $f''(x)$ of $f(x) = 3x e^{-x}$?
5. Why is $e^x$ its own derivative, and what is unique about exponential functions?

Tip: When differentiating products, always identify each component and use the product rule step-by-step to avoid mistakes.