Let f(x) = x² e^x. Find f′(1), rounded to 3 decimal places.

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Let $f (x) = x^{2} e^{x}$ . Find $f^{'} (1)$ , rounded to $3$ decimal places.

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Hint

Use the product rule: each factor contributes a term.

Solution

By the product rule, $f^{'} (x) = 2 x e^{x} + x^{2} e^{x} = (x^{2} + 2 x) e^{x}$ .
At $x = 1$ the bracket equals $1 + 2 = 3$ , so $f^{'} (1) = 3 e$ .
With $e \approx 2.71828$ , this is $3 \times 2.71828 = 8.15484$ .
Rounded to $3$ decimal places, $f^{'} (1) \approx 8.155$ .

Answer: $\approx 8.155$