Product rule with polynomial and exponential factors

Let $h(x) = (x^2 - 4)e^{-x}$ . Find $h'(0)$ .

Hint

Write $h(x) = f(x)g(x)$ with $f(x) = x^2-4$ and $g(x) = e^{-x}$ . Use the product rule $h'(x) = f'(x)g(x) + f(x)g'(x)$ , remembering that $\frac{d}{dx}[e^{-x}] = -e^{-x}$ by the chain rule. Then substitute $x=0$ .

Step-by-step solution

Identify $f(x) = x^2 - 4$ and $g(x) = e^{-x}$ , so $f'(x) = 2x$ and $g'(x) = -e^{-x}$ (chain rule applied to the exponent $-x$ ). By the product rule, $h'(x) = f'(x)g(x) + f(x)g'(x) = 2x\,e^{-x} + (x^2-4)(-e^{-x})$ .

This simplifies to $h'(x) = e^{-x}\left(2x - (x^2-4)\right) = e^{-x}(-x^2+2x+4)$ . Substituting $x=0$ : $h'(0) = e^{0}(-0+0+4) = 4$ .

Product rule with polynomial and exponential factors

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