The n^{th} derivative of x e^x vanishes when | vedclass

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(C) Let $f(x) = x e^x$. By using the product rule for differentiation: $f'(x) = \frac{d}{dx}(x) e^x + x \frac{d}{dx}(e^x) = e^x + x e^x = (1 + x) e^x$

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$x = -n$

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(C) Let $f(x) = x e^x$. By using the product rule for differentiation: $f'(x) = \frac{d}{dx}(x) e^x + x \frac{d}{dx}(e^x) = e^x + x e^x = (1 + x) e^x$. $f''(x) = \frac{d}{dx}(1 + x) e^x + (1 + x) \frac{d}{dx}(e^x) = e^x + (1 + x) e^x = (2 + x) e^x$. $f'''(x) = \frac{d}{dx}(2 + x) e^x + (2 + x) \frac{d}{dx}(e^x) = e^x + (2 + x) e^x = (3 + x) e^x$. Following this pattern,the $n^{th}$ derivative is given by: $f^{(n)}(x) = (n + x) e^x$. For the $n^{th}$ derivative to vanish,we set $f^{(n)}(x) = 0$: $(n + x) e^x = 0$. Since $e^x$ is never zero for any real $x$,we must have: $n + x = 0 \implies x = -n$.

The $n^{th}$ derivative of $x e^x$ vanishes when

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