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The nth derivative of $x e^{x}$ vanishes when

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x = –n

x = 0

x = –1

x = n

answer is C.

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Detailed Solution

$f (x) = x e^{x} \Rightarrow f^{'} (x) = e^{x} + x e^{x} \Rightarrow f^{''} (x) = e^{x} + e^{x} + x e^{x}$

$f^{n} (x) = n e^{x} + x e^{x} . \Rightarrow Now, f^{n} (x) = 0 \Rightarrow x = - n$

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