Answer
$f^{1000}(x)=-1000e^{-x}+xe^{-x}$
Work Step by Step
$f(x)=xe^{-x}$
$f'(x)=\frac{d}{dx}xe^{-x}$
$f'(x)=e^{-x}\frac{d}{dx}(x)+x\frac{d}{dx}(e^{-x})$
$f'(x)=e^{-x}-x(e^{-x})$
Again differentiate with rest to x, we get
$f''(x)=-2e^{-x}+x(e^{-x})$
Take third derivative of the function.
$f'''(x)=3e^{-x}-x(e^{-x})$
Proceeding this process in the same way up to nth derivative, we observe the expression for nth derivative can be calculated as follows:
$f^{n}(x)=(-1)^{(n-1)}(ne^{-x})+(-1)^{n}(xe^{-x})$
Take n = 1000, therefore,
$f^{1000}(x)=(-1)^{(1000-1)}(1000e^{-x})+(-1)^{1000}(xe^{-x})$
$f^{1000}(x)=(-1)^{(999)}(1000e^{-x})+(-1)^{1000}(xe^{-x})$
Hence, $f^{1000}(x)=-1000e^{-x}+xe^{-x}$
