Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 56

Determine the thousandth derivative of $f(x) = xe^{-x}$.


$
\begin{equation}
\begin{aligned}

\text{if } f(x) =& xe^{-x}, \text{ then by using Product Rule}
\\
\\
f'(x) =& xe^{-x} (-1) + e^{-x} = -xe^{-x} + e^{-x}
\\
\\
\text{then } &
\\
\\
f''(x) =& -2e^{-x} + xe^{-x}
\\
\\
f'''(x) =& 3e^{-x} - xe^{-x}
\\
\\
f^4(x) =& -4e^{-x} + xe^{-x}


\end{aligned}
\end{equation}
$



So, by looking to these pattern we can formulate the derivative of the function for any number $n$.


$
\begin{equation}
\begin{aligned}

f^{(n)} (x) =& (-1)^{n + 1} ne^{-x} + (-1)^n xe^{-x}
\\
\\
f^{(n)} (x) =& (-1)^n (x - n)e^{-x}

\end{aligned}
\end{equation}
$


Therefore,


$
\begin{equation}
\begin{aligned}

f^{(1000)} (x) =& (-1)^{1000} (x - 1000) e^{-x}
\\
\\
f^{(1000)} (x) =& (1)(x - 1000) e^{-x}
\\
\\
f^{(1000)} (x) =& xe^{-x} - 1000 e^{-x}

\end{aligned}
\end{equation}
$