Help with computing the $n$-th derivative of $f(x)=e^{-cx}\sum_{k=1}^{\infty}\frac{c^k}{(k+m)(k-1)!}(x-x_u)^k$ I've been dealing with a problem for sometime and after several attempts, I'm still not able to solve it so here it goes:
Define function
\begin{align}
f(x)=e^{-cx}\sum_{k=1}^{\infty}\frac{c^k}{(k+m)(k-1)!}(x-x_u)^k.
\end{align}
$f$ is defined for all $0\leq x \leq x_u$. From data and observations, $c>0$ and $m$ is a positive integer.
We are interested in having a closed-form of the $n$-th derivative of $f(x)$.
We tried different things but perhaps a good approach would be to write
\begin{align}
\partial_x^{(n)}f(x)=\sum_{k=1}^{\infty}\frac{c^k}{(k+m)(k-1)!}\frac{d^n}{dx^n}\left[(x-x_u)^ke^{-cx}\right].
\end{align}
Then I tried to use the general Leibniz rule combined with what I learned from here and here to deal with the last part but I cannot find a closed-from expression for $\partial_x^{(n)}f(x)$.
Any suggestions and/or answers would be much appreciated.

I inserted your series into WolframAlpha and it gave me this output (I need to check if it is true):
$$
f(x)=e^{-cx}\sum_{k=1}^{\infty}\frac{c^k(x-x_u)^k}{(k+m)(k-1)!} = \frac{(-1)^{m+1}e^{-cx}}{c^m(x-x_u)^m}\gamma(m+1, -c(x-x_u)),
$$
where $\gamma(a, x)$ is the lower incomplete gamma function.
For computing the $n$-th derivative of a product of functions, you can use the generalized Leibniz rule, where
$$
(f_1f_2\dots f_m)^{(n)} = \sum_{k_1+k_2+\dots +k_m=n}\binom{n}{k_1, k_2, \dots, k_m}\prod_{1 \leq t \leq m}f_t^{(k_t)}
$$
being $\sum_{t=1}^m k_t = n$ and
$$
\binom{n}{k_1, k_2, \dots, k_m} = \frac{n!}{k_1!k_2!\dots k_m!}.
$$
It will be useful to consider:
$$
\begin{aligned}
&\frac{\mathrm{d}^k\phantom{x}}{\mathrm{d}x^k}\left(e^{-cx}\right) = (-c)^ke^{-cx}, \\
&\frac{\mathrm{d}^k\phantom{x}}{\mathrm{d}x^k}\left(\frac{1}{(x-x_u)^m}\right) = \frac{(m+k-1)!}{(m-1)!}\frac{(-1)^k}{(x-x_u)^{m+k}}.
\end{aligned}
$$
For the incomplete gamma function, as $m+1$ is an integer, we can compute it (following this source) as
$$
\gamma(m+1, -c(x-x_u)) = m!\left[1-e^{c(x-x_u)}\sum_{l=0}^m\frac{(-1)^lc^l(x-x_u)^l}{l!}\right].
$$
Its $k$-th derivative can be computed using again the generalized Leibniz rule, where
$$
\begin{aligned}
&\frac{\mathrm{d}^k\phantom{x}}{\mathrm{d}x^k}\left(e^{c(x-x_u)}\right) = c^ke^{c(x-x_u)}, \\
&\frac{\mathrm{d}^k\phantom{x}}{\mathrm{d}x^k} \left(\sum_{l=0}^m\frac{(-1)^lc^l(x-x_u)^l}{l!}\right) = \sum_{l=k}^m\frac{(-1)^lc^ll!(x-x_u)^{l-k}}{(l-k)!}.
\end{aligned}
$$
If you want, I can go further with the expressions. Hope it is clear (and do not have any typos)!
