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.

Consider this answer, and write your function not as you did but as
$$\partial^{n}f(x) = \sum_{k = 1}^{+\infty} \frac{1}{(k+m)(k-1)!}\frac{\text{d}^n}{\text{d}x^n}\color{red}{\left[e^{-cx}(c(x - x_u))^k\right]}$$
The red term is nothing but
$$(-1)^k\color{blue}{e^{-cx}(c(x_u-x))^k)}$$
Where the blue term is the one analysed in the liked question above.
Follow the same procedure, and then you will get the $nth$ derivative expression.
Further Details
Copying from the other answer:
$$\begin{align}
& \text{when}\quad n\le k: \frac{d^nf_k(x)}{dx^n} = (-1)^n\sum_{i=0}^{n} \binom n i (c)^{n-i}[k]_if_{k-i}(x) \\
\end{align}$$
This ensures you the closed expression for that blue term, whence the derivation is ended. You can then write down glueing all the pieces:
$$\partial^nf(x) = \sum_{k=1}^{+\infty}\frac{1}{(k+m)(k-1)!}(-1)^{k+n}\sum_{i=0}^{n} \binom n i (c)^{n-i}[k]_if_{k-i}(x)$$
Where we define $[k]_m = k(k-1)\cdots(k-m)$ for convenience in notations, also $[k]_0 = 1$ and $[k]_1 = k$).
