Let $f(x)=erfi(a+x)$ and $g(x)=e^{cx}$ with
\begin{align\*}
f^{(n)}(x)=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum\_{m=0}^{n-1}\sum\_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-n+1}}{m!}\\
\prod\_{p=1}^{n-1}(2m-2j-p+1).
\end{align\*}
and
\begin{align\*}
g^{(n)}(x)=c^n e^{cx}
\end{align\*}
I want to find an expression for $ (f(x).g(x))^{(n)}$. I tried Leibniz rule for differentiation
\begin{align\*}
&(f(x).g(x))^{(n)} =\sum\_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)\\
&\quad=\sum\_{k=0}^{n}\binom{n}{k}\left(\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum\_{m=0}^{k-1}\sum\_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-k+1}}{m!} \prod\_{p=1}^{k-1}(2m-2j-p+1)\right)\\
&\times\left(c^{(n-k)} e^{cx}\right)
\end{align\*}
I have no other way to find $ f^{(n)}$, the reason is that we need to find first derivative of error function to get rid of integral and then generalize derivative. While applying Leibniz rule $f^{(k)}=0 $ for $ k=0$. How to overcome this problem?



nth derivative of error function
To avoid "extended discussion", here is what I said.
\begin{align} (f(x).g(x))^{(n)} &=\sum\_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)\\
&=f(x)g^{(n)}+\sum\_{k=1}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x) \\
&=\frac2{\sqrt{\pi}}c^ne^{cx}\int\_0^{a+x}e^{t^2}dt
+\sum\_{k=1}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x).
\end{align}
Now use your formulas for $f^{(k)}(x)$, for $k\geq1$, and the obvious one $g^{(k)}(x)$ for any $k$.


there is a simplification of your bouble-sum. For $n\geq1$, we have
\begin{align\*}
f^{(n)}(x)&=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum\_{m=0}^{n-1}\sum\_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-n+1}}{m!}
\prod\_{p=1}^{n-1}(2m-2j-p+1) \\
&=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}(a+x)^{2m-n+1}\sum\_{m=0}^{n-1}\frac{(n-1)!}{m!}\sum\_{j=0}^{m}(-1)^j\binom{m}j\binom{2m-2j}{n-1}\\
&=\frac2{\sqrt{\pi}}e^{(a+x)^2}(a+x)^{2m-n+1}(n-1)!\sum\_{m=0}^{n-1}\frac1{m!}\binom{m}{n-m-1}\frac1{2^{n-2m-1}}.
\end{align\*}
So, here is the identity I used:
$$\sum\_{j=0}^{m}(-1)^j\binom{m}j\binom{2m-2j}n=\binom{m}{n-m}\frac1{2^{n-2m}}.$$
As far as proving this and similar identities goes, you may follow the procedure I outlined [in my answer here](https://mathoverflow.net/questions/193611/binomial-coefficient-identity/257970#257970).


