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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. It should be $ f^{(0)}=f $ for applying Leibniz rule while $f^{(0)}=0 $ in this case. How to overcome this problem?

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    $g^{(n-k)}$ is correct and you put $f^{(k)}(x)$ and $g^{(n-k)}(x)$ together corecctly. If there is something wrong, then the problem is with $f^{(n)}.$ – 2017-01-25
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    Yes...that is the issue. I took first derivative of error function and then generalize the rest $ n-1$ derivatives. To apply Leibniz rule, we need $ k=0$, then answer is 0. – 2017-01-25
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    The expression $ f^{(n)}$ is correct, I checked many times. The issue is to adjust with Leibniz rule. That creates problems or apply it for $ n=0$, then its answer is $0$. 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 sign and then generalize the derivative. – 2017-01-25

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