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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.

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.

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: $$\sum_{j=0}^{m}(-1)^j\binom{m}j\binom{2m-2j}n=\binom{m}{n-m}\frac1{2^{n-2m}}.$$

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.

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 $k$.

I notice that your derivatives $f^{(n)}(x)$ are off by an index; more importantly, there is a simplification. For $n\geq1$, we have \begin{align*} f^{(n)}(x)&=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum_{m=0}^n\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-n}}{m!} \prod_{p=1}^n(2m-2j-p+1) \\ &=\frac2{\sqrt{\pi}}e^{(a+x)^2}(a+x)^{2m-n}\sum_{m=0}^n\binom{m}{n-2m}\frac{m!n!}{2^{n-2m}}. \end{align*}

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: $$\sum_{j=0}^{m}(-1)^j\binom{m}j\binom{2m-2j}n=\binom{m}{n-m}\frac1{2^{n-2m}}.$$