Is there a formula for the successive derivatives of $e^{x^2/2}$? I am trying to see if there is a formula for the successive derivatives of $e^{x^2/2}$. I was able to compute up to the first 10, and the 10th derivative is:
$$
\frac{\partial^{10}(e^{x^2/2})}{\partial x^{10}} = e^{x^2/2}(945+4725 x^2+3150 x^4+630 x^6+45 x^8+x^{10})
$$
However, no matter how I look at it, each successive derivative keeps generating more and more powers, so I am not sure how I can write a simple closed form answer to this. Would anyone have any ideas? Thanks!
 A: $$ \frac{d}{dx} e^{x^2/2} = x e^{x^2/2} = P_1(x) e^{x^2/2}$$
$$ \frac{d^n}{dx^n} e^{x^2/2} = \frac{d}{dx} (P_{n-1}(x) e^{x^2/2}) = (x P_{n-1}(x) + \frac{dP_{n-1}}{dx})e^{x^2/2} = P_n(x) e^{x^2/2}$$
Hence $P_n$ can be defined using recursion:
$$ P_n(x) = xP_{n-1} + \frac{dP_{n-1}}{dx}, P_0(x) = 1
$$
This is the Hermite polynomials. According to Wikipedia,
$$ P_n(x) = n! \sum_{k=0}^{\lfloor n/2\rfloor} \frac{(-1)^k}{2^k k!(n-2k)!} x^{n-2k}
$$
A: This is coming directly from my blog post linked in my comment to the question here.

It will be obvious after differentiating $e^{x^{2}/2}$ a few number of times that the result will contain a factor $e^{x^{2}/2}$. Thus let's assume that $$\frac{d^{n}}{dx^{n}}\{e^{x^{2}/2}\} = e^{x^{2}/2}P_{n}(x)\tag{1}$$ where $P_{n}(x)$ is some function of $x$ dependent on $n$. It will turn out to be a polynomial.
To evaluate $P_{n}(x)$ explicitly we need to find another representation of $e^{x^{2}/2}$. To that end we start with the famous integral $$\frac{\sqrt{\pi}}{2} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$ Putting $x = x\sqrt{2}$ we get $$\sqrt{\frac{\pi}{2}} = \int_{0}^{\infty}e^{-z^{2}/2}\,dz = \frac{1}{2}\int_{-\infty}^{\infty}e^{-z^{2}/2}\,dz$$ Putting $z = t - x$ (with $x$ as a constant here) we get $$\sqrt{2\pi} = e^{-x^{2}/2}\int_{-\infty}^{\infty}e^{xt}e^{-t^{2}/2}\,dt$$ and therefore we have $$e^{x^{2}/2} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{xt}e^{-t^{2}/2}\,dt\tag{2}$$ Differentiating the above equation $n$ times we get
\begin{align}
P_{n}(x)\,&= \frac{e^{-x^{2}/2}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}t^{n}e^{tx}e^{-t^{2}/2}\,dt\notag\\ 
&= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}t^{n}e^{-(t - x)^{2}/2}\,dt\notag\\ 
&= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(u + x)^{n}e^{-u^{2}/2}\,du\notag\\ 
&= \sum_{k = 0}^{n}\binom{n}{k}x^{n - k}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u^{k}e^{-u^{2}/2}\,du\right)\notag\\ 
&= \sum_{0 \leq 2k \leq n}\binom{n}{2k}x^{n - 2k}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u^{2k}e^{-u^{2}/2}\,du\right)\notag\\ 
&= \sum_{0 \leq 2k \leq n}\binom{n}{2k}x^{n - 2k}\left(\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}u^{2k}e^{-u^{2}/2}\,du\right)\notag\\ 
&= \sum_{0 \leq 2k \leq n}\binom{n}{2k}x^{n - 2k}\left(\frac{2^{k}}{\sqrt{\pi}}\int_{0}^{\infty}t^{k - 1/2}e^{-t}\,dt\right)\text{ (by putting }u^{2} = 2t^{2})\notag\\ 
&= \sum_{0 \leq 2k \leq n}\binom{n}{2k}x^{n - 2k}\cdot\dfrac{2^{k}\Gamma\left(k + \dfrac{1}{2}\right)}{\sqrt{\pi}}\notag\\ 
&= \sum_{0 \leq 2k \leq n}\binom{n}{2k}x^{n - 2k}\cdot\dfrac{(2k)!}{2^{k}k!}\notag
\end{align} and we thus obtain $$P_{n}(x) = \sum_{0 \leq 2k \leq n}\binom{n}{2k}\frac{(2k)!}{2^{k}k!}x^{n - 2k}\tag{3}$$
