Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$ I need the closed-form for the $n$-th derivative ($n\geq0 $):
$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)$$
Thanks!
By following the suggestion of Hermite polynomials:
$$H_n(x)=(-1)^ne^{x^2}\frac{\partial^n}{\partial x^n}e^{-x^2}$$
and doing the variable change $x=\pi a y^{-\frac{1}{2}}$, I obtain:
$$\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n\frac{\partial^n}{\partial y^n}$$
and therefore
$$H_n(\pi a y^{-\frac{1}{2}})=(-1)^{n+1}e^{\frac{\pi^2a^2}{y}}2\left(\frac{\pi a}{y^{\frac{3}{2}}}\right)^n\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}$$
Finally
$$\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}=\frac{1}{2}e^{-\frac{\pi^2a^2}{y}}(-1)^{n+1}H_n(\pi a y^{-\frac{1}{2}})\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n$$
Is this correct?

Here we show that OPs approach is regrettably not correct and provide an alternative to answer the problem.
This answer is based upon the $n$-th derivative of the composite of two functions. It is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule
Let $D_x$ represent differentiation with respect to $x$ and $z=z(x)$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$.  The following is valid for $n\geq 0$:
\begin{align*}
D_x^n f(z)=\sum_{k=0}^nD_z^kf(z)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_x^nz^j\tag{1}
\end{align*}

Let's denote for convenience only $c:=\pi^2a^2$.

The following is valid for $n\geq 1$
\begin{align*}
D_x^ne^{-\frac{c}{x}}=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{n-1}{k}k!c^{n-k}x^{-2n+k}\tag{2}
\end{align*}

Using Hoppes formula (1) we obtain by setting
\begin{align*}
f(z)=e^{z}\qquad\quad\text{and}\quad\qquad z=z(x)=-\frac{c}{x}
\end{align*}
the following for $n\geq 1$

\begin{align*}
D_x^ne^{-\frac{c}{x}}&=\sum_{k=0}^nD_z^ke^z\frac{(-1)^k}{k!}
\sum_{j=0}^k(-1)^j\binom{k}{j}\left(-\frac{c}{x}\right)^{k-j}D_x\left(-\frac{c}{x}\right)^j\tag{3}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}
\left(-\frac{c}{x}\right)^{k-j}n!c^j\binom{n+j-1}{j-1}\frac{(-1)^{n+j}}{x^{n+j}}\tag{4}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{n!}{k!}(-1)^nc^k\sum_{j=0}^k(-1)^j\binom{k}{j}\binom{n+j-1}{j-1}x^{-n-k}\tag{5}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{n!}{k!}(-1)^nc^k\binom{n-1}{k-1}(-1)^kx^{-n-k}\tag{6}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n-1}{k-1}c^kx^{-n-k}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^{k}\binom{n}{k}\binom{n-1}{k}c^{n-k}x^{-2n+k}\tag{7}\\
\end{align*}
and the claim follows.

Comment:

*

*In (3) we apply Hoppes formula (1)


*In (4) we use
\begin{align*}
D_z^ke^z=e^z=e^{-\frac{c}{x}}\quad\text{and}
\quad D_x\left(-\frac{c}{x}\right)^j=n!c^j\binom{n+j-1}{j-1}(-1)^{n+j}
\end{align*}


*In (5) we do some rearrangement


*In (6) we use the identity
\begin{align*}
\sum_{j=0}^k(-1)^j\binom{k}{j}\binom{n+j-1}{j-1}=\binom{n-1}{k-1}(-1)^k
\end{align*}

*

*In (7) we replace $k$ with $n-k$ and set the upper index limit to $n-1$

Small examples: $n=1,2,3$

Using formula (2) we obtain for $n=1,2,3$
\begin{align*}
D_x^1e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^2}\\
D_x^2e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^4}\left(c-2x\right)\tag{8}\\
D_x^3e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^6}\left(c^2-6cx+6x^2\right)\\
\end{align*}

Hermite polynomials:
If we consider Hermite polynomials in the form
\begin{align*}
H_n(x)=-e^{x^2}D_xe^{-x^2}
\end{align*}
and apply a substitution
\begin{align*}
x=x(y)=\sqrt{\frac{c}{y}}=\sqrt{c}y^{-\frac{1}{2}}
\end{align*}
in order to obtain
\begin{align*}
H_n(x(y))&=-e^{(x(y))^2}D_ye^{-(x(y))^2}\\
&=-e^{\frac{c}{y}}D_ye^{-\frac{c}{y}}
\end{align*}
we are in the same situation as above, since we have to calculate the $n$-th derivative of a composition of functions, which is not feasible with OPs approach. It could be done with Hoppes formula instead.

Let's look at a small example by taking $n=2$. In the following I omit most of the time  the argument $y$ and write $xx^\prime$ instead of $x(y)x^\prime(y)$ to ease reading.

\begin{align*}
H_2(x(y))&=e^{x^2}D_y^2\left(e^{-x^2}\right)\\
&=-e^{x^2}D_y\left(-2xx^\prime e^{-x^2}\right)\\
&=-e^{x^2}\left(D_y\left(-2xx^\prime\right)e^{-x^2}+(-2xx^\prime)D_y\left(e^{-x^2}\right)\right)\\
&=-e^{x^2}\left(\left(-2{x^\prime}^2-2xx^{\prime\prime}\right)e^{-x^2}+4x^2{x^\prime}^2e^{-x^2}\right)\\
&=-2{x^\prime}^2-2xx^{\prime\prime}+4x^2{x^\prime}^2
\end{align*}
Since
\begin{align*}
x&=x(y)=\sqrt{c}\,y^{-\frac{1}{2}}\\
x^\prime&=x^\prime(y)=-\frac{1}{2}\sqrt{c}\,y^{-\frac{3}{2}}\\
x^{\prime\prime}&=x^{\prime\prime}(y)=\frac{3}{4}\sqrt{c}\,y^{-\frac{5}{2}}\\
\end{align*}
we finally obtain
\begin{align*}
H_2(x(y))&-2{x^\prime}^2-2xx^{\prime\prime}+4x^2{x^\prime}^2\\
&=-2\frac{1}{4}c\,y^{-3}-2\sqrt{c}\,y^{-\frac{1}{2}}\frac{3}{4}\sqrt{c}\,y^{-\frac{5}{2}}
+4cy^{-1}\frac{1}{4}c\,y^{-3}\\
&=-\frac{1}{2}c\,y^{-3}-\frac{3}{2}c\,y^{-3}+c^2\,y^{-4}\\
&=-2c\,y^{-3}+c^2\,y^{-4}\\
&=\frac{c}{y^4}\left(c^2-2y\right)
\end{align*}
in accordance with (8). So, we can apply Hoppes formula or perform some equivalent calculation.

