$n$-th Derivative $\frac{d^{n}}{d x^{n}} e^{-\sqrt{x} |\omega|}$ via Recursive Product Rule Let $g(x) = x^{-\frac{1}{2}}$ and $f(x) = e^{-\sqrt{x} |\omega|}$. I am trying to find an expression for the $M$-th derivative of their product:
\begin{align}
    \frac{d^M}{dx^M} \left[ f(x) g(x) \right] = \sum_{k=0}^{M} \binom{M}{k} f^{(M-k)}(x) g^{(k)}(x)
\end{align}
where $M$ is an even integer and $\omega \in \mathbb{R}$ is a constant parameter.
The derivative for $g(x)$ is easy:
\begin{align}
    \frac{d^{k}}{d x^{k}} g(x) = \frac{(2k-1)!!}{2} x^{-\frac{2k+1}{2}}
\end{align}
However, for $f(x)$, it is not so easy, and this is where I get stuck.
For the $M$-th derivative of $f(x)$, Mathematica gives me:
M! DifferenceRoot[
   Function[{\[FormalY], \[FormalN]}, {(2 + 6 \[FormalN] + 
          4 \[FormalN]^2) \[FormalY][1 + \[FormalN]] + 
       4 (1 + \[FormalN]) (2 + \[FormalN]) x \[FormalY][
         2 + \[FormalN]] - \[FormalY][\[FormalN]] Abs[w]^2 == 
      0, \[FormalY][0] == 
      E^(-Sqrt[x] Abs[w]), \[FormalY][1] == -((
       E^(-Sqrt[x] Abs[w]) Abs[w])/(2 Sqrt[x]))}]][M]

But I have no idea what any of that means, and their site is not very helpful...
It seems that the solution will involve some kind of recursive application of the product rule, however, I can't seem to find a pattern after I do a handful of layers...

Question:

*

*What is $\frac{d^{n}}{d x^{n}} f(x) = \frac{d^{n}}{d x^{n}} e^{-\sqrt{x} |\omega|}$ when $n$ is an even integer?

 A: It's sufficient to consider the case $\omega=1$: if we let
$$
\color{DarkBlue}{f_n(x):=e^{\sqrt{x}}\sqrt{x}\frac{d^n}{dx^n}\frac{e^{-\sqrt{x}}}{\sqrt{x}}},
$$
then the needed expression is
$$
\frac{d^n}{dx^n}\frac{e^{-|\omega|\sqrt{x}}}{\sqrt{x}}=|\omega|^n\frac{e^{-|\omega|\sqrt{x}}}{\sqrt{x}}f_n(\omega^2 x).
$$
To compute $f_n(x)$, we consider (for $x>0$ and $|t|<x$)
$$
f(x,t):=\sum_{n=0}^\infty f_n(x)\frac{(-t)^n}{n!}=e^{\sqrt{x}}\sqrt{x}\frac{e^{-\sqrt{x-t}}}{\sqrt{x-t}}=\frac{e^{\sqrt{x}-\sqrt{x-t}}}{\sqrt{1-t/x}}.
$$
Now we use the formula (for $|z|<1$) $$\frac{(1-\sqrt{1-z})^k}{\sqrt{1-z}}=2^k\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{z}{4}\right)^n.$$ To prove it, one may use the known formula
$$
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^k=\sum_{n=0}^\infty\binom{2n+k}{n}\frac{k}{2n+k}z^n;
$$
here $k/(2n+k)$ is interpreted as $1$ when $n=k=0$ (recipe: multiply this by $z^k$, replace $z$ with $z/4$ and $k$ with $k+1$, and take the derivative).
Returning to the computation, we have
$$
f(x,t)=\sum_{k=0}^\infty\frac{1}{k!}\frac{(\sqrt{x}-\sqrt{x-t})^k}{\sqrt{1-t/x}}
=\sum_{k=0}^\infty\frac{(2\sqrt{x})^k}{k!}\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{t}{4x}\right)^n,
$$
which, after (exchanging the summations and) extracting the coefficient of $t^n$, results in
$$
f_n(x)=(-1)^n\sum_{k=0}^n\frac{(2n-k)!}{k!(n-k)!}(2\sqrt{x})^{k-2n}
\color{DarkBlue}{=(-1)^n\sum_{k=0}^n\frac{(n+k)!}{k!(n-k)!}(2\sqrt{x})^{-n-k}}.
$$

Another approach is to use the following representation:
$$
\frac{e^{b(s-\sqrt{a^2+s^2})}}{\sqrt{a^2+s^2}}=\int_0^\infty e^{-st}J_0\left(a\sqrt{t^2+2bt}\right)dt,
$$
where $J_0(z)=\sum_{n=0}^\infty(-z^2/4)^n/n!^2$ is a Bessel function.
Here we put $s=\sqrt{x}$, $a=\sqrt{y}$ and $b=1$; the LHS is then $f(x,-y)/\sqrt{x}$, and the RHS may be "evaluated" by substituting the power series for $J_0$ directly, and integrating the result termwise, using the binomial formula for $(t+2)^n$. The outcome is the same.
