$\displaystyle k^{th}$ derivative of a Gaussian function with zero mean

Question

The gaussian function is: $$f(x,\mu,\sigma)=\dfrac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\dfrac{(x-\mu)^2}{\sigma^2}\right)$$ Putting $\mu=0$, we can get the $\displaystyle k^{th}$ derivative of this function. It is: $$\dfrac{d^k}{dx^k}f(x,\sigma)=\dfrac{2^k\left(\left(-\dfrac{x}{\sigma^2}\right)^kU\left(-\dfrac{k}{2},\dfrac{1}{2},\dfrac{x^2}{\sigma^2}\right)\right)}{\exp\left(\dfrac{x^2}{\sigma^2}\right)\left(\dfrac{x^2}{\sigma^2}\right)^{k/2}}$$ where $U(a,b,x)$ is the confluent hypergeometric function of th second kind. I'm stuck in proving the previous formula. How can it be proven? Thanks

Answer 1

Hint:

In general $$f\left(x,\mu,\sigma\right)=\sigma^{-1}\phi\left(\frac{x-\mu}{\sigma}\right)$$ where $\phi\left(x\right)=f\left(x,0,1\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^{2}}$. Consequently: $$f^{\left(n\right)}\left(x,\mu,\sigma\right)=\sigma^{-n}\phi^{\left(n\right)}\left(\frac{x-\mu}{\sigma}\right)$$ This equality allows you to concentrate on the $\phi^{\left(n\right)}$.

The annoying parameters are placed outside.