The $2n^{\text{th}}$-order derivative of $\exp(-x^2)$

Question

It is known that the $n^{\text{th}}$-order derivative of the Gaussian function $\exp(-x^2)$ is given by $$ \frac{d^n}{dx^n}\exp(-x^2)=(-1)^n\exp(-x^2)H_n(x), $$ where $H_n(x)$ denotes the $n^{\text{th}}$-order Hermite polynomial. My question is how do we find the $2n^{\text{th}}$-order derivative of the function $\exp(-x^2)$ (i.e. every derivative is even)? Do we simply replace $n$ with $2n$ to obtain $$ \frac{d^{2n}}{dx^{2n}}\exp(-x^2)=\exp(-x^2)H_{2n}(x)? $$ Intuitively this makes sense, but I do not know if anything subtle is being missed.

Answer 1

The derivatives start for $n=0,1,2,\ldots$ $$ \frac{d^0}{dx^0} e^{-x^2} = e^{-x^2} ; $$ $$ \frac{d^2}{dx^2} e^{-x^2} = [-2+4x^2]e^{-x^2} ; $$ $$ \frac{d^4}{dx^4} e^{-x^2} = [12-48x^2+16x^4]e^{-x^2} ; $$ $$ \frac{d^6}{dx^6} e^{-x^2} = [-120+720x^2-480x^4+64x^6]e^{-x^2} ; $$ $$ \frac{d^{2n}}{dx^{2n}} e^{-x^2} = H_{2n}(x) e^{-x^2} ; $$ where $H$ are the Hermite Polynomials; see https://oeis.org/A185296 for an extended list.