Prove for general form of function at -x containing derivatives of order n

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I have stumbled across multiple casses of functions (explicitly Hermit and Legendre polynomials) for which I wanted to prove the symmetry. While doing so I always ended up with the following equations:

\begin{equation} H_n(-x)=a *\frac{d^n}{dy^n}f(y)|_{y=-x}=a*(-1)^n*\frac{d^n}{dx^n}f(x) \end{equation}

Where $f(x)$ is an even function ($f(x) = f(-x)$)

Now my question: Is there a formal way to prove that I can in any case extract the factor of $(-1)^n$ from this derivative? Maybe using total differentials? And if so, how can I do it? For the specific cases I was able to prove it using induction but as far as solutions to these problems go, this was not necessary and the above step was just performed with no additional comments.

Thank you for the help in advance

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I fail to understand your question, really. The formal definition of Hermites, however, cannot be what you wrote, but, instead, is $$ H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}~~~\leadsto \\ \quad H_n(x) = \left(2x - \frac{d}{dx} \right)^n \cdot 1~~~.$$

It is then evident that the minus signs may be absorbed into x, s.t. $$ H_n(-x) = e^{x^2}\frac{d^n}{dx^n}e^{-x^2} = (-1)^n \left(2x - \frac{d}{dx} \right)^n \cdot 1~~~.$$