Derivatives of Gaussian functions are orthogonal using the inverse of the Gaussian function as the weight function, is that a coincidence?

Question

I just noticed that for a Gaussian function, e.g. $G(x) = e^{-x^2}$, we can calculate its derivatives $$\frac{d^n}{d x^n} G(x) = (-1)^n H_n(x) e^{-x^2}$$ where $H_n(x)$ is the so-called Hermite polynomials satisfying $$ \int_{-\infty}^{\infty} H_n(x) H_m(x) e^{-x^2} dx = 2^{m+n} n! \sqrt{\pi} \delta_{mn},$$ or we can write $$ \int_{-\infty}^{\infty} \frac{1}{G(x)}\frac{d^n G(x)}{d x^n}\frac{d^m G(x)}{d x^m} dx = 0$$ when $m\neq n$. I found this property quite interesting. How can we prove it? Is the property also true for other functions $G(x)$?