= Solution
Let $G(k)=\widetilde g(k)$. Since $g'=-xg$, <differentiation> under the <Fourier transform> integral and <integration by parts> give
$$
G'(k)=-i\,\widetilde{xg}(k)=i\,\widetilde{g'}(k)=i(ik)G(k)=-kG(k).
$$
Thus $G(k)=G(0)e^{-k^2/2}$. The <Gaussian integral> gives $G(0)=\sqrt{2\pi}$, proving the <Fourier transform of a Gaussian> formula
$$
\boxed{\widetilde g(k)=\sqrt{2\pi}\,e^{-k^2/2},\qquad c=\sqrt{2\pi}.}
$$
For the higher <derivatives>, start with $r_0=1$. If $g^{(m)}=r_mg$, the <product rule> gives
$$
g^{(m+1)}=(r_m'-xr_m)g,\qquad\boxed{r_{m+1}=r_m'-xr_m.}
$$
If $r_m$ has degree $m$ and leading coefficient $(-1)^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1)^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1)^m\operatorname{He}_m$, where $\operatorname{He}_m$ is the <Probabilists' Hermite polynomial>.
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