Assume that $p(x)$ is a polynomial, $D=\frac{d}{dx}$ and $a$ is a constant value. It is strightforward to show that $p(D) e^{ax}=p(a)e^{ax}$ and $p(D) (e^{ax}f(x))=e^{ax}p(D+a)f(x)$.
Question: What can be said about $p(D) (e^{iaf(x)})$?
Note: I understand that, in general, $p(D) f(x)$ can be expressed in Fourier domain as $\int p(i\omega) F(\omega) e^{i \omega x} d\omega$, where $F(\omega)=Fourier(f(x))$. I, however, was wondering whether any simiplification/modification can be done to $p(D) (e^{iaf(x)})$ using above (or any other) operator methods.