Does there exist a formula to expand
\begin{align} \left[h(p)\frac{d^m}{d p^m}\right]^n\Big(f(p)g(p)\Big)? \end{align}
I know that the Leibniz identity is
\begin{align} \frac{d^n}{d p^n}fg=\sum_{k=0}^n{n\choose k}\frac{d^k f}{dp^k}\frac{d^{n-k} g}{dp^{n-k}}\,, \end{align}
but I am not how to extend it. I would expect that I would have the same sort of form, but instead of ${n \choose k}$, I would have a function of $p$, $a_k(p)$, such that
\begin{align} \left[h(p)\frac{d^m}{d p^m}\right]^n\Big(f(p)g(p)\Big)=\sum_{k=0}^na_k(p)\frac{d^k f}{dp^k}\frac{d^{n-k} g}{dp^{n-k}}\,. \end{align}
Is there a way to evaluate what $a_k(p)$ would be?
