I know that there is a general formula to compute the $n_{th}$ derivative of a composition as it is mentioned in this post. The formula is named Faà di Bruno's formula. However, it is really ugly and complicated! :)
Now, I want to find the $n_{th}$ derivative of $f(x)^m$ where $m$ is some positive integer. So, I am interested in
$${\left( {f{{(x)}^m}} \right)^{(n)}} \equiv f{(x)^{\left( {m,n} \right)}} = ?$$
Is there a tidy formula for this?
My Computations
$$\matrix{ {f{{(x)}^{\left( {m,1} \right)}} = } \hfill & {{{m!} \over {\left( {m - 1} \right)!}}f{{(x)}^{\left( {m - 1,0} \right)}}f{{(x)}^{\left( {1,1} \right)}}} \hfill & {} \hfill & {} \hfill & {} \hfill \cr {f{{(x)}^{\left( {m,2} \right)}} = } \hfill & {{{m!} \over {\left( {m - 2} \right)!}}f{{(x)}^{\left( {m - 2,0} \right)}}{{\left( {f{{(x)}^{\left( {1,1} \right)}}} \right)}^2}} \hfill & + \hfill & {{{m!} \over {\left( {m - 1} \right)!}}f{{(x)}^{\left( {m - 1,0} \right)}}f{{(x)}^{\left( {1,2} \right)}}} \hfill & {} \hfill \cr {f{{(x)}^{\left( {m,3} \right)}} = } \hfill & {{{m!} \over {\left( {m - 3} \right)!}}f{{(x)}^{\left( {m - 3,0} \right)}}{{\left( {f{{(x)}^{\left( {1,1} \right)}}} \right)}^3}} \hfill & + \hfill & {{{m!} \over {\left( {m - 2} \right)!}}{{3!} \over {2!}}f{{(x)}^{\left( {m - 2,0} \right)}}f{{(x)}^{\left( {1,2} \right)}}f{{(x)}^{\left( {1,1} \right)}}} \hfill & + \hfill \cr {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + } \hfill & {{{m!} \over {\left( {m - 1} \right)!}}f{{(x)}^{\left( {m - 1,0} \right)}}f{{(x)}^{\left( {1,3} \right)}}} \hfill & {} \hfill & {} \hfill & {} \hfill \cr . \hfill & . \hfill & . \hfill & . \hfill & . \hfill \cr . \hfill & . \hfill & . \hfill & . \hfill & . \hfill \cr . \hfill & . \hfill & . \hfill & . \hfill & . \hfill \cr } $$
It should be something like
$$f{(x)^{\left( {m,n} \right)}} = \sum\limits_{i = 1}^n {{C_i}(n,x)f{{(x)}^{\left( {m - i,0} \right)}}} $$
however, determining ${{C_i}(n,x)}$ is not easy as it seems! :)