General Leibniz rule for triple products I have a question regarding the General Leibniz rule which is the rule for the $n^{th}$ derivative of a product and reads: 
$$
(f g)^{(n)}=\sum_{k=0}^{n} {n \choose k} \,f^{(k)} g^{(n-k)}.
$$ 
However, what about if there is a triple product instead of just a product. (i.e. $(f \cdot g \cdot  h)^{(n)}$)? Is there a comprahensive formula for such a derivative? I have yet to find one, but perhaps someone knows it. 
 A: Yes, it is. Just change binomial coefficient to trinomial coefficient. Namely
$$(f\cdot g\cdot h)^{(n)}=\sum_{k_1+k_2+k_3=n}{n\choose {k_1,k_2,k_3}}f^{(k_1)}g^{(k_2)}h^{(k_3)}$$
The proof is quite straightforward from Leibniz rule.
$$\begin{align}(fgh)^{(n)}&=\sum_{k=0}^n\frac{n!}{k!(n-k)!}f^{(k)}(gh)^{(n-k)}\\&=\sum_{k=0}^n\frac{n!}{k!(n-k)!}f^{(k)}\sum_{l=0}^{n-k}\frac{(n-k)!}{(n-k-l)!l!}g^{(l)}h^{(n-k-l)}\\&=\sum_{k+l\leq n}^n\frac{n!}{k!l!(n-k-l)!}f^{(k)}g^{(l)}h^{(n-k-l)}\end{align}$$
As you can see, this can be generalized to product of any $m$ functions via multinomial coefficient by induction.
A: In general
$$
\bigg(\prod_{j=1}^m f_j\bigg)^{\!(n)}=\sum_{\substack{k_1,\ldots,k_m\ge 0\\ k_1+\cdots+k_m=n}} \frac{n!}{k_1!\cdots k_m!} f_1^{(k_1)}\cdots f_m^{(k_m)}.
$$
A: Using the standard Leibniz rule for products twice:
$$
\begin{align}
((f\cdot g)\cdot h)^{(n)}
&=\sum_{k=0}^n\binom{n}{k}(f\cdot g)^{(k)}h^{(n-k)}\\
&=\sum_{k=0}^n\binom{n}{k}\sum_{j=0}^k\binom{k}{j}f^{(j)}g^{(k-j)}h^{(n-k)}\\
&=\sum_{k=0}^n\sum_{j=0}^k\frac{n!}{j!(k-j)!(n-k)!}f^{(j)}g^{(k-j)}h^{(n-k)}\\
&=\sum_{j=0}^n\sum_{k=j}^n\frac{n!}{j!(k-j)!(n-k)!}f^{(j)}g^{(k-j)}h^{(n-k)}\\
&=\sum_{j=0}^n\sum_{k=0}^{n-j}\frac{n!}{j!k!(n-k-j)!}f^{(j)}g^{(k)}h^{(n-k-j)}\\
&=\sum_{\substack{i+j+k=n\\i,j,k\ge0}}\frac{n!}{i!j!k!}f^{(j)}g^{(k)}h^{(i)}
\end{align}
$$
