= Solution
For functions with the required <derivatives>, the <Leibniz rule> is
$$
(fg)^{(n)}=\sum_{k=0}^n\binom nk f^{(k)}g^{(n-k)}.
$$
At $n=0$ it is immediate. If it holds at $n$, differentiating each summand gives two sums. The coefficient of $f^{(k)}g^{(n+1-k)}$ after regrouping is $\binom nk+\binom n{k-1}=\binom{n+1}k$ by <Pascal's identity>, including the two endpoint terms. This proves the formula by <mathematical induction>.
Apply it to $(fg)h$ and then expand the derivatives of $fg$. For $p+q+r=n$, the resulting coefficient is $\binom n{p+q}\binom{p+q}p=n!/(p!q!r!)$. Consequently
$$
\boxed{(fgh)^{(n)}=\sum_{p+q+r=n}\frac{n!}{p!q!r!}f^{(p)}g^{(q)}h^{(r)},}
$$
where $p,q,r$ range over nonnegative integers. This is the three-factor <generalized Leibniz rule>.
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