proof of generalized Leibniz rule

The generalized Leibniz rule can be derived from the plain Leibniz rule by induction on $r$ .

If $r=2$ , the generalized Leibniz rule reduces to the plain Leibniz rule. This will be the starting point for the induction. To complete the induction, assume that the generalized Leibniz rule holds for a certain value of $r$ ; we shall now show that it holds for $r+1$ .

Write $\\prod_{i=1}^{r+1}f_{i}(t)=\\left(f_{r+1}(t)\\right)\\left(\\prod_{i=1}^{r+1}f_{i}(%t)\\right)$ . Applying the plain Leibniz rule,

{d^{n}\\over dt^{n}}\\left(f_{r+1}(t)\\right)\\left(\\prod_{i=1}^{r+1}f_{i}(t)%\\right)=\\sum_{n_{r+1}=0}^{n}\\left({n\\atop n_{r+1}}\\right)\\left({d^{n_{r+1}}%\\over dn^{n_{r+1}}}f_{r+1}(t)\\right)\\left({d^{n-n_{r+1}}\\over dn^{n-n_{r+1}}}%\\prod_{i=1}^{r+1}f_{i}(t)\\right)

By the generalized Leibniz rule for $r$ (assumed to be true as the induction hypothesis), this equals

\\sum_{n_{r+1}=0}^{n}\\sum_{n_{1}+\\cdots+n_{r}=n-n_{r+1}}\\left({n-n_{r+1}\\atop n%_{1},n_{2},\\ldots n_{r}}\\right)\\left({n\\atop n_{r+1}}\\right)\\left({d^{n_{r+1}}%\\over dn^{n_{r+1}}}f_{r+1}(t)\\right)\\left(\\prod_{i=1}^{r}{d^{n_{i}}\\over dt^{n%_{i}}}f_{i}(t)\\right)

Note that

\\left({n-n_{r+1}\\atop n_{1},n_{2},\\ldots n_{r}}\\right)\\left({n\\atop n_{r+1}}%\\right)=\\left({n-n_{r+1}\\atop n_{1},n_{2},\\ldots n_{r},n_{r+1}}\\right)

This is an immediate consequence of the expression for multinomial coefficients as quotients of factorials. Using this identity, the quantity can be written as

\\sum_{n_{1}+\\cdots+n_{r}+n_{r+1}=n}\\left({n-n_{r+1}\\atop n_{1},n_{2},\\ldots n_%{r},n_{r+1}}\\right)\\prod_{i=1}^{r+1}{d^{n_{i}}\\over dt^{n_{i}}}f_{i}(t)

which is the generalized Leibniz rule for the case of $r+1$ .