# proof of generalized Leibniz rule

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

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

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

Title proof of generalized Leibniz rule ProofOfGeneralizedLeibnizRule 2013-03-22 14:34:14 2013-03-22 14:34:14 rspuzio (6075) rspuzio (6075) 7 rspuzio (6075) Proof msc 26A06