I'm trying to generalize the product rule to more than the product of two functions using the fact that I can treat the product of $n$-1 functions as a single one. Here is an example of what I mean:
[f(x)g(x)h(x)]' = [f(x)p(x)]' where $p(x) = g(x)h(x)$
[f(x)p(x)]' = f'(x)p(x) + f(x)p'(x) = f'(x)p(x) + f(x)[g(x)h(x)]'
f'(x)p(x) + f(x)[g(x)h(x)]' = f'(x)g(x)h(x) + f(x)[g'(x)h(x) + g(x)h'(x)]'
which equals f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)
I generalized this as follows:
\Big[\prod_{i=1}^{n}f_i(x)\Big]'= f_1'(x)g_1(x) + f_1(x)g'_1(x)
where $g_m(x)=\prod_{i=1}^{n-m}f_i(x)$, and $g'_{m-1}=[f_m(x)g_m(x)]'=f'_m(x)g_m(x) + f_m(x)g'_m(x)$.
Now, I do realize that this is a generalization, and there is really nothing to prove, but say I wanted to prove that
\Big[\prod_{i=1}^{n}f_i(x)\Big]'=\sum_{i=1}^{n}f'_i(x)h_i(x)$
where h_i(x)=\frac{1}{f_i(x)}\prod_{j=1}^nf_j(x)$, how would I go about doing this (using the generalization above)? I apologize if my notation is hard to understand. Thank you.