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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.

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    @yoyo +1 your comment was exactly the generalization, I was looking for. Did you use it in a question/answer? Why not extending the WP page: [Product_rule#Generalizations](http://en.wikipedia.org/wiki/Product_rule#Generalizations)...2012-08-03

2 Answers 2

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You can use induction on $n$, the number of functions. if $n = 1$, there is nothing to prove. if $n = 2$, then you just get the product rule. Assume the claim is true for $n$ functions, and prove it for $n+1$. Write $f_1f_2...f_{n+1}$ = $f_1g$ where $g = f_2..f_{n+1}$. Now differentiate $f_1g$ using the product rule and apply the induction hypothesis to g'. Note that $g$ is a product of $n$ functions, so the induction hypothesis tells you what g' is.

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Simplest way to establish this is by induction on $n$.

The case $n=1$ is immediate; the case $n=2$ is the usual product rule. Assuming you have established the desired formula $\left(\prod_{i=1}^n f_i(x)\right)' = \sum_{i=1}^n \left(f_i'(x)\prod_{\stackrel{1\leq j\leq n}{i\neq j}}f_j(x)\right)$ for $n$, then to get the $n+1$ case we have: $$\begin{align*} \left(\prod_{i=1}^{n+1}f_i(x)\right)' &= \left(\left(\prod_{i=1}^n f_i(x)\right)f_{n+1}(x)\right)'\\ &= \left(\prod_{i=1}^nf_i(x)\right)'f_{n+1}(x) + \left(\prod_{i=1}^nf_i(x)\right)f_{n+1}^{'}(x)\\ &= \left(\sum_{i=1}^nf_i'(x)\prod_{\stackrel{1\leq j\leq n}{i\neq j}}f_j(x)\right)f_{n+1}(x) + \left(\prod_{i=1}^nf_i(x)\right)f_{n+1}'(x)\\ &= \sum_{i=1}^nf_i'(x)\prod_{\stackrel{1\leq j\leq n+1}{i\neq j}}f_j(x) + \left(\prod_{i=1}^nf_i(x)\right)f_{n+1}'(x)\\ &=\sum_{i=1}^{n+1} f_i'(x)\prod_{\stackrel{1\leq j\leq n+1}{i\neq j}}f_j(x), \end{align*}$$ as desired.

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    @Hautdesert: In my opinion (and that of many in this site), promptness should not, in and of itself, go into accepting an answer. You should accept an answer only (i) After you are satisfied with the answer; (ii) you understand the answer; and then (iii) accept *whichever* answer you found the most helpful and/or interesting. If you don't understand my answer, then it doesn't matter whether you "are sure" it is correct or not, it's *not helpful to you* and as such should not be accepted. If you still don't understand it, then don't accept it.2011-07-13