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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


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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This problem is perfectly suited for a proof using mathematical induction. – Rasmus Jul 11 '11 at 21:44
Possibly an easy way to remember: $\log \prod f = \sum \log f$ and $(\log f)' = f'/f$ – Aryabhata Jul 11 '11 at 21:58
$$\Big(\prod_{i=1}^kf_i\Big)^{(n)}=\sum_{n=j_1+...+j_k}{n\choose j_1,...,j_k}\prod_{i=1}^kf_i^{(j_i)}$$ – yoyo Jul 11 '11 at 23:35
@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... – draks ... Aug 3 '12 at 18:21
up vote 4 down vote accepted

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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I'm sorry, I don't seem to understand your answer, although I'm sure it's correct. Does the notation "1≤j≤n; i≠j" mean that j is all the positive integers from 1 to n inclusive, excluding i? I'm just not familiar with that notation. – Hautdesert Jul 12 '11 at 18:44
@Hautdesert: Yes; I'm multiplying over all indices $j$ that satisfy $1\leq j\leq n$, and that satisfy $i\neq j$. It's just a compact way of writing what you did without having to define the $h_i$. – Arturo Magidin Jul 12 '11 at 18:55
I don't follow the transition from the third line to the fourth line. What happens to $f_{n+1}(x)$? – Hautdesert Jul 12 '11 at 19:24
@Hautdesert: If you look carefully at the index of the product in the first summand, in the third line it goes up to $n$, in the fourth line it goes all the way to $n+1$; the $f_{n+1}(x)$ has been absorbed into that product. – Arturo Magidin Jul 12 '11 at 19:26
@Hautdesert: P.S. You should not accept an answer if you are having trouble following it! Why did you accept my answer, only to say later that you don't understand it? First understand the answers, then decide which one is the most helpful and you can accept that one then. Right now, you might want to un-accept this, since you are having trouble following it. – Arturo Magidin Jul 12 '11 at 19:26

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