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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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    This problem is perfectly suited for a proof using [mathematical induction](http://en.wikipedia.org/wiki/Mathematical_induction).2011-07-11
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    Possibly an easy way to remember: $\log \prod f = \sum \log f$ and $(\log f)' = f'/f$2011-07-11
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    $$\Big(\prod_{i=1}^kf_i\Big)^{(n)}=\sum_{n=j_1+...+j_k}{n\choose j_1,...,j_k}\prod_{i=1}^kf_i^{(j_i)}$$2011-07-11
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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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    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.2011-07-12
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    @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 $h_i$.2011-07-12
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    I don't follow the transition from the third line to the fourth line. What happens to $f_{n+1}(x)$?2011-07-12
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    @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.2011-07-12
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    @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.2011-07-12
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    Like I said, I'm sure your answer is correct. My acceptance of your answer is not only based on the correctness of your response, but also on the "quickness" of your reply. That is why I accepted your answer. But, if you honestly insist on me un-accepting your answer, and then re-accepting it, I will grant you your request.2011-07-12
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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