The derivative of a product of more than two functions

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Perhaps the confusion is stemming from the complicated notation. Here's something I wrote that might be easier to follow:

Let $f_1,\cdots,f_n$ be functions. Then the derivative of the product $f_1\cdots f_n$ is the following: $$(f_1\cdots f_n)'=\sum_{j=1}^n\left[f_j'\prod_{i=1,\;i\neq j}^n\Bigg[f_i\Bigg]\right].$$ The indices for the $\Pi$ mean to multiply through from $i=1$ to $n$ while skipping over the $j$th index.

Proof: We proceed by mathematical induction on $n$. Consider the case $n=1$: $$(f_1)'=\sum_{j=1}^1\Bigg[f_j'\prod_{i=1,\;i\neq j}^1\Big[f_i\Big]\Bigg]\stackrel{\checkmark}{=}f_1'.$$ Suppose the statement holds for $n=k$, $k>1$. That is, $$(f_1\cdots f_k)'=\sum_{j=1}^k\Bigg[f_j'\prod_{i=1,\;i\neq j}^k\Big[f_i\Big]\Bigg]=(f_1'f_2\cdots f_k)+(f_1f_2'f_3\cdots f_k)+\cdots+\left(f_1\cdots f_{k-1} f_k'\right).$$ We then have that, via the product rule and our supposition, \begin{align*} (f_1\cdots f_kf_{k+1})'&=f_{k+1}\sum_{j=1}^k\Bigg[f_j'\prod_{i=1,\;i\neq j}^k\Big[f_i\Big]\Bigg]+f_{k+1}'\prod_{i=1}^k\Big[f_i\Big]\\ \\ &=f_{k+1}\bigg[(f_1'f_2\cdots f_k)+(f_1f_2'f_3\cdots f_k)+\cdots+\left(f_1\cdots f_{k-1} f_k'\right)\bigg]+f_{k+1}'\bigg[f_1\cdots f_k\bigg]\\ \\ &=\bigg[(f_1'f_2\cdots f_{k+1})+(f_1f_2'f_3\cdots f_{k+1})+\cdots+\left(f_1\cdots f_{k} f_{k+1}'\right)\bigg]\\ \\ &\stackrel{\checkmark}{=}\sum_{j=1}^{k+1}\Bigg[f_j'\prod_{i=1,\;i\neq j}^{k+1}\Big[f_i\Big]\Bigg].\;\;\;\;\;\;\;\;\;\;\blacksquare \end{align*}