Proving the product rule for n functions I am trying to prove that the product rule works for $n$ many functions, where $n$ is an integer greater than $2$. I am able to prove it for two functions, where the rule states if $k(x)=f(x)g(x)$ , then $k'(x)=f'(x)g(x) + f(x)g'(x)$. However, I have surfed the Internet for a proof for $n$ many functions and found none; and, I believe that If I were to prove it for $n$ functions directly it would take  a very long time - given I even know how to. But, I read somewhere about an induction proof of it but it assumed that the rule works for $n$ functions then proved that if it works for $n$ functions than it also works for $n+1$ functions. So, my question is, can anyone possibly link me to a rigid proof of the product rule for $n$ functions or possibly write one since I have no idea how to write a quick proof for it.
 A: This is a case where logarithmic differentiation makes life much easier.
Consider $$F(x)=\prod_{i=1}^n f_i(x)$$ Taking logarithms of both sides $$\log\big(F(x)\big)=\sum_{i=1}^n \log\big(f_i(x)\big)$$ Differentiating $$\frac 1{F(x)}\frac{dF(x)}{dx}=\sum_{i=1}^n \frac 1{f_i(x)}\frac{df_i(x)}{dx}$$  $$\frac{dF(x)}{dx}=\Big(\sum_{i=1}^n \frac 1{f_i(x)}\frac{df_i(x)}{dx}\Big) \times\Big(\prod_{i=1}^n f_i(x)\Big)=\sum_{i=1}^n \frac{\prod_{j=1}^n f_j(x) }{f_i(x)}\frac{df_i(x)}{dx}$$ $$\frac{dF(x)}{dx}=\sum_{i=1}^n \frac{df_i(x)}{dx}\prod_{j=1,j\ne i}^n f_j(x) $$ 
A: You can use induction on $n$, the number of functions. 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 $f_1\cdot f_2 \cdots f_{n+1}$ = $f_1g$ where $g = f_2\cdots f_{n+1}$.
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.
The way mathematical induction works is that if you've proved it for a base case, here it is $n=2$ and you prove that if your result is true for some integer $n$ then it must be true for some integer $n+1$ as well. Then because you've shown your result in the case $n=2$ is true, you've also shown that your result must be true in the case that $n=2 + 1= 3$ is true, and hence you've shown that $n=3+1 = 4$ is true, and hence $n=4+1 = 5$, etc...
