1. ## Induction proof?

If $\displaystyle \prod_{n = 1}^{n} x_{n}$ denotes $\displaystyle x_{1}\cdot x_{2}\cdot x_{3}\cdot ...\cdot x_{n}$ then prove, by induction, that:
$\displaystyle \left\{\prod_{n = 1}^{n} f_{n}(x)\right\}' = \sum_{i = 1}^{n} \{f_{1}(x)\cdot f_{2}(x)\cdot ...\cdot f_{i}'(x)\cdot ...\cdot f_{n}(x)\}$

where $\displaystyle '$ denotes derivative w.r.t $\displaystyle x$

2. Have you tried ?

It's not difficult, you know that $\displaystyle (f_1f_2)'=f_1'f_2+f_1f_2'$ (initialisation).

So, suppose this egality true for $\displaystyle n$ functions, and show it's true for $\displaystyle n+1$ functions too :

$\displaystyle \left(\prod_{i=1}^{n+1}f_i\right)'=\left(f_{n+1}\p rod_{i=1}^{n}f_i\right)'=f_{n+1}'\prod_{i=1}^{n}f_ i+f_{n+1}{\color{red}\left(\prod_{i=1}^{n}f_i\righ t)'}$

Now substitute term in red (hypothesis of induction) :

$\displaystyle \left(\prod_{i=1}^{n+1}f_i\right)'=\left(f_{n+1}\p rod_{i=1}^{n}f_i\right)'=f_{n+1}'\prod_{i=1}^{n}f_ i+f_{n+1}\sum_{i=1}^{n}\left\{f_1...f_i'...f_n\rig ht\}$

i.e : $\displaystyle \left(\prod_{i=1}^{n+1}f_i\right)'=f_{n+1}'\prod_{ i=1}^{n}f_i+\sum_{i=1}^{n}\left\{f_1...f_i'...f_nf _{n+1}\right\}$

And finally : $\displaystyle \color{blue}\boxed{\left(\prod_{i=1}^{n+1}f_i\righ t)'=\sum_{i=1}^{{\color{red}n+1}}\left\{f_1...f_i' ...f_{n+1}\right\}}$

What ends the prove by induction.