# Product rule of 3 functions

1. Oct 21, 2011

### Notwen7

Product rule generally seems straight forward but what if one comes across a scenario involving 3 functions instead of 2?

For example:

d/dx[x*e^(x^2)*f(x)]

f(x) is just some generic function

So there three functions of x are:
-x
- e^(x^2)
-f(x)

I am personally lost about how to solve this problem. I was considering doing product rule on the first 2 functions and then using that to do another product rule by involving the third function. If only f(x) was known than this problem could be much more predictable.

If anyone can help steer me in the right direction I would greatly appreciate it. Hopefully this problem isn't more simple than I thought.

2. Oct 21, 2011

### bp_psy

You just apply the product rule as usual. If you have some product of functions F=fghu...v then F'=f'ghu...v+fg'hu...v+fgh'u...v)...fghu...v'.

3. Oct 21, 2011

### HallsofIvy

For three functions, fgh, you can think of it as f(gh) and apply the product rule for that: f'(gh)+ f(gh)'. And, now, of course, apply the product rule to (gh)': (fgh)'= f'gh+ f(g'h+ gh')= f'gh+ fg'h+ fgh'.

You can use induction to prove that the formula bp psy gives works for the product of "n" functions or use logarithmic differentiation: if F= (fghu...v) then ln(F)= ln(f)+ ln(g)+ ln(h)+ ln(u)...+ ln(v) so that F'/F= f'/f+ g'/+ h'/h+ u'/u+ ...+ v'/v. Now, multiply through by F= fghu...v and the functions in the denominator cancel leaving
F'= f'ghu..v+ fg'hu..v+ fgh'u...v+ fghu'...v+ fghu...v'.

4. Oct 21, 2011

### Notwen7

That makes a lot of sense. Thanks bp_psy and HallsofIvy for the input!