Derivatives: Product Rule? f(x) * g(x) * s(x) =?

Question

I am familiar with the product rule for problems like:
d[f(x)*g(x)] /dx = f(x) * g '(x) + g(x) * f '(x)

However, I don't know what to do when I have a problem like:
d[f(x)*g(x)*s(x)] /dx =

Answer 1

You'd use the associative and commutative property of multiplication and group together the functions, then apply the product rule repeatedly. So (I'll suppress the variable and use the Newton ' notation riather than the Leibnitz d/dx notation to make this easier to read and write ...
(f*g*h)' =
f'(g*h) + f(g*h)' =
f'(g*h) + f[g'*h + g*h']
Depending on your functions, you might want to group them in such a way as to make your work easier, the commutative and associative property of multiplicaition allows you to do this as long as you're in a system that has that has these properties (that is, real or complex value functions, not matrix functions, which do not have the commutative property)
Also if you have a lot of functions in play, make a table keeping track of part of your work so that you don't have to keep re-writing (and taking a chance on making an error in copying) the parts that you've already figured out.

Answer 2

all combinations with one different derivative in each.

d/dx = f'gs + fg's + fgs'

Answer 3

Just apply the product rule twice.

d[f(x) * g(x) * s(x)] /dx
= f'(x) * (g(x) * s(x)) + f(x) * (g(x) * s'(x) + g'(x) * s(x))
= f'(x) * g(x) * s(x) + f(x) * g'(x) * s(x) + f(x) * g(x) * s'(x)

Answer 4

Same thing, f'(x)g(x)s(x) + f(x)g'(x)s(x) + f(x)g(x)s'(x)