# Thread: Derivative of more Products than 2?

1. ## Derivative of more Products than 2?

I'm not sure where to begin with finding the derivative of this function:

f(x) = x * e^x * cscx

I understand the product rule and how it applies to problems that have two functions, but how would one incorporate a third function?

I've using the product rule on the first two, x & e^x, and then used that derivative and putting it in as one of the two functions in the product rule, along with cscx, but this seems like a sloppy/invalid method.

Thank you!

2. I would do what you have suggested

$\displaystyle f(x) = x \times e^x \times \csc(x)$

Make $\displaystyle u = x \times e^x$ and $\displaystyle v = \csc(x)$

To find $\displaystyle u'$ you will need to use the product rule again.

3. Oh, well thank you! That's two you've answered for me.

4. Think of (uvw)'= ((uv)w)' and treat (uv) as a single function:
((uv)w)'= (uv)'w+ (uv)w'. Now use the product rule on (uv). (uv)'= u'v+ uv" so (uvw)'= (u'v+ uv')w+ (uv)w'= u'vw+ uv'w+ uvw'.

That extends very easily to any number of functions: (uvw...z)'= u'vw...z+ uv'w...z+ uvw'...z+ ...+ uvw...z'.

Another way to get that is to use "logarithmic differentiation". If f= uvw...z then ln(f)= ln(uvw...z)= ln(u)+ ln(v)+ ln(w)+ ...+ ln(z). (ln(f))'= (1/f)f'= u'/u+ v'/v+ w'/w+ ... z'/z. Multiplying on both sides by f= uvw...z gives f'= u'vw...z+ uv'w...z+ uvw'...z+...+ uvw...z'