Derivative of product of three functions: product rule I am trying to find the derivative of $f(x)= xe^x \csc x$, and I am not too sure how to even start. 
Is it two terms or three? $xe^x$ and $\csc x$ or is it $x$, $e^x$ and $\csc x$? I can't get a proper answer either way. 
With two terms I get $xe^x(-\csc x\cot x) + \csc x(e^x)$.
 A: As J.M. said,
one easy way yo handle
arbitrary products and quotients
is to use the
logarithmic derivative:
$(\ln f(x))'
=\dfrac{f'(x)}{f(x)}
$.
Suppose $f(x)$
contains a mixture of
products and quotients:
$f(x)
=\dfrac{a_1(x)a_2(x) ... a_m(x)}{b_1(x)b_2(x)...b_n(x)}
=\dfrac{\prod_{i=1}^m a_i(x)}{\prod_{j=1}^n b_j(x)}
$.
Taking the log,
$\ln f(x)
=\sum_{i=1}^m \ln a_i(x)-\sum_{j=1}^n \ln b_j(x)
$.
Differentiating both sides,
$\begin{array}\\
\dfrac{f'(x)}{f(x)}
&=(\ln f(x))'\\
&=(\sum_{i=1}^m \ln a_i(x)-\sum_{j=1}^n \ln b_j(x))'\\
&=\sum_{i=1}^m (\ln a_i(x))'-\sum_{j=1}^n (\ln b_j(x))'\\
&=\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\\
\end{array}
$.
Therefore
$\begin{array}\\
f'(x)
&=f(x)\left(\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\right)\\
&=\dfrac{\prod_{i=1}^m a_i(x)}{\prod_{j=1}^n b_j(x)}\left(\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\right)\\
\end{array}
$.
Here are the two most common cases:
$f(x) = a_1(x) a_2(x)$ - the product with $m=2$ and $n=0$.
$f'(x)
=a_1(x) a_2(x)\left(\dfrac{a_1'(x)}{a_1(x)}+\dfrac{a_2'(x)}{a_2(x)}\right)
=a_1'(x) a_2(x)+a_1(x) a_2'(x)
$.
$f(x) = \dfrac{a(x)}{b(x)}$ - the quotient
with $m=n=1$.
$f'(x)
= \dfrac{a(x)}{b(x)}\left(\dfrac{a'(x)}{a(x)}-\dfrac{b'(x)}{b(x)}\right)
= \dfrac{a'(x)}{b(x)}-\dfrac{a(x)b'(x)}{b^2(x)}
= \dfrac{a'(x)b(x)-a(x)b'(x)}{b^2(x)}
$.
Finally, your case - 
the product of three functions:
$\begin{array}\\
f(x)
&=a(x)b(x)c(x)\left(\dfrac{a'(x)}{a(x)}+\dfrac{b'(x)}{b(x)}+\dfrac{c'(x)}{c(x)}\right)\\
&=a'(x)b(x)c(x)+a(x)b'(x)c(x)+a(x)b(x)c'(x)\\
\end{array}
$
A: To find the derivative of $(abc)'$ you use repeated application of the product rule:
$$
(abc)' = (ab)'c+abc' = (ab'+a'b)c+abc' = a'bc+ab'c+abc'.
$$
In your case $a(x) = x$, $b(x) = \mathrm e^x$ and $c(x) = \operatorname{csc}(x)$, so
$$
a' = 1, b' = \mathrm e^x \text{ and }c' = -\cot x\csc x.
$$
To make it more clear: in $x \mathrm e^x\csc x$ you have three function rather than two, but $x\mathrm e^x$ is also a product of two functions, so
$$
(x\mathrm e^x\csc x)' = (x\mathrm e^x)'\csc x+x\mathrm e^x(\csc x)'.
$$
We can calculate the latter term, but what about $(x\mathrm e^x)'$? You again apply the product rule: 
$$
(x\mathrm e^x)' = x'\mathrm e^x+x(\mathrm e^x)'.
$$
A: $$
Let\quad y = \prod_1^n f_i(x)\quad and \quad thus\quad\ln(y) = \sum_1^n \ln(f_i(x))$$
recalling that$$ y'=y\space(\ln(y))'$$
then we obtain$$y'=\prod_1^n f_i(x)\space[\sum_1^n \ln(f_j(x))]'=\prod_1^n f_i(x)\space[\sum_1^n \dfrac {f'_j(x)}{f_j(x)} ]=\sum_1^n f'_i(x)[\prod_{j\neq i}  f_j(x)]
$$ 
and setting n=3
$$ (f_1 f_2 f_3)'=f'_1 f_2 f_3 +f_1 f'_2 f_3 + f_1 f_2 f'_3$$
A: From first principles, the product rule operation is invariant under the product of any number of functions. This can be proved through induction but it's interesting to see it using the definition of the derivative.
$$\lim_{dx \to 0}\frac{f_1(x+dx)...f_n(x+dx) - f_1(x)...f_n(x)}{dx}$$
is  
$$\lim_{dx \to 0}\frac{(f_1(x)'dx+f_1(x))...(f_n'(x)dx+f_n(x)) - f_1(x)...f_n(x)}{dx}$$
Note that the only term in the expansion that doesn't end up with a $dx$ multiplied by it is the derivative of some $f_q$ multiplied by the product of the remaining functions. The product $f_1(x)f_2(x)...f_n(x)$ disappears in the numerator. It follows that the derivative is 
$$ {\sum_{p=1}^n\prod_{i=1}^n f_i(x)}\frac{f'_p(x)}{f_p(x)}$$.
for any product of functions $$\prod_{i=1}^n f_i(x)$$
