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}
$