product rule for multiple products Consider the following derivative
$$
\frac{d}{dx}\left[\prod_i\left(x\prod_jf_{ij}(x)\right)\right]
$$
I am not certain how to perform this. Can I apply the product rule as follows: Let $g_i=x\prod_j f_{ij}(x)$ then
$$
\frac{d}{dx}\left[\prod_i\left(x\prod_jf_{ij}(x)\right)\right] = \frac{d}{dx}\left[\prod_ig_i(x)\right] = \left(\prod_i g_i(x)\right)\left(\sum_i\frac{g'_i(x)}{g_i(x)}\right)
$$
where
$$
g'_i(x)  = \frac{d}{dx}\left(x\prod_j f_{ij}(x)\right) = \left[\prod_jf_{ij}(x) + x\frac{d}{dx}\left(\prod_jf_{ij}(x)\right)\right]
$$
with
$$
\frac{d}{dx}\left(\prod_jf_{ij}(x)\right) = \left(\prod_j
f_{ij}(x)\right)\left(\sum_j\frac{f_{ij}'(x)}{f_{ij}(x)}\right)
$$
Putting all of this back together I arrive at
$$
\frac{d}{dx}\left[\prod_i\left(x\prod_jf_{ij}(x)\right)\right] = \prod_i\left(x\prod_j f_{ij}(x)\right)\left[\sum_i\left(x\prod_jf_{ij}(x)\right)^{-1} \left[\prod_jf_{ij}(x) + x\left(\prod_j
f_{ij}(x)\right)\left(\sum_j\frac{f_{ij}'(x)}{f_{ij}(x)}\right)\right]\right]
$$
However, I am not certain that I am applying the product rule correctly for higher-order multiplications. Any help is appreciated!
 A: Your calculation is fine. We derive from
\begin{align*}
\frac{d}{dx}&\left(f_1(x)f_2(x)f_3(x)\right)\\
&=f_1^{\prime}(x)\left(f_2(x)f_3(x)\right)+f_1(x)\left(f_2(x)f_3(x)\right)^{\prime}\\
&=f_1^{\prime}(x)\left(f_2(x)f_3(x)\right)+f_1(x)\left(f_2^{\prime}(x)f_3(x)+f_2(x)f_3^{\prime}(x)\right)\\
&=f_1(x)f_2(x)f_3(x)\left(\frac{f_1^{\prime}(x)}{f_1(x)}+\frac{f_2^{\prime}(x)}{f_2(x)}+\frac{f_3^{\prime}(x)}{f_3(x)}\right)
\end{align*}
the formula for general $n$:
\begin{align*}
\color{blue}{\frac{d}{dx}\prod_{i=1}^nf_i(x)=\prod_{i=1}^nf_i(x)
\sum_{k=1}^n\frac{f_k^{\prime}(x)}{f_k(x)}}\tag{1}
\end{align*}

We obtain using (1)
\begin{align*}
\color{blue}{\frac{d}{dx}}&\color{blue}{\prod_{i=1}^n\left(x\prod_{j=1}^mf_{ij}(x)\right)}\\
&=\prod_{i=1}^n\left(x\prod_{j=1}^mf_{ij}(x)\right)\sum_{k=1}^n
\frac{\left(x\prod_{j=1}^nf_{kj}(x)\right)^{\prime}}{\left(x\prod_{j=1}^nf_{kj}(x)\right)}\tag{2}
\end{align*}

Since again using (1) and the product formula we get
\begin{align*}
\left(x\prod_{j=1}^nf_{kj}(x)\right)^{\prime}
&=\prod_{j=1}^mf_{kj}(x)+x\prod_{j=1}^mf_{kj}(x)
\sum_{l=1}^m\frac{f_{kl}^{\prime}(x)}{f_{kl}(x)}\\
&=\prod_{j=1}^mf_{kj}(x)\left(1+x\sum_{l=1}^m\frac{f_{kl}^{\prime}(x)}{f_{kl}(x)}\right)\tag{3}
\end{align*}
we continue with (2) using (3) and obtain

\begin{align*}
\prod_{i=1}^n&\left(x\prod_{j=1}^mf_{ij}(x)\right)\sum_{k=1}^n
\left(x\prod_{j=1}^nf_{kj}(x)\right)^{-1}\left(x\prod_{j=1}^nf_{kj}(x)\right)^{\prime}\\
&\,\,\color{blue}{=\prod_{i=1}^n\left(x\prod_{j=1}^mf_{ij}(x)\right)\sum_{k=1}^n
\left(x\prod_{j=1}^nf_{kj}(x)\right)^{-1}
\prod_{j=1}^mf_{kj}(x)\left(1+x\sum_{l=1}^m\frac{f_{kl}^{\prime}(x)}{f_{kl}(x)}\right)}
\end{align*}
in accordance with OPs calculation.

