product rule for multiple products

Question

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!

Answer 1

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.