Funes el memorioso

\documentclass[12pt]{article}
\pagestyle{empty}
\usepackage{amsmath,amssymb,amsthm}
\usepackage[papersize={100mm, 130mm}, text={90mm, 120mm}]{geometry}
 
% order: \order[st]{1}, \order{k}
\newcommand{\order}[2][th]{\ensuremath{{#2}^{\mathrm{#1}}}}
 
\newcommand{\Abf}{\ensuremath{\mathbf{A}}}
\newcommand{\Bbf}{\ensuremath{\mathbf{B}}}
\newcommand{\fpd}[2]{\ensuremath{\frac{\partial{#1}}{\partial{#2}}}}
 
\begin{document}
We prove the product rule of matrix derivative.
\begin{equation*}
\fpd{}{t}\Abf\Bbf = \fpd{\Abf}{t}\Bbf + \Abf\fpd{\Bbf}{t}
\end{equation*}
Let $[\boldsymbol{*}]_{ij}$ be the element of $\boldsymbol{*}$ in the
$\order{i}$ row and $\order{j}$ column. Then it's enough to prove
\begin{equation*}
\fpd{}{t}\left[\Abf\Bbf\right]_{ij} = 
\left[\fpd{\Abf}{t}\Bbf + \Abf\fpd{\Bbf}{t}\right]_{ij.}
\end{equation*}
Note that $[\Abf\Bbf]_{ij=}=\sum_ka_{ij}b_{kj}$.
 
\begin{align*}
  \fpd{}{t}[\Abf\Bbf]_{ij} &= \fpd{}{t}\sum_ka_{ik}b_{ij} \\
  &= \sum_k\left(
  \fpd{a_{ik}}{t}b_{kj} + a_{ik}\fpd{b_{kj}}{t}
  \right) \\
  &= \sum_k\left(\fpd{a_{ik}}{t}b_{kj}\right) +
  \sum_k\left(a_{ik}\fpd{b_{kj}}{t}\right) \\
  &= \left[\fpd{\Abf}{t}\Bbf\right]_{ij} + 
  \left[\Abf\fpd{\Bbf}{t}\right]_{ij} \\
  &= \left[ \fpd{\Abf}{t}\Bbf + \Abf\fpd{\Bbf}{t} \right]_{ij}
\end{align*}
\end{document}