我可以将第一行与其他行对齐
\begin{eqnarray*}
\dfrac{\partial^2 P_{ijl}}{\partial \tau_l \partial \phi} & = &\dfrac{\partial}{\partial \tau_l}\left(P_{ijl}(1-P_{ijl})\frac{\partial \eta_{ijl}}{\partial \phi}\right) \\
& = &\dfrac{\partial P_{ijl}}{\partial \tau_l}\left(1-P_{ijl}\right)\frac{\partial \eta_{ijl}}{\partial \phi}+\frac{\partial \left(1-P_{ijl}\right)}{\partial \tau_l}P_{ijl}\frac{\partial \eta_{ijl}}{\partial \phi}+P_{ijl}(1-P_{ijl})\frac{\partial }{\partial \tau_l}\left(\frac{\partial \eta_{ijl}}{\partial \phi}\right) \\
& = & ${P_{ijl}}{(1-P_{ijl})^2}$\dfrac{\partial \eta_{ijl}}{\partial \phi}-P^2_{ijl}\left(1-P_{ijl}\right)\frac{\partial \eta_{ijl}}{\partial \phi}+0 \\
& = & ${P_{ijl}}{(1-P_{ijl})}(1-{2P}_{ijl})$\dfrac{\partial \eta_{ijl}}{\partial \phi}
\end{eqnarray*}
答案1
% arara: pdflatex
\documentclass{article}
\usepackage{mathtools}
\newcommand\pd[3][]{\frac{\partial^{#1}#2}{\partial#3^{#1}}}
\newcommand{\md}[6]{\frac{\partial^{#2}#1}{\partial#3^{#4}\partial#5^{#6}}}
\begin{document}
\begin{align*}
\md{P_{ijl}}{2}{\tau_l}{}{\phi}{} &=\pd{}{\tau_l}\bigg(P_{ijl}(1-P_{ijl})\pd{\eta_{ijl}}{\phi}\bigg) \\
&=\pd{P_{ijl}}{\tau_l}(1-P_{ijl})\pd{\eta_{ijl}}{\phi}+\pd{(1-P_{ijl})}{\tau_l}P_{ijl}\pd{\eta_{ijl}}{\phi}+P_{ijl}(1-P_{ijl})\pd{}{\tau_l}\bigg(\pd{\eta_{ijl}}{\phi}\bigg) \\
&=P_{ijl}(1-P_{ijl})^2\pd{\eta_{ijl}}{\phi}-P^2_{ijl}(1-P_{ijl})\pd{\eta_{ijl}}{\phi}+0 \\
&= P_{ijl}(1-P_{ijl})(1-2P_{ijl})\pd{\eta_{ijl}}{\phi}
\end{align*}
\end{document}
