问题：我想在我的文档中按系统顺序呈现动画效果。可以吗？
提前致谢
我目前做了什么
\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{parskip}
\usepackage[left=1.30cm, right=1.30cm, top=1.30cm, bottom=1.30cm]{geometry}
\begin{document}
\large
\textbf{Example: Obtain first order derivative of the function} $\boldsymbol{\dfrac{x^2+1}{x^2-1}}$
\begin{align*}
\dfrac{d}{dx} \left[\dfrac{x^2+1}{x^2-1}\right]&=\dfrac{(x^2-1)\cdot \dfrac{d}{dx} \left\{x^2+1\right\} - (x^2+1)\cdot \dfrac{d}{dx} \left\{x^2-1\right\}}{(x^2-1)^2}&\\[16pt]
&=\dfrac{(x^2-1)\cdot\left\{2x+0\right\}-(x^2+1)\cdot\left\{2x-0\right\}}{(x^2-1)^2}&\\[16pt]
&=\dfrac{(x^2-1)\cdot\left\{2x\right\}-(x^2+1)\cdot\left\{2x\right\}}{(x^2-1)^2}&\\[16pt]
&=\dfrac{2x\left[(x^2-1)-(x^2+1)\right]}{(x^2-1)^2}&\\[16pt]
&=\dfrac{2x\left[x^2-1-x^2-1\right]}{(x^2-1)^2}&\\[16pt]
&=\dfrac{2x\left(-2\right)}{(x^2-1)^2}&\\[16pt]
&=\dfrac{-4x}{(x^2-1)^2}&\\[16pt]
\end{align*}
\end{document}
答案1
这是一个很好的答案但它在多行数学公式中不起作用。这是我的有点尴尬的技巧：
\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{parskip}
\usepackage[left=1.30cm, right=1.30cm, top=1.30cm, bottom=1.30cm]{geometry}
\begin{document}
\large
\textbf{Example: Obtain first order derivative of the function} $\boldsymbol{\dfrac{x^2+1}{x^2-1}}$
\begin{align*}
\dfrac{d}{dx} \left[\dfrac{x^2+1}{x^2-1}\right]&=\dfrac{(x^2-1)\cdot \dfrac{d}{dx} \left\{x^2+1\right\} - (x^2+1)\cdot \dfrac{d}{dx} \left\{x^2-1\right\}}{(x^2-1)^2}\\[16pt]
\end{align*}\newpage
\large
\textbf{Example: Obtain first order derivative of the function} $\boldsymbol{\dfrac{x^2+1}{x^2-1}}$
\begin{align*}
\dfrac{d}{dx} \left[\dfrac{x^2+1}{x^2-1}\right]&=\dfrac{(x^2-1)\cdot \dfrac{d}{dx} \left\{x^2+1\right\} - (x^2+1)\cdot \dfrac{d}{dx} \left\{x^2-1\right\}}{(x^2-1)^2}\\[16pt]
&=\dfrac{(x^2-1)\cdot\left\{2x+0\right\}-(x^2+1)\cdot\left\{2x-0\right\}}{(x^2-1)^2}\\[16pt]
\end{align*}
\end{document}
