\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{multicol}
\usepackage{pifont}
\usepackage{tasks}
\usepackage{empheq}
\usepackage{tikz,tkz-tab}
\usetikzlibrary{calc,positioning}
\usepackage{multirow}
\usepackage{makecell}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert} % Commande pour obtenir la valeur absolue.
\DeclarePairedDelimiter{\openintvl}{]}{[}
\renewcommand{\arraystretch}{2}
\usetheme{Madrid}
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\begin{document}
\begin{frame}[allowframebreaks]
\renewcommand\arraystretch{0.1}
\begin{displaymath}
\begin{array}{|c|c|c|c|}
\hline
\mbox{Fonction }f &D_f &\mbox{Fonction d\'eriv\'ee }f'&D_{f'} \\
\hline
x^n,\ n\in{\mathbb N}& {\mathbb R} &nx^{n-1}&{\mathbb R} \\
\hline
\rule[-2.5ex]{0pt}{6ex}\displaystyle\frac{1}{x^n}=x^{-n}, n\in{\mathbb N}\setminus\{0\} &{\mathbb R}\setminus\{0\} &-\displaystyle\frac{n}{x^{n+1}}=-nx^{-n-1}&{\mathbb R}\setminus\{0\} \\
\hline
x^{\alpha},\ \alpha\in]0,+\infty[&]0,+\infty[ &\alpha x^{\alpha-1}&]0,+\infty[ \\
\hline\hline
\mbox{e}^x &{\mathbb R} &\mbox{e}^x&{\mathbb R} \\
\hline
\rule[-2.5ex]{0pt}{6ex} \ln\vert x\vert &{\mathbb R}\setminus\{0\} &\displaystyle\frac{1}{x}&{\mathbb R}\setminus\{0\} \\
\hline\hline
\cos x &{\mathbb R} &-\sin x&{\mathbb R} \\
\hline
\sin x &{\mathbb R} &\cos x&{\mathbb R} \\
\hline
\rule[-2.5ex]{0pt}{6ex}\tan x &{\mathbb R}\setminus\{\frac{\pi}{2}+\pi{\mathbb Z}\} &1+\tan^2( x)= \displaystyle\frac{1}{\cos^2(x)}&{\mathbb R}\setminus\{\frac{\pi}{2}+\pi{\mathbb Z}\} \\
\hline
\hline
\cosh x &{\mathbb R} &\sinh x&{\mathbb R} \\
\hline
\sinh x &{\mathbb R} &\cosh x&{\mathbb R} \\
\hline
\tanh x &{\mathbb R} &1-\tanh^2(x)= \displaystyle\frac{1}{\cosh^2(x)}&{\mathbb R} \\
\hline\hline
\arccos x &[-1,1] &-\displaystyle \displaystyle\frac{1}{\sqrt{1-x^2}}&]-1,1[ \\
\hline
\arcsin x &[-1,1] &\displaystyle\frac{1}{\sqrt{1-x^2}}&]-1,1[ \\
\hline
\arctan x &{\mathbb R} &\displaystyle\frac{1}{1+x^2}&{\mathbb R} \\
\hline
\end{array}
\end{displaymath}
\end{frame}
\end{document}
答案1
我会用定制的长桌来解决您的问题。
由于我不喜欢表格中的垂直规则,因此我根据 booktabs 指南切换了布局。
\documentclass{beamer}
\usepackage{booktabs}
\usepackage[column=O]{cellspace}
\usepackage{longtable}
\usetheme{Madrid}
\setlength\cellspacebottomlimit{0.37em}
\setlength\cellspacetoplimit{0.37em}
\begin{document}
\begin{frame}[allowframebreaks]
\scriptsize
\begin{longtable}{*{4}{>{$\displaystyle}O{c}<{$}}}
\toprule
\mbox{Fonction }f & D_f & \mbox{Fonction d\'eriv\'ee }f' & D_{f'} \\ \midrule
x^n,\ n\in{\mathbb N} & \mathbb{R} & nx^{n-1} & \mathbb{R} \\
\frac{1}{x^n}=x^{-n}, n\in{\mathbb N}\setminus\{0\} & \mathbb{R}\setminus\{0\} & -\frac{n}{x^{n+1}}=-nx^{-n-1} & \mathbb{R}\setminus\{0\} \\
x^{\alpha},\ \alpha\in]0,+\infty[ & ]0,+\infty[ & \alpha x^{\alpha-1} & ]0,+\infty[ \\ \midrule
\mbox{e}^x & \mathbb{R} & \mbox{e}^x & \mathbb{R} \\
\ln\vert x\vert & \mathbb{R}\setminus\{0\} & \frac{1}{x} & \mathbb{R}\setminus\{0\} \\ \midrule
\cos x & \mathbb{R} & -\sin x & \mathbb{R} \\
\sin x & \mathbb{R} & \cos x & \mathbb{R} \\
\tan x & \mathbb{R}\setminus\{\frac{\pi}{2}+\pi{\mathbb Z}\} & 1+\tan^2(x)= \frac{1}{\cos^2(x)} & \mathbb{R}\setminus\{\frac{\pi}{2}+\pi{\mathbb Z}\} \\ \bottomrule
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\multicolumn{4}{O{r}}{continue on next slide...} \\ \pagebreak
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\toprule
\mbox{Fonction }f & D_f & \mbox{Fonction d\'eriv\'ee }f' & D_{f'} \\ \midrule
\cosh x & \mathbb{R} & \sinh x & \mathbb{R} \\
\sinh x & \mathbb{R} & \cosh x & \mathbb{R} \\
\tanh x & \mathbb{R} & 1-\tanh^2(x)= \frac{1}{\cosh^2(x)} & \mathbb{R} \\ \midrule
\arccos x & [-1,1] & - \frac{1}{\sqrt{1-x^2}} & ]-1,1[ \\
\arcsin x & [-1,1] & \frac{1}{\sqrt{1-x^2}} & ]-1,1[ \\
\arctan x & \mathbb{R} & \frac{1}{1+x^2} & \mathbb{R} \\ \bottomrule
\end{longtable}
\end{frame}
\end{document}