你能帮我一下吗？我需要把这个表格放到我的文件中，但它不能被破坏，而且会导致公式触及单元格的边缘。我希望公式在单元格中居中（公式在左侧），并且我希望单元格的尺寸相等。非常感谢。
\documentclass{book}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath, mathtools, amsthm, amssymb}
\begin{document}
\begin{tabular}{|l|l|}
\hline
$y = \text{ costante }$& $y' = 0$\\
\hline
$y=x$ & $ y' = 1$\\
\hline
$y=x^{m} $ & $y' = mx ^{m-1} $\\
\hline
$ y= \displaystyle\frac{1}{x^{n} } $ & $y'=- \displaystyle\frac{n}{x^{n+1} } $\\
\hline
$ y = \displaystyle \frac{1}{x} $ & $ y' = \displaystyle\frac{1}{x^2 } $\\
\hline
$ y = \sqrt{n}{x} $ & $ y' = \displaystyle\frac{1}{n \sqrt{n}{x ^{n-1} }} $\\
\hline
$ y = \sqrt{}{x} $ & $ y' = \displaystyle\frac{1}{2 \sqrt{}{x}} $\\
\hline
$ y = \displaystyle\frac{1}{\sqrt{}{x}} $ & $ - \displaystyle\frac{1}{2x \sqrt{}{x}} $\\
\hline
$ y = \sqrt{n}{x ^{m} } $ & $ \displaystyle\frac{m}{n \sqrt{n}{x ^{n-m} }} $ \\
\hline
$ y= \displaystyle\frac{1}{\sqrt{n}{x ^{m} }} $ & $ - \displaystyle\frac{m}{nx \sqrt{n}{x ^{m} }} $\\
\hline
$ y = \sin_{}^{} (x) $ & $ y' = \cos_{}^{} (x) $ \\
\hline
$ y = \cos_{}^{} (x) $ & $y' = - \sin_{}^{} (x) $ \\
\hline
$ y = \tan_{}^{} (x) $ & $ y' = \displaystyle\frac{1}{\cos_{}^{2} (x) }= 1 + \tan_{}^{2} (x) $ \\
\hline
$ y = \cot_{}^{} (x) $ & $ y'= \displaystyle\frac{1}{\sin_{}^{2} (x) } = 1 - \cot_{}^{2} (x) $\\
\hline
$ y = \sinh_{}^{} (x) $ & $ y'= \cosh_{}^{} (x) $ \\
\hline
$ y = \cosh_{}^{} (x) $ & $ y' = \sinh_{}^{} (x) $ \\
\hline
$ y = \tanh_{}^{} (x) $ & $ y'= \displaystyle\frac{1}{\cosh_{}^{2} (x) } = 1- \tanh_{ }^{2} (x) $ \\
\hline
$ y = \coth_{}^{} (x) $ & $ y' = \displaystyle\frac{1}{\sinh_{}^{2} (x) }= 1- \coth_{}^{2} (x) $\\
\hline
$ y = \mbox{arccosh}_{}^{} (x) $ & $ y' = \displaystyle\frac{1}{\sqrt{}{x ^2 -1}} $\\
\hline
$ y = \mbox{arcsinh}_{}^{} (x) $ & $ y' = \displaystyle\frac{1}{\sqrt{}{x ^2 + 1}} $\\
\hline
$y = \mbox{arctanh}_{}^{} (x)$&$y'= \displaystyle\frac{1}{1- x ^2 }$\\
\hline
$y = \log_{a}^{} (x)$&$y' = \displaystyle\frac{1}{x} \log_{a}^{} (e) = \displaystyle\frac{1}{x \ln^{} (a) } $\\
\hline
$y = \ln^{} (x) $&$y'= \displaystyle\frac{1}{x} $\\
\hline
$y = a ^{x} $&$y'= a ^{x} \log_{}^{} (a)$\\
\hline
$y = e^{x} $&$y' = e^{x}$ \\
\hline
$y = e^{-x}$&$ y' = - e^{-x} $\\
\hline
$y = \arcsin_{}^{} (x)$& $y'= \frac{1}{\sqrt{}{a- x ^2 }}$\\
\hline
$y = \arccos_{}^{} (x) $&$- \displaystyle\frac{1}{\sqrt{}{1- x ^2 }}$\\
\hline
$y = \arctan_{}^{} (x) $&$ y'= \displaystyle\frac{1}{1 + x^2 }$\\
\hline
$ y = \mbox{arccot}_{}^{} (x) $ & $- \displaystyle\frac{1}{1 + x ^2 } $\\
\hline
\end{tabular}
\end{document}
答案1
- 你的表格很长，所以应该是长表类型，可以跨越两页
- 由于表格很窄，您可以考虑将其分成两个平行的部分，以便放在一页上。
- 对于等高单元格，我建议使用
tabularray包，它可以简单地定义行高 - 我将从列中删除
y =并y' =插入列标题。
如果使用长表，结果可能是：
\documentclass{book}
\usepackage[T1]{fontenc}
\usepackage{amssymb}
\usepackage{tabularray}
\UseTblrLibrary{amsmath}
\DeclareMathOperator{\arccosh}{arccosh}
\DeclareMathOperator{\arcsinh}{arcsinh}
\DeclareMathOperator{\arctanh}{arctanh}
\begin{document}
\begin{longtblr}[
label=none,
entry=none,
]{hlines, vlines,
colspec={Q[c, mode=dmath] Q[c, mode=dmath]},
rowsep=3pt,
row{2-Z} = {ht=2\baselineskip},
rowhead=1
}
y & y' \\
\text{ costante } & 0 \\
y=x & 1 \\
y=x^{m} & mx^{m-1} \\
\frac{1}{x^{n} } & y'c = - \frac{n}{x^{n+1}} \\
\frac{1}{x} & \frac{1}{x^2 } \\
\sqrt{n}{x} & \frac{1}{n \sqrt{n}{x^{n-1} }} \\
\sqrt{}{x} & \frac{1}{2 \sqrt{}{x}} \\
\frac{1}{\sqrt{x}} & -\frac{1}{2x \sqrt{x}} \\
\sqrt{n}{x ^{m} } & \frac{m}{n \sqrt{n}{x ^{n-m} }} \\
\frac{1}{\sqrt{n}{x^{m} }}
& - \frac{m}{nx \sqrt{n}{x ^{m} }} \\
\sin (x) & \cos(x) \\
\cos (x) & -\sin(x) \\
\tan (x) & \frac{1}{\cos^{2}(x) } = 1 + \tan^{2}(x) \\
\cot (x) & \frac{1}{\sin^{2}(x) } = 1 - \cot^{2}(x) \\
\sinh (x) & \cosh(x) \\
\cosh (x) & \sinh(x) \\
\tanh (x) & \frac{1}{\cosh^{2}(x) } = 1- \tanh^{2} (x) \\
\coth (x) & \frac{1}{\sinh^{2}(x) }= 1- \coth^{2} (x) \\
\arccosh(x) & \frac{1}{\sqrt{}{x^2 - 1}} \\
\arcsinh(x) & \frac{1}{\sqrt{}{x^2 + 1}} \\
\arctanh(x) & \frac{1}{1 - x^2} \\
\log_{a}(x) & \frac{1}{x} \log_{a}(e) = \frac{1}{x \ln(a) } \\
\ln(x) & y'= \frac{1}{x} \\
a ^{x} & a^{x} \log (a) \\
e^{x} & e^{x} \\
e^{-x} & -e^{-x} \\
\arcsin(x) & \frac{1}{\sqrt{}{a- x^2}} \\
\arccos(x) & -\frac{1}{\sqrt{}{1- x^2}} \\
\arctan(x) & \frac{1}{1 + x^2} \\
\mbox{arccot} (x) & -\frac{1}{1 + x^2} \\
\end{longtblr}
\end{document}
如果使用两个并行表，结果可能是：
\documentclass{book}
\usepackage[T1]{fontenc}
\usepackage{amssymb}
\usepackage{tabularray}
\UseTblrLibrary{amsmath}
\DeclareMathOperator{\arccosh}{arccosh}
\DeclareMathOperator{\arcsinh}{arcsinh}
\DeclareMathOperator{\arctanh}{arctanh}
\begin{document}
\noindent%
\begin{tblr}{hlines, vline{1-Y}=solid,vline{Z}=1pt,
colspec={Q[c, mode=dmath] Q[c, mode=dmath]},
rowsep=3pt,
row{2-Z} = {ht=2.4\baselineskip},
}
y & y' \\
\text{ costante } & 0 \\
y=x & 1 \\
y=x^{m} & mx^{m-1} \\
\frac{1}{x^{n} } & y'c = - \frac{n}{x^{n+1}} \\
\frac{1}{x} & \frac{1}{x^2 } \\
\sqrt{n}{x} & \frac{1}{n \sqrt{n}{x^{n-1} }} \\
\sqrt{}{x} & \frac{1}{2 \sqrt{}{x}} \\
\frac{1}{\sqrt{x}} & -\frac{1}{2x \sqrt{x}} \\
\sqrt{n}{x ^{m} } & \frac{m}{n \sqrt{n}{x ^{n-m} }} \\
\frac{1}{\sqrt{n}{x^{m} }}
& - \frac{m}{nx \sqrt{n}{x ^{m} }} \\
\sin (x) & \cos(x) \\
\cos (x) & -\sin(x) \\
\tan (x) & \frac{1}{\cos^{2}(x) }¸= 1 + \tan^{2}(x) \\
\cot (x) & \frac{1}{\sin^{2}(x) } = 1 - \cot^{2}(x) \\
\sinh (x) & \cosh(x) \\
\end{tblr}%
\begin{tblr}{hlines, vline{2-Z}=solid,
colspec={Q[c, mode=dmath] Q[c, mode=dmath]},
rowsep=3pt,
row{2-Z} = {ht=2.4\baselineskip},
}
y & y' \\
\cosh (x) & \sinh(x) \\
\tanh (x) & \frac{1}{\cosh^{2}(x) } = 1- \tanh^{2} (x) \\
\coth (x) & \frac{1}{\sinh^{2}(x) }= 1- \coth^{2} (x) \\
\arccosh(x) & \frac{1}{\sqrt{}{x^2 - 1}} \\
\arcsinh(x) & \frac{1}{\sqrt{}{x^2 + 1}} \\
\arctanh(x) & \frac{1}{1 - x^2} \\
\log_{a}(x) & \frac{1}{x} \log_{a}(e) = \frac{1}{x \ln(a) } \\
\ln(x) & y'= \frac{1}{x} \\
a ^{x} & a^{x} \log (a) \\
e^{x} & e^{x} \\
e^{-x} & -e^{-x} \\
\arcsin(x) & \frac{1}{\sqrt{}{a- x^2}} \\
\arccos(x) & -\frac{1}{\sqrt{}{1- x^2}} \\
\arctan(x) & \frac{1}{1 + x^2} \\
\mbox{arccot} (x) & -\frac{1}{1 + x^2} \\
\end{tblr}
\end{document}