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答案1
答案2
使用 更简单mathtools
。我添加了一个用于旧式导数的命令:
\documentclass{article}
\usepackage{fourier, erewhon}
\usepackage{textcomp}
\usepackage{geometry}%
\usepackage{mathtools}
\newcommand\osder[2]{#2^{\textsc{\scriptsize(\romannumeral #1)}}}
\begin{document}
\[ \begin{matrix*}[l]
x^2 & x^3 & x^4 & x^5 & x^6 & x^7 \\
f''(x)=2 & f'''(x)=6 & f^{(\textrm{iv})}(x)=24 & f^{(\textrm{v})}(x)=120 & f^{(\textrm{vi})}(x)=720 & f^{(\textrm{vii})}(x)=5040\\[1ex]
\multicolumn{2}{l}{\text{or in old style: }} & \osder{4}{f}(x)=24 & \osder{5}{f}(x)=120 & \osder{6}{f}(x)=720 & \osder{7}{f}(x)=5040
\end{matrix*} \]
\end{document}
答案3
在第二次调用中,我使用罗马数字来表示高阶导数,这是我见过的常见风格。
\documentclass{article}
\usepackage{stackengine}
\renewcommand\stackalignment{l}
\renewcommand\stacktype{L}
\setstackgap{L}{1.2\baselineskip}
\stackMath
\begin{document}
\[
\stackunder{x^2}{f''(x)=2}\quad
\stackunder{x^3}{f'''(x)=6}\quad
\stackunder{x^4}{f''''(x)=24}\quad
\stackunder{x^5}{f'''''(x)=120}\quad
\stackunder{x^6}{f''''''(x)=720}\quad
\stackunder{x^7}{f'''''''(x)=5040}\quad
\]
\bigskip
\[
\stackunder{x^2}{f''(x)=2}\quad
\stackunder{x^3}{f'''(x)=6}\quad
\stackunder{x^4}{f^\mathrm{(iv)}(x)=24}\quad
\stackunder{x^5}{f^\mathrm{(v)}(x)=120}\quad
\stackunder{x^6}{f^\mathrm{(vi)}(x)=720}\quad
\stackunder{x^7}{f^\mathrm{(vii)}(x)=5040}\quad
\]
\end{document}
使用 TABstack 可以实现相同的结果:
\documentclass{article}
\usepackage{tabstackengine}
\renewcommand\stackalignment{l}
\renewcommand\stacktype{L}
\setstacktabbedgap{1em}
\stackMath
\begin{document}
\[
\tabbedstackunder[1.2\baselineskip]{x^2 & x^3 & x^4 & x^5 & x^6 & x^7}{
f''(x)=2 & f'''(x)=6 & f^\mathrm{(iv)}(x)=24 & f^\mathrm{(v)}(x)=120 &
f^\mathrm{(vi)}(x)=720 & f^\mathrm{(vii)}(x)=5040}
\]
\end{document}