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新的微分 d 定义在论坛测试

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kuing Posted 2014-12-27 22:02 |Read mode
旧:\newcommand\rmd[1]{\mathop{\mathrm{d}#1}}
新:\newcommand\rmd{\mathop{}\!\mathrm{d}}

\(\newcommand\oldrmd[1]{\mathop{\mathrm{d}#1}}\)
\(\newcommand\newrmd{\mathop{}\!\mathrm{d}}\)
$1+\oldrmd{x^2}+1$ wrong

${1+\oldrmd{x^2}}+1$ right

$1+\newrmd x^2+1$

$\int f(x)\oldrmd x$

$\int f(x)\newrmd x$

$\displaystyle
\frac{\oldrmd x}{\oldrmd y}+\frac{\oldrmd{^2x}}{\oldrmd{y^2}}$

$\displaystyle
\frac{\newrmd x}{\newrmd y}+\frac{\newrmd^2x}{\newrmd y^2}$

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 Author| kuing Posted 2014-12-27 22:02
OK,没问题

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