乳胶中含有长颗粒的部分

Question

$$\frac{\partial \mathcal{L}}{\partial \eta_1}= -\sum_{i=1}^{n}\frac{\frac{\eta_1}{c}-\cos(y_i-\beta_i X_i)}{c-\eta_1\cos(y_i-\beta_i X_i)-\eta_2\sin(y_i-\beta_i X_i)} $$

我需要一个大的分母和分子。

Answer 1

您可以使用 \dfrac ( displaystylefraction) 命令amsmath或 \mfrac（medium-sizedfraction，大约 80% 的 displaystyle）（来自nccmath）：

\documentclass{article}
\usepackage[utf8]{inputenc}}
\usepackage{mathtools, nccmath}


\newcommand*{\e}{\mathrm{e}}

\begin{document}

\[ \frac{\partial \mathcal{L}}{\partial \eta_1}= -\sum_{i=1}^{n}\frac{\dfrac{\eta_1}{c}-\cos(y_i-\beta_i X_i)}{c-\eta_1\cos(y_i-\beta_i X_i)-\eta_2\sin(y_i-\beta_i X_i)} \]%

 \[ \frac{\partial \mathcal{L}}{\partial \eta_1}= -\sum_{i=1}^{n}\frac{\mfrac{\eta_1}{c}-\cos(y_i-\beta_i X_i)}{c-\eta_1\cos(y_i-\beta_i X_i)-\eta_2\sin(y_i-\beta_i X_i)} \]%

  \[ \frac{\partial \mathcal{L}}{\partial \eta_1}= -\sum_{i=1}^{n}\frac{\frac{\eta_1}{c}-\cos(y_i-\beta_i X_i)}{c-\eta_1\cos(y_i-\beta_i X_i)-\eta_2\sin(y_i-\beta_i X_i)} \]%

\end{document}