# Subscript to fractions

How do I write this in Latex. I've looked everywhere and to be honest I don't even know how to describe it in google search bar.

I have

$$(\frac{\partial u}{\partial t})_(t=0)$$


but it's the last part that does work properly

• Hi, welcome. Note that you shouldn't be using $$..$$ for display math, see Why is $…$ preferable to $$? – Torbjørn T. Mar 8 '18 at 12:23 • Welcome to TeX.SX! Try _{t=0} instead of _(t=0). – Bobyandbob Mar 8 '18 at 12:25 • If you are still learning, a good way is to use this Latex equation editor. Since it has pre-made code it was easy for me to quickly learn the syntax for representing complex equations by examining common components and symbols. It's a good place to experiment. – syntonicC Mar 8 '18 at 17:48 ## 2 Answers The underscore character (_) will make only one math atom a subscript. In the case of your code, it was the (. To make it work you have to group the t=0 using braces ({...}). Also, use \left( and \right) instead of just ( and ) to make the parentheses taller for the fraction. Finally, use $ and $ to delimit the display math instead of $$ (reference).

\documentclass{article}
\usepackage{amsmath}
\begin{document}
$\left(\frac{\partial u}{\partial t}\right)_{t=0}$
\end{document}

• That is perfect thank you. I meant to use the curly brackets instead of the normal ones but I would have never of figured t out. And I was just going to ask you how to make the brackets taller. Also, I will try the [ instead of  It all works, thank you very much and hope you have a great day. – Jerry Mar 8 '18 at 12:27

I suggest using a dedicated package like esdiff to simplify typing. It defined a \diffp{function}{variables} command. In addition, the order of crossed partial derivatives is automatically calculated. The evaluation point of partial derivatives is added in the \diffp* command:

\documentclass{article}%

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage[thinc]{esdiff}

\begin{document}

$\diffp*{u}{t}{t = 0}\qquad \diffp{u}{t x}\qquad \diffp{u}{{t^2} x}$%

\end{document}


• Very nice. I not known this solution. +1 for all. – Sebastiano Mar 8 '18 at 20:46