# Parentheses changing size depending on what's in them?

Very simple problem, but very confusing. When I type

\left( \frac{\partial f}{\partial x }  \right)^2


I get smaller parentheses but when I type the same thing but with y instead of x, i.e.

\left( \frac{\partial f}{\partial y }  \right)^2


I get larger ones... what's up with that? Is it a known problem or do I have to supply more information?

This difference isn't present if the expression is written by itself, but only if it's in the denominator, i.e. when I write

\frac{...something...} { \left( \frac{\partial f}{\partial x }  \right)^2 + \left( \frac{\partial f}{\partial y}  \right)^2}


I get parentheses of different size!

• Simply y adds more vertical space and the parenthesis in bigger. Are you looking for a solution to make them the same size? – karlkoeller Aug 14 '13 at 8:27
• Thanks, that makes sense! A solution would be nice, you can post it as an answer if you want :) – lel Aug 14 '13 at 8:30
• To manually control parentheses size, you can use \big( and \big), instead of \left( and \right. See tex.stackexchange.com/a/38870/18228 for more size options. – Herr K. Aug 14 '13 at 8:41
• The method linked by @KevinC is the right one in these cases (\biggl and \biggr seem to be the more appropriate in your case). As an alternative you can put \vphantom{y} next to the x in your first equation. – karlkoeller Aug 14 '13 at 8:51
• More than interested, excellent! I really hope it grows into a high-quality community similar to mathoverflow! – lel Aug 27 '14 at 14:07

Here is an great old post that provide thorough explanation on how \left and \right work:

Incorrect delimiter size in math mode

Now back to your question, as karlkoeller explained, letter y adds more vertical space so the parenthesis become bigger. It can be observed as the following

There are basically three ways to solve the problem, we can choose among redefining \left and \right (which is strongly recommended) as the post above suggested, increase the size of df/dx, and decrease the size of df/dy. The size changing can be accomplished by using \rule or \raisebox:

\documentclass{article}
\usepackage{amsmath}

\newlength{\dyfoo}
\newlength{\hyfoo}
\newlength{\dxfoo}
\newlength{\hxfoo}
\settodepth{\dyfoo}{$$\displaystyle\frac{\partial f}{\partial y}$$}
\settoheight{\hyfoo}{$$\displaystyle\frac{\partial f}{\partial y}$$}
\settodepth{\dxfoo}{$$\displaystyle\frac{\partial f}{\partial x}$$}
\settoheight{\hxfoo}{$$\displaystyle\frac{\partial f}{\partial x}$$}

\begin{document}
$\left(\frac{\partial f}{\partial x}\right) \left(\frac{\partial f}{\partial \vphantom{y}x}\right) \left(\frac{\partial f}{\partial y}\right) \left(\raisebox{0pt}[\hyfoo][\dyfoo]{$$\displaystyle\frac{\partial f}{\partial x}$$}\right) \left(\raisebox{0pt}[\hxfoo][\dxfoo]{$$\displaystyle\frac{\partial f}{\partial y}$$}\right)$
\end{document}


Compare the result:

Remark:

\left and \right delimiters are supposed to be symmetric around math axis. Hence the vertical difference in both delimiter are no larger than 2pt. So there shouldn't be any necessity to make the adjustment.