# Adding underbrace on $\left(\frac{dB}{dA}\right)^2\right$

I am trying to add underbrace to $\left(\frac{dB}{dA}\right)^2\right$. I have \underbrace{\left(\frac{dB}{dA}\right)^2\right}_\text{(-)}.

However, LaTeX is giving me an error. Do I have to install a package for this?

EDIT: I have the following packages:

\usepackage{graphicx}

\usepackage{amssymb,amsthm}

\usepackage{natbib}

\bibpunct{(}{)}{;}{;}{,}{,}

\usepackage{xr-hyper}

\usepackage[
citecolor=blue,
urlcolor=blue,
]{hyperref}

\usepackage{bm}

\usepackage{fullpage}

\usepackage{ amssymb }

\pagestyle{plain}

\setlength{\parskip}{\baselineskip}
\setlength{\parindent}{0pt}

\setcounter{secnumdepth}{2}

\allowdisplaybreaks[4]

% Commenting/debugging
\let\IG\iffalse
\let\ENDIG\fi

%% Shortcuts
\newcommand{\td}[2]{\dfrac{d #1}{d #2}}
\newcommand{\std}[2]{\dfrac{d^2 #1}{d {#2}^2}}
\newcommand{\ctd}[3]{\dfrac{d^2 #1}{d #2 d #3}}

\newcommand{\pd}[2]{\dfrac{\partial #1}{\partial #2}}
\newcommand{\spd}[2]{\dfrac{\partial^2 #1}{\partial {#2}^2}}
\newcommand{\cpd}[3]{\dfrac{\partial^2 #1}{\partial #2 \partial #3}}

\newcommand{\pdi}[2]{\partial #1/\partial #2}

\newcommand{\LR}{\Leftrightarrow}
\newcommand{\Lg}{\mathcal{L}}
\newcommand{\half}{\tfrac{1}{2}}
\newcommand{\eqp}{\phantom{=}}
\newcommand{\eqs}{\buildrel s \over =}

\begin{document}
\end{document}

\thispagestyle{empty}
\frac{d^2B}{dA^2}=-\frac{1}{\underbrace{(1-p)U'(B)}_\text{(+)}}\left[\underbrace{p U''(A)}_\text{(-)} + \underbrace{(1-p)U''(B)}_\text{(-)}\left(\frac{dB}{dA}\right)^2\right]>0.


I want to add braces to \left(\frac{dB}{dA}\right).

• See if $\underbrace{\left\{\left(\frac{dB}{dA}\right)^2\right\}_\text{(-)}}$ does what you want. Please always post a fully compilable MWE including \documentclass and the appropriate packages. – Peter Grill Sep 29 '14 at 3:51
• @PeterGrill I'm not getting a negative sign below with this. – OGC Sep 29 '14 at 3:53

How about a vertical alignment of the \underbraces using a strut:

\documentclass{article}

\usepackage{amsmath}

\begin{document}
$\newcommand{\tempstrut}{\rule[-1.2\baselineskip]{0pt}{1.5\baselineskip}} \frac{\mathrm{d}^2B}{\mathrm{d}A^2} = - \underbrace{\tempstrut\frac{1}{(1-p)U'(B)}}_{(+)} \Bigg[\underbrace{\tempstrut p U''(A)}_{(-)} + \underbrace{\tempstrut (1-p)U''(B)}_{(-)} \underbrace{\tempstrut\Bigg(\frac{\mathrm{d}B}{\mathrm{d}A}\Bigg)^2}_{(+)}\Bigg] > 0.$
\end{document}

\documentclass{article}
\usepackage{graphicx}

\usepackage{amsmath,amssymb}

\begin{document}
$\frac{d^2B}{dA^2}=-\frac{1}{\underbrace{(1-p)U'(B)}_{\text{(+)}}}\Biggl[\underbrace{p U''(A)}_{\text{(-)}} + \underbrace{(1-p)U''(B)}_{\text{(-)}} \underbrace{\left(\frac{dB}{dA}\right)^2}_{\text{(+)}}\Biggr]>0.$
\end{document}


• The last term is positive though, and do not want a double square on it. – OGC Sep 29 '14 at 4:25