How do I make the fraction look more presentable and not so small?

Question

\begin{align*}
\dv[2]{SA}{r} &=\frac{{\left({\pi^{2}r^{10}+9V^{2}r^{4}}\right)^{\frac{1}{2}}}(12\pi^{2}r^{5})
-\left({\frac{2\pi^{2}r^{6}-9V^{2}}{2\sqrt{\pi^{2}r^{10}+9V^{2}r^{4}}}}\right)}
\end{align*}

I am using amsmath. Additionally, I am not sure why the word "align" comes before the equal sign but I can figure that out later. I mostly confused as to why the fraction is so small.

Answer 1

I think that using the \mfrac command (medium-sized fraction), from nccmath looks better. I took the opportunity to remove some unnecessary pairs of braces and replaced the \left( \right) around the nested fraction with better sized \Bigl( \Bigr).

\documentclass{article}
\usepackage{amsmath, nccmath}
\usepackage{esdiff}
\begin{document}
\begin{align}
    \diff[2]{SA}{r} &=\frac{\left(\pi^2 r^{10}+9V^2 r^4\right)^{\frac{1}{2}}(12\pi^2 r^5)- \Bigl(\mfrac{2\pi^2 r^6 -9V^2} {2\sqrt{\pi^2 r^{10}+9V^2 r^4}}\Bigr)}{\text{denominator}}\
\end{align}
\end{document} 

Answer 2

I can think of two suggestions:

• Use \dfrac instead of \frac (while eliminating the unnecessary parentheses) for the second additive term in the numerator.

• Replace the repeated term \sqrt{\pi^{2}r^{10}+9V^{2}r^{4}} with a symbol, say, W.

I dare say the second expression is a bit more readable.

Either way, do please also supply the missing denominator term.

\documentclass{article}
\usepackage{amsmath}
\newcommand\dv[3][]{\frac{\mathrm{d}^{#1}\mathit{#2}}{\mathrm{d}#3^{#1}}}
\begin{document}
Set $W\equiv{(\pi^{2}r^{10}+9V^{2}r^{4})}^{1/2}$. Then
\begin{align}
\dv[2]{SA}{r} 
&=\frac{{({\pi^{2}r^{10}+9V^{2}r^{4}})^{\frac{1}{2}}}(12\pi^{2}r^{5})-
%%\left({
\dfrac{2\pi^{2}r^{6}-9V^{2}}{2\sqrt{\pi^{2}r^{10}+9V^{2}r^{4}}}
%%}\right)
}{\text{(missing denominator??)}} \[2\jot]
&= \frac{12\pi^{2}r^{5}W-(\pi^{2}r^{6}-4.5V^{2})/W%
}{\text{(missing denominator??)}}
\end{align}
\end{document}

Answer 3

It is possible to use \dfrac instead of \frac in the last fraction

\documentclass{article}
\usepackage{physics}
\begin{document}
\begin{align}
    \dv[2]{SA}{r} &=\frac{{\left({\pi^{2}r^{10}+9V^{2}r^{4}}\right)^{\frac{1}{2}}}(12\pi^{2}r^{5})-\left({\frac{2\pi^{2}r^{6}-9V^{2}}{2\sqrt{\pi^{2}r^{10}+9V^{2}r^{4}}}}\right)}{}\
    \dv[2]{SA}{r} &=\frac{{\left({\pi^{2}r^{10}+9V^{2}r^{4}}\right)^{\frac{1}{2}}}(12\pi^{2}r^{5})-\left({\dfrac{2\pi^{2}r^{6}-9V^{2}}{2\sqrt{\pi^{2}r^{10}+9V^{2}r^{4}}}}\right)}{}
\end{align}
\end{document}

Answer 4

Using \displaystyle, you can do something like

\documentclass{article}
\usepackage{ physics}
\begin{document}
\begin{align*}
\dv[2]{SA}{r} &=\frac{{\left({\pi^{2}r^{10}+9V^{2}r^{4}}\right)^{\frac{1}{2}}}(12\pi^{2}r^{5})-\left({\frac{2\pi^{2}r^{6}-9V^{2}}{2\sqrt{\pi^{2}r^{10}+9V^{2}r^{4}}}}\right)}{}\\
\dv[2]{SA}{r} &=\frac{\displaystyle{\left({\pi^{2}r^{10}+9V^{2}r^{4}}\right)^{\frac{1}{2}}}(12\pi^{2}r^{5})-\left({\frac{2\pi^{2}r^{6}-9V^{2}}{2\sqrt{\pi^{2}r^{10}+9V^{2}r^{4}}}}\right)}{}
\end{align*}
\end{document}