Anonymous 14035
I tried this code with two examples:
```
\documentclass{article}
\usepackage{fourier-otf}
\usepackage{luacas}
\usepackage{amsthm}
\theoremstyle{definition}
\newtheorem{ex}{Example}
\usepackage[top=15mm,bottom=20mm,left=2cm,right=2cm]{geometry}
\begin{document}
\begin{ex}
\begin{CAS}
vars('x')
u = topoly(x^2 - 5*x + 6)
du = u:derivative()
v = du/2
f = diff(sqrt(u),x)
\end{CAS}
Let be given the function
\[f(x) = \sqrt{\print{u}}\]
Using by hand
\[y' = \dfrac{\print{du}}{2\sqrt{\print{u}}}.\]
Using luacas
\[ y' = \print*{f}.\]
How can I write the output of luacas in the form?
\[y' = \dfrac{x-2}{\sqrt{x^2 - 4x + 6}}.\]
\end{ex}
\begin{ex}
\begin{CAS}
vars('x')
u = topoly(x^2 - 4*x + 6)
du = u:derivative()
v = du/2
f = diff(sqrt(u),x)
\end{CAS}
Let be given the function
\[f(x) = \sqrt{\print{u}}\]
Using by hand
\[y' = \dfrac{\print{du}}{2\sqrt{\print{u}}} = \dfrac{\print{v}}{\sqrt{\print{u}}}.\]
Using luacas
\[ y' = \print*{f}.\]
How can I write the output of luacas in the form?
\[y' = \dfrac{x-2}{\sqrt{x^2 - 4x + 6}}.\]
\end{ex}
\end{document}
```
