Deriving the Lagrangian of Catenoid using Lagrange-Euler Equation I was trying to derive the Largangian of $F(x, y, y')\;=\;y\sqrt{1\;+\;\left(y'\right)^2}$. Which I did it with not much help. This is my working out: 
$\frac{\partial F}{\partial y}\;=\;\sqrt{1\;+\;\left(y'\right)^2}$
$
\begin{align}
\frac{\partial F}{\partial y'}\;&=\;\frac{\mathrm{d}}{\mathrm{d}\,y'}\left[y\sqrt{1\;+\;\left(y'\right)^2}\right] \\
&=\;0\;+\;\frac{2yy'}{2\sqrt{1\;+\;\left(y'\right)^2}} \\
&=\;\frac{yy'}{\sqrt{1\;+\;\left(y'\right)^2}}
\end{align}
$
$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y'}\;&=\;\frac{\mathrm{d}}{\mathrm{d} x}\left[y\sqrt{1\;+\;\left(y'\right)^2}\right] \\
&=\;\frac{\left(y'y'\;+\;yy''\right)\sqrt{1\;+\;\left(y'\right)^2}}{\left(\sqrt{1\;+\;\left(y'\right)^2}\right)^2} \;-\;\frac{yy'\;\times\;\frac{y'y''}{\sqrt{1\;+\;\left(y'\right)}}}{\left(\sqrt{1\;+\;\left(y'\right)^2}\right)^2} \\
&=\;\frac{y'y'\;+\;yy''}{\sqrt{1\;+\;\left(y'\right)^2}} \;-\;\frac{yy'y'y''}{\left(\sqrt{1\;+\;\left(y'\right)^2}\right)^3} 
\end{align}
$
I got up to the Lagnrangian, which is 
$$\mathcal{L}\;=\;\frac{y'y'\;+\;yy''}{\sqrt{1\;+\;\left(y'\right)^2}} \;-\;\frac{yy'y'y''}{\left(\sqrt{1\;+\;\left(y'\right)^2}\right)^3}\;-\;\sqrt{1\;+\;\left(y'\right)^2}\;=\;0$$
So my question comes after I saw this MathSE answer. I wonder how can one simplify the Lagrangian $\mathcal{L}$ to the following: 
$$ 1\;+\;(y')^2\;-\;yy''\;=\;0$$
 A: If you convert the denominators with their LCM and expand the terms you will get the ODE. Here are the steps: 
$
\begin{align}
\frac{\left(y'y'\;+\;yy''\right)\left(1\;+\;\left(y'\right)^2\right)\;-\;y\,y'y'y''
\;-\;\left[1\;+\;\left(y'\right)^2\right]^2}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber\\
\frac{y'y'\left(1\;+\;\left(y'\right)^2\right)\;+\;yy''\left(1\;+\;\left(y'\right)^2\right)\;-\;y\,y'y'y''
\;-\;\left[1\;+\;\left(y'\right)^2\right]^2}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber\\
\frac{y'y'\left(1\;+\;\left(y'\right)^2\right)\;+\;yy''\;+\;yy''y'y'\;-\;y\,y'y'y''
\;-\;\left[1\;+\;\left(y'\right)^2\right]^2}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber\\
\frac{y'y'\left(1\;+\;\left(y'\right)^2\right)\;+\;yy''
\;\color{red}{-}\;\left[1\;+\;\left(y'\right)^2\right]^2}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber\\
\frac{y'y'\left(1\;+\;\left(y'\right)^2\right)\;+\;yy''
\;\color{red}{-}\;1\;\color{red}{-}\;2\left(y'\right)^2\;\color{red}{-}\;\left(y'\right)^4}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber\\
\frac{y'y'+\;\left(y'\right)^4\;+\;yy''
\;\color{red}{-}\;1\;\color{red}{-}\;2y'y'\;\color{red}{-}\;\left(y'\right)^4}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber\\
\frac{yy''
\;-\;1\;-\;y'y'}{\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3} \;&=\;0 \nonumber
\end{align}
$
Since the denominator $\left[\sqrt{1\;+\;\left(y'\right)^2}\right]^3$ will never be $0$, then we have 
$$yy''\;-\;1\;-\;y'y'\;=\;0$$
