It was given in an introductory example that using the Euler-Lagrange equation for $\ J[y] = \int_{a}^{b}({C(x,y) \cdot \sqrt{1+(y'(x))^2}})dx $ (where $\ C(x,y) $ is assumed to be known for all x,y) reduces to $\ C_y - C_xy' - C \cdot \frac{y''}{1+(y')^2} = 0$, but it didn't show the method to obtain that result. I understand that the Euler-Lagrange case to use here is: $\ L_f - \frac{d}{dx}L_{f'} = 0 $
since $\ L $ is a Lagrangian of $\ x $, $\ f(x) $, and $\ f'(x) $. So:
$\ L = C \cdot \sqrt{1+(y')^2} $
I think it follows that $\ L_f = C_y \cdot \sqrt{1+(y')^2} $ and $\ L_{f'} = \frac{C \cdot f'}{\sqrt{1+(y')^2}} $ and $\ \frac{d}{dx}L_{f'} = \frac{C_x \cdot y''}{(1+(y')^2)^{3/2}} $
Putting that into the Euler-Lagrange:
$\ 0 = C_y \cdot \sqrt{1+(y')^2} - \frac{C_x \cdot y''}{(1+(y')^2)^{3/2}}$
Assuming I did the partial derivatives correctly, I'm not sure how to proceed from here?
Probably best to slow down when it comes to these. Your equation should be: \begin{eqnarray} \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'}&=&0\\ \frac{\partial L}{\partial y} &=& C_y (x,y)\sqrt{1+(y')^2}\\ \frac{\partial L}{\partial y'} &=& C(x,y) \frac{y'}{\sqrt{1+(y')^2}}\\ \frac{d}{dx} \frac{\partial L}{\partial y'} &=& \frac{dC(x,y)}{dx}\frac{y'}{\sqrt{1+(y')^2}}+C(x,y)\frac{y''}{\sqrt{1+(y')^2}}-C(x,y)\frac{(y')^2 y''}{(1+(y')^2)^{3/2}}\\&=&\frac{C_x(x,y)y'(1+y'^2)+C_y(x,y)y'^2(1+y'^2)+C(x,y)y''(1+y'^2)-C(x,y)y'^2 y''}{(1+y'^2)^{3/2}}\\ \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'}&=& \frac{C_y (x,y)(1+y'^2)^2-(C_x(x,y)y'(1+y'^2)+C_y(x,y)y'^2(1+y'^2)+C(x,y)y'')}{(1+y'^2)^{3/2}}\\ &=&\frac{(C_y(x,y)-C_x(x,y)y')(1+y'^2)-C(x,y)y''}{(1+y'^2)^{3/2}}\\ &=&\frac{C_y(x,y)-C_x(x,y) y' - \frac{C(x,y)y''}{1+y'^2}}{\sqrt{1+y'^2}} \end{eqnarray}
This looks familiar...