Explaining Example of Euler-Lagrange Equation

Question

It seems quite clear to me that for a function $y=f(x)$, $\frac{d(y')}{dy}=\frac{d(y')}{dx}\frac{dx}{dy}=\frac{y''(x)}{y'(x)}$. This exploration of the Euler Lagrange theorem claims that, for $F(x,y,y')=\sqrt{1+(y')^2}$, we find $$\frac{d}{dy}F(x,y,y')=0$$However, it seems to me that $$ \begin{align} \frac{d}{dy}F(x,y,y')&=\frac{d}{dy}\sqrt{1+(y')^2}\\ &=\frac{1}{\sqrt{1+(y')^2}}\Big{(}\frac{d}{dy}(1+(y')^2)\Big{)}\\ &=\frac{1}{\sqrt{1+(y')^2}}\Big{(}2y'\frac{d(y')}{dy}\Big{)}\\ &=\frac{1}{\sqrt{1+(y')^2}}\Big{(}2y'\frac{y''(x)}{y'(x)}\Big{)}\\ &=\frac{2y''(x)}{\sqrt{1+(y')^2}}\neq{0} \end{align} $$ What am I misunderstanding?

Answer 1

The problem is the (convenient but awful) notation that is used in the calculus of variations.

The $y,y'$ in the expression are (from the point of view of the expression) unrelated.

Write $F(a,b,c) = \sqrt{1+c^2}$, then ${\partial F(a,b,c) \over \partial b} = 0$.