Q:

Explanation of Lagrange Equation with Chain Rule

I am just reading through some lecture notes explaining the Lagrange Equation, and I am a bit confused with some chain rule stuff,
I get to the part with:
$$\frac{\partial F}{\partial y} = \frac{d}{dx}(\frac{\partial F}{\partial y'})$$ Here is where I am confused, it says, expanding the total derivative on the right using the chain rule:
$$\frac{\partial F}{\partial y} = \frac{\partial }{\partial x} (\frac{\partial F}{\partial y'}) + \frac{\partial }{\partial y} (\frac{\partial F}{\partial y'})y' + \frac{\partial }{\partial y'} (\frac{\partial F}{\partial y'})y''   $$
Could someone give me a 'dummies' explanation of the intermediate step here(perhaps with diagram would help)? Just a bit confused.
Many Thanks

A:

The main thing to keep in mind is that $y=y(x)$ and $y'=\dfrac{dy}{dx}$ are both to be understood as independent functions of $x$ in the Euler-Lagrange equation. To emphasize this, let me write $v=y'$. Then he total derivative $\dfrac{d}{dx}$ must take them all into account:
\begin{align}
\frac{d}{dx}
&=\frac{dx}{dx}\frac{\partial}{\partial x}+\frac{dy}{dx}\frac{\partial}{\partial y}+\frac{dv}{dx}\frac{\partial}{\partial v}\\
&=\frac{\partial}{\partial x}+y'\frac{\partial}{\partial y}+v'\frac{\partial}{\partial v}
\end{align}
Since $v=y'$ and $v'=y''$, we get the proper total derivative for the Euler-Lagrange equation.

