$\frac {\partial} {\partial y'}\bigg(\frac {dF}{dx}\bigg)$ In a part about the Euler - Langrange equation it is said that for a function $F(x,y,y')$
$$ \frac {dF}{dx} = \frac {\partial F}{\partial x}+ \frac {\partial F}{\partial y}y'+ \frac{\partial F}{\partial y'}y''$$
Now we wish to apply this to:
$$\frac {\partial} {\partial y'}\bigg(\frac {dF}{dx}\bigg)$$
Hence in my opinion: 
$$\frac {\partial} {\partial y'}\bigg(\frac {\partial F}{\partial x}+ \frac {\partial F}{\partial y}y'+ \frac{\partial F}{\partial y'}y''\bigg)$$
Which tmho would become:
$$\frac {\partial^2F}{\partial x\partial y'}+\frac {\partial^2F}{\partial y\partial y'}y'+\frac {\partial^2F}{\partial y'^2}y''$$
But the answer tells me: 
$$\frac {\partial^2F}{\partial x\partial y'}+\frac {\partial^2F}{\partial y\partial y'}y'+\frac {\partial^2F}{\partial y'^2}y''+\frac{\partial F}{\partial y}$$
Could somebody explain where I am going wrong? I can't see where the last fraction $\frac {\partial F}{\partial y}$ comes from...
 A: The error is due to a misinterpretation of the partial derivative and full derivative. Computationally, the calculation would be 
\begin{align}\frac {\partial} {\partial y'}\bigg(\frac {dF}{dx}\bigg)&=\frac {\partial} {\partial y'}\bigg(\frac {\partial F}{\partial x}+ \frac {\partial F}{\partial y}y'+ \frac{\partial F}{\partial y'}y''\bigg) \\&=
\frac {\partial} {\partial y'}\bigg(\frac {\partial F}{\partial x}\bigg)+\frac {\partial} {\partial y'}\bigg( \frac {\partial F}{\partial y}y'\bigg)+\frac {\partial} {\partial y'}\bigg(\frac{\partial F}{\partial y'}y''\bigg) \\&=
\frac {\partial^2F}{\partial x\partial y'}+\bigg(\frac {\partial^2F}{\partial y\partial y'}y'+\frac{\partial F}{\partial y}\bigg)+\frac {\partial^2F}{\partial y'^2}y''
\end{align}
where the last equality is true because
$$\frac {\partial} {\partial y'}\bigg( \frac {\partial F}{\partial y}y'\bigg)=\frac {\partial^2F}{\partial y\partial y'}y'+\frac{\partial F}{\partial y}\tag{1}$$
$$\frac {\partial} {\partial y'}\bigg(\frac{\partial F}{\partial y'}y''\bigg)=\frac {\partial^2F}{\partial y'^2}y''\tag{2}$$
As noted in the comments, $(1)$ follows from a direct application of the product rule. However, $(2)$ follows from the definition of the partial derivative. When you defined $F=F(x,y,y')$ as a function of $x,y,y'$ and then wrote
$$ \frac {dF}{dx} = \frac {\partial F}{\partial x}+ \frac {\partial F}{\partial y}y'+ \frac{\partial F}{\partial y'}y''\tag{3}$$
you took the full derivative of $\frac {dF}{dx}$. However
$$\frac {\partial} {\partial y'}\bigg(\frac{\partial F}{\partial y'}y''\bigg)$$
isn't a full derivative. It is a partial derivative in which you assume that both $x$ and $y''$ are fixed (i.e. constants). This is why $(2)$ isn't
$$\frac {\partial} {\partial y'}\bigg(\frac{\partial F}{\partial y'}y''\bigg)=\frac {\partial^2F}{\partial y'^2}y''+ \bigg(\frac{\partial F}{\partial y'}\bigg)\frac {\partial} {\partial y'} y''$$
A good summary of the difference between the two is shown inside this post. The key difference is that when you take a partial derivative, you operate under a sort of assumption that you hold one variable fixed while the other changes. When computing a total derivative, you no longer have this assumption. As shown inside $(3)$, changes in one variable can affect the other two variables.
