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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...

Joe Goldiamond
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    Product rule on the middle term $\frac{\partial}{\partial y'} (\frac{\partial F}{\partial y}y') = (\frac{\partial}{\partial y'}\frac{\partial F}{\partial y})y' + (\frac{\partial F}{\partial y})\underbrace{\frac{\partial}{\partial y'}y'}_{=1}$ – Dayton Aug 08 '19 at 22:54
  • I understand what you are doing, but applied to the last term:$\frac{\partial}{\partial y'} (\frac{\partial F}{\partial y'}y'') = (\frac{\partial}{\partial y'}\frac{\partial F}{\partial y'})y'' + (\frac{\partial F}{\partial y'}) {\frac{\partial}{\partial y'}y''}$.Which is not correct I think – Joe Goldiamond Aug 08 '19 at 23:25
  • Does the last term vanish? Because it is equal to zero? – Joe Goldiamond Aug 09 '19 at 13:53
  • Remember you are taking a PARTIAL derivative $\frac{\partial }{\partial y'}$ NOT a full derivative like you did for $\frac{d}{dx}$, therefor you are holding $x$ and $y''$ constant. – Dayton Aug 09 '19 at 16:14

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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.

Axion004
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  • Remark: Some people use total derivative while other people say full derivative. I have always thought that total derivative means full derivative. – Axion004 Aug 10 '19 at 14:33