Is this property of differentiable functions only to be used when independent?

Question

I am seeing in my textbook the following:

In the problem statement it says , assume $F$ is independent of $x$ , and $F$ is a function of $x, y$ and $y$.

And then in the solution for the problem it implies this without any explanation:

$\frac{dF}{dx}=(\frac{\partial}{\partial x}+y'\frac{\partial}{\partial y}+y''\frac{\partial}{\partial y'})F$

I'm wondering , is it just because $F$ is independent of $x$ that we are allowed to say this? OR, is this something that we can always do?

PS, shortly after, it says $\frac{\partial F}{\partial x}=0$, which is expected because it is independent of $x$ as it says in the problem statement.

Answer 1

Well...It seems that F doesn't explicitly depend on x while y does and ${\partial F}\over{\partial x}$ can be reformulated as according to $y^{'}{{\partial F}\over{\partial y}}$

Answer 2

It's just the chain rule, but it's kind of hidden. In the calculus of variations, like in a lot of calculus, people are really lax about specifying what depends on what. To clarify what's going on here I'm going to introduce some intermediate variables $X$, $Y_1$, and $Y_2$ (which are all functions of $x$) defined as

$X(x) = x$

$Y_1(x) = y(x)$

$Y_2(x) = y'(x)$

And instead of writing $F(y, y', x)$ we will write it as $F(Y_1, Y_2, X)$. Now let's look at how $F$ depends on $x$. $x$ maps to a triple $(Y_1, Y_2, X)$ by the above rules, and these map to F by whatever rule we have for $F$. Symbolically $$F(x) = F(Y_1(x), Y_2(x), X(x)).$$ So writing out the chain rule for a function with three intermediate variables we get $$\frac{d F}{d x} = \frac{\partial F}{\partial X} \frac{d X}{d x} + \frac{\partial F}{\partial Y_1} \frac{d Y_1}{d x} + \frac{\partial F}{\partial Y_2} \frac{d Y_2}{d x}$$

Now looking back at the way that we defined our intermediate variables we can make following simplifications

$$\frac{d X}{d x} = 1$$

$$\frac{d Y_1}{d x} = \frac{d}{d x} y = y'$$

$$\frac{d Y_2}{d x} = \frac{d}{d x} y' = y''.$$

Substitute these into the chain rule and go back to treating $F$ as a fuction of $x$, $y$, and $y'$ instead of $X$, $Y_1$, and $Y_2$ and you get formula in your question.

(The notes make it a bit harder to see by using one order when writing $F$ as a function of $x$, $y$, and $y'$, and a different order for writing the terms of the chain rule, but the order of the terms added together for the chain rule actually doesn't matter.)

Oh, and you can always do this; it doesn't depend on $F$ being independent of $x$. That's used in the later part of the solution.