Why is it that the derivative of $f(y(x))$ with respect to $x$ vanishes?

Question

I have been messing around with the Euler-Lagrange-Equation lately and found that one could use the Beltrami Identity with a function like this:

$$f(y,y') = \sqrt{\frac{1+y'^2}{-2gy}}~, ~~where~ y = y(x)$$

But the Beltrami Identity

$$ \frac{d}{dx}\bigg( f - y' \frac{\partial f}{\partial y'}\bigg) = 0 $$

is only applicable when $\frac{\partial f}{\partial x} = 0$, or as I've red "f is not explicitly dependent on x".

But why is it possible for the derivitive to vanish in that case? $x$ still could change and it is not a constant, right?

Thank you for your help.

Regards,

Answer 1

Indeed, "$f$ is not explicitly dependent on $x$". The key here is explicitly, that is, you can write $f$ as a function of $y$ and $y'$ without once mentioning $x$, even though $y$ is of course dependent on $x$, otherwise $y$ would just be a constant.

This all hinges on understanding the difference between a partial derivative $\partial f / \partial x$ and the total derivative $df /dx$. The first is zero in your case, while the second is not.