I wasn't really sure how to phrase the title. Suppose I have a purely time-dependent function $x(t)$, and I want to know its time derivative $\dot{x}:=\dfrac{dx}{dt}$. Then I ask how the time derivative of $x$ changes as $x$ changes. So I'm considering $\dfrac{d\dot{x}}{dx}$. Does this make sense? Because in all applied courses I've taken you're told it's not "proper maths" but you can basically divide through by a differential if it appears both in the numerator and denominator of a fraction. So $\dfrac{d\dot{x}}{dx}=\dfrac{d}{dx} \dfrac{dx}{dt}=\dfrac{d}{dt}.$ This doesn't really make sense to me - how can we have taken a derivative and arrived at a differential operator?
The motivation for this question is that I'm doing an applied course where I'm calculating the Euler-Lagrange equations for some Lagrangian, and the Lagrangian is explicitly a function of $\dot{x}$, and the E-L equations involve taking the partial $x$ derivative of the Lagrangian. I understand that in taking the partial derivative with respect to $x$, you hold all other variables constant. Does $\dot{x}$ count as a separate variable? I'm pretty sure that when I did an introductory dynamics course a year or two ago, I was told that the derivative $\dfrac{\partial{\dot{x}}}{\partial{x}}$ is zero, but I don't think it was ever explained.
It's occurred to me more recently that you only ever build upon what is taught previous courses, and that there's not really very many opportunities to go over old concepts that you should already know, so it might be a stupid question but if I don't ask I'm stuck with the misunderstanding. Thanks for any help!
I dont' think it is necessary redefining $\dot x= v$ and introduce any EulerLagrage equation here. It seams for me that you want to estimate a sensitivity of "velocity" with respect to "coordinate". Basically you need a ratio of "acelleration" $\ddot x$ and "velocity" $\dot x$. And indeed $$ \frac {d (dx/dt)} {dx}=\frac {d (dx/dt)} {dt (dx/dt)} =\frac{\ddot x}{\dot x} $$ This is how I understand the problem. I hope it might help you.