Step in derivation of Euler-Lagrange equations of motion

From http://www.mathpages.com/home/kmath523/kmath523.htm

Variations in $x,y,z$ and $X$ at constant $t$ are independent of $t$ (since each of these variables is strictly a function of $t$), so we have
$${\frac{\partial x}{\partial X}=\frac{\partial{\dot x}}{\partial{\dot X}}}$$

(This is just after equation (5) on the page.)

I’m having trouble making sense of this. If each of these variables is strictly a function of $t$, and $t$ is held constant, how does a time derivative (${\dot x}$) make sense?

If it were to mean that $t$ is replaced with a constant after the differentiation, then I could take the example of

$$ x(y)=X^{3/4}=t^{3}$$
$$ X(t)=t^{4}$$

I could then calculate

$$ {\frac{\partial x}{\partial X}=\frac{\dot x}{\dot X} = \frac{3}{4t}}$$

and

$$ {\frac{\partial {\dot x}}{\partial {\dot X}}=\frac{\ddot x}{\ddot X}=\frac{1}{2t}}$$

…and those are not equal for any $t$.

What is the justification for this step?

Note: This is a follow up to my first question about this, but that error was resolved. Rates of change for functions dependent on same variable

Solutions Collecting From Web of "Step in derivation of Euler-Lagrange equations of motion"

  • Between eq. (4) and eq. (5)

By definition

$$x=x(X(t),Y(t)),$$

and not $x=x(X(t),Y(t),t)$; so

$$\frac{dx}{dt}=\frac{\partial x}{\partial X}\frac{dX}{dt}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},$$
from which follows
$$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right)=
\frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt},$$

as $\frac{\partial }{\partial X}\frac{dX}{dt}=\frac{\partial }{\partial X}\frac{dY}{dt}=0.$ On the other hand

$$\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right):=
\frac{d}{dt}g(X(t),Y(t))=
\frac{\partial g}{\partial X}\frac{dX}{dt}+ \frac{\partial g}{\partial Y}\frac{dY}{dt} = \frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt}, $$

i.e.

$$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right) =\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right).$$

  • Just after eq. (5)

As $$x=x(X(t),Y(t)),$$ then $\frac{\partial x }{\partial \dot{X}}=0$ and $\frac{\partial x }{\partial \dot{X}}\left(\frac{\partial x}{\partial X} \right)=0,$ as well.
Using the notation $\dot{x}=\frac{dx}{dt} $ it follows that

$$\frac{\partial\dot{x} }{\partial \dot{X}}=
\frac{\partial }{\partial \dot{X}}\left(\frac{\partial x}{\partial X}\dot{X}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},\right)=\frac{\partial x}{\partial X},$$

as claimed.