Derivating in respect of a function that's a derivative of another function

I have a problem understanding this equation (I encountered it in a mechanics problem). Given:

$x = x(t)$

$q = q(t)$

$v = dx/dt$

$w = dq/dt$

Then:

$\frac {\partial v}{ \partial w} = \frac{\partial x}{\partial q}$

Is that true? If yes, why is that true? How can I proof that?

EDIT: Actually it's a passage in the derivation of the Lagrange equations in Analytical Mechanics:

$v = dOP/dt$

$\dot{q} = dq/dt$

$\partial v / \partial \dot{q} = \partial OP/ \partial q$

Solution 1:

If you have a parametrized curve $(x,y)=(f(t),g(t))$, then the tangent vector is $(f'(t),g'(t))$. Provided that the tangent vector isn't pointing straight up or straight down (that is, provided that $f'(t)\neq 0$), the curve looks locally like the graph of a function $y=H(x)$, and its slope $dy/dx=H'(x)$ at some point $(x,y)=(a,H(a))$ is just $g'(t_0)/f'(t_0)$ where $t_0$ is the parameter value corresponding to that point on the curve (that is, $a=f(t_0)$ and $H(a)=g(t_0)$). To see this, think of the right triangle with the tangent vector as its hypotenuse; it has $\Delta y=g'(t_0)$ and $\Delta x=f'(t_0)$.

So what the statement $dy/dx=(dy/dt)/(dx/dt)$ really means is $H'(a)=g'(t_0)/f'(t_0)$.