# 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:

$v = dOP/dt$

$\dot{q} = dq/dt$

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

-
I don't think that conclusion is true. – anonymous Oct 27 '10 at 13:11
From an old school (pre-rigour, that is) point of view, the conclusion is perfectly reasonable: $\frac{dv}{dw} = \frac{dx/dt}{dq/dt}$. Now we just cancel the $dt$'s and we have $\frac{dx}{dq}$, right? (Of course showing that this sillyness actually works is the hard part) – kahen Oct 27 '10 at 13:32
@kahen exactly, my question is in fact, why this "simplification" works? – pygabriel Oct 27 '10 at 13:41

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 $H'(a)=g'(t_0)/f'(t_0)$.