Simplification of $\frac{d}{dx}\frac{dx}{dt}$? How can $\frac{d}{dx} \frac{dx}{dt}$ be simplified, where $x$ is a function of $t$?
My guess is that since partial derivatives commute, that $\frac{d}{dx} \frac{dx}{dt}=\frac{d}{dt}\frac{dx}{dx}=\frac{d}{dt}(1)=0$, but am not sure if I'm missing something.
Additionally would this mean that $\frac{d}{dx} \frac{dx^2}{dt}=\frac{d}{dt}\frac{dx^2}{dx}=\frac{d}{dt}(2x)=2\dot{x}$?
 A: Suppose $x=f(t)$ gives the position of a particle at time $t$.  Then $\frac{dx}{dt} = f'(t)$ is the velocity of the particle at time $t$.  Let's write $v(t) = f'(t)$ for shorthand.
Your question is asking:  What's $\frac{dv}{dx}$?
Now the issue seems to be that $v$ does not explicitly depend on $x$ at all, which would seem to imply that $\frac{dv}{dx} = 0$.  This, however, oversimplifies the situation.
Am example might help:  Suppose we have a particle whose position is given by $x = t^2$ for $t\ge 0$.  Then $v = \frac{dx}{dt} = f'(t) = 2t$, which looks (as described above) like it doesn't depend on $x$.  But appearances can be deceiving!  In fact, the velocity does depend on position.  Just think about the moving object:  certainly, at different positions, the velocity will be different.  How do we reveal this in the equation of motion?
The key is that the formula $x = t^2$ does two things:  it explicitly defines the position as a function of time, but it also implicitly defines the time as a function of position.  That is, if we solve for $t$, we have $t = \sqrt{x}$ (remember, we are assuming $t \ge 0$).  Then substituting this into the velocity formula, we have
$$ v = 2\sqrt{x}$$
and then $\frac{dv}{dx} = \frac{1}{\sqrt{x}}$.
Now let's step away from the specific example and go back to the general case.  We have
$$x = f(t)$$
whose derivative with respect to $t$ is
$$\frac{dx}{dt} = f'(t)$$
But also, inverting the original equation, we have
$$t = f^{-1}(x)$$
which, on substituting into the equation above, gives
$$\frac{dx}{dt} = f'(f^{-1}(x))$$
Now to take the derivative of this with respect to $x$, use the chain rule:
$$\frac{d}{dx} \frac{dx}{dt} = f''(f^{-1}(x)) \cdot (f^{-1})'(x)$$
Finally, use the Inverse Function Theorem, which says that $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$, to conclude that
$$\frac{d}{dx} \frac{dx}{dt} = f''(f^{-1}(x)) \cdot \frac{1}{f'(f^{-1}(x))}$$
Going back to our example, we have $f^{-1}(x) = \sqrt{x}$, $f'(t) = 2t$, and $f''(t) = 2$, so the formula above gives 
$$\frac{d}{dx} \frac{dx}{dt} = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$$
as found above.
