multiplying derivatives I have a question about multiplying derivatives. In the expression, $$ \frac{d^{2}v}{dxdt}$$ if I multiply by $\frac{dt}{dt}$ will it then become equal to $$\frac{d^{2}v}{dt^{2}}\frac{dt}{dx} \mbox{ or } \frac{d^{2}v}{d^{2}t^{2}}\frac{dt}{dx} $$
 A: I think that this is just a confusion about notation. So first a disclaimer, some might disagree with the following.
Note that $$\frac{dt}{dt} = \frac{d}{dt}t = 1.$$
Note that 
$$
\frac{d^2v}{dxdt} = \frac{d}{dx}\frac{d}{dt}v(x,t).
$$
Since you are using the hard $d$'s as opposed to the partial $\partial$'s here we are saying that we are in 
$$
\frac{d}{dt}v(x,t)
$$
considering $v$ of course as a function of $x$ and $t$, but we are also considering $x$ as a function of $t$. We sometimes call the derivatives with hard $d$'s the total derivatives.  So you have by the chain rule
$$
\frac{d}{dt}v(x,t) = \frac{\partial v}{\partial x}\frac{dx}{dt} + \frac{\partial v}{\partial t}\frac{dt}{dt}.
$$
I wanted to write this because you do actually see a $\frac{dt}{dt}$ some up sometimes.
As another sidenote: We usually don't write things like
$$
\frac{d^2v}{d^{\color{red} 2}v^2}.
$$
We don't have a square on the $d$ on the bottom.
A: As a general rule, it is risky to treat $dt$ as if it were a number. There are some cases where it works, but lots of cases where it doesn't. It is better to go back to definitions.
In your above equation, $x$ might or might not be a function of $t$. For instance, it might be the starting location of a particle at time $t=0$, and $v(x,t)$ is the velocity of said particle at time $t$ given that starting position. Then what does $\frac{dt}{dx}$ even mean? They are both free variables, so $\frac{dt}{dx}=0$ and $\frac{dx}{dt}=0$.
A: you've already got good answers, so I just want to say some words about notation that can help you out.
This notation was introduced by Liebniz because he thought of the derivative as being the ratio of two infinitely small numbers called infinitesimals. This is the intuition behind the derivative, however these notions were reformulated by Cauchy and Weirstrass in terms of the formal definition of limit. You probably agree that if they were reformulated it's because something wrong was happening.
And indeed that's the case, there are cases in which interpreting the derivative as the ratio of infinitesimal $dy$ by infinitesimal $dx$ leads to the correct result, and some cases in which does not as pointed out by Thomas Andrews.
The main point is: if you want to be "protected" from being mislead, understand the notation of Liebniz for the derivative as just a notation and nothing more. Personaly I don't even use it, I prefer the prime notation to avoid confusion with the real formal $dx$ and $dy$ called differential forms.
Differential forms are on the other hand linear functionals, in other words, they are like "machines" that receives vectors as inputs and outputs real numbers. These things are dual to vectors, in the sense that in the view of linear algebra they are vectors too. And of course you don't divide vectors.
So, don't let the notation mislead you: Liebinz's notation is just a notation, stop thinking of it as infinitesimals that you can manipulate at will and you'll be fine to go.
As an example, look at these two versions of writing the chain rule: Let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R}\to\mathbb{R}$, then you have the composition $f \circ g: \mathbb{R} \to \mathbb{R}$. In Liebniz notation you first let $u = g(x)$ and then you write:
$$\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}$$
Well, isn't that confusing ? The left hand side contains the derivative of $f$ with respect to $x$, not the derivative of the composition. It's like, you know what function you're working by looking at the letter. In more complexes cases this can simply make you think you're doing some other thing and get totally confused. 
No look the modern notation:
$$(f\circ g)'(x)=f'(g(x))g'(x)$$
Isn't much clearer ? It's explicit that the lhs is the derivative of the composition and th rhs is the derivative of $f$ at $g(x)$ times the derivative of $g$ at $x$.
I hope this shows you a little of how can this notion of infinitesimals mislead you. I hope it makes those concepts clear to you. Good luck!
