$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}$, for $x(t),y(t)$ I have a hopefully simple question. I am guessing that this has been asked before but I wasn't sure how to search for it.
Suppose we have two functions: $x(t)$, $y(t)$.
What is $\frac{dx}{dt}/\frac{dy}{dt}$? And why?
Note that this is not the setup for the standard chain rule (or at least its not obvious to me that it is). I talked to some friends and some of them say "just cancel out the $dt$s." Obviously, $\frac{dx}{dt}$, for example, is just the notation for a limit and not a ratio of numbers and so there is no reason as such that we should be able to "just cancel out."
Working with the limits directly, I can get up to:
$$ \lim_{\triangle t \to 0} \frac{x(t+ \triangle t)-x(t)}{y(t+\triangle t)-y(t)} $$
Thinking in terms of infinitesimals, this "feels" a lot like $dx/dy$ but I don't see any formal reason for why it is.
Thanks for your time.
 A: The ratio you post is a consequence of the following, where we have $f(g(t))$, with $x = f(t)$, $y= g(t)$. Then by the chain-rule,
$$\dfrac{dx}{dt} = \dfrac{dx}{dy}\cdot \dfrac{dy}{dt} \iff \dfrac{dx}{dy} = \dfrac{\frac{dx}{dt}}{\frac{dy}{dt}}$$
A: If you ask me what $$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}$$ is, my answer is simple: it is $\frac{x'(t)}{y'(t)}$. As your friends say, $\frac{d}{dt}$ is the operator that acts as the derivative (with respect to $t$). 
All other interpretations require additional assumptions: if you assume that $t$ can be expressed as a differentiable function of $x$, then
$$
\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx},
$$
and if $t=t(x)$ is invertible, then $\frac{dt}{dx} = \left(\frac{dx}{dt} \right)^{-1}$, and therefore
$$
\frac{dy}{dx} = \frac{dy}{dt} \left(\frac{dx}{dt} \right)^{-1}.
$$
But you need to know that you can invert a function, at least.
Your idea about taking a limit as $\Delta t \to 0$ is very dangerous, since $y(t+\Delta t)=y(t)$ also for $\Delta t \neq 0$, and you are not allowed to divide by zero. This is the popular but wrong proof of the chain rule.
A: The formal reason for why $$\lim_{\triangle t \to 0} \frac{x(t+ \triangle t)-x(t)}{y(t+\triangle t)-y(t)} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ is because the very definition of dy/dt is
$$\frac{dy}{dt} = \lim\limits_{\Delta t\rightarrow\infty} \frac{y(t+\Delta t) - y(t)}{\Delta t} $$.
Basically,
$$\lim_{\triangle t \to 0} \frac{x(t+ \triangle t)-x(t)}{y(t+\triangle t)-y(t)} =\lim_{\triangle t \to 0} \frac{\frac{x(t+ \triangle t)-x(t)}{\Delta t}}{\frac{y(t+\triangle t)-y(t)}{\Delta t}} =  \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx}$$
To clarify the final step,
If we we want to find a tangent line on to a curve where y is a differentiable function of x, then the Chain Rule gives $\frac{dy}{dt} = \frac{dy}{dt}\cdot\frac{dx}{dt}$, solving for $\frac{dy}{dx}$ gives us $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
If you want a better way to think about it rather than the $dt$'s cancelling out, think of it this way:
$$ 
\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{dy}{dx}\cdot\frac{dt}{dt} = \frac{dy}{dx}\cdot 1 = \frac{dy}{dx}
$$
As for $\frac{dt}{dt} = 1$, even though every one on this site is about to kill me for this, the above expression can be thought of as:
$$
\frac{dt}{dt} \rightarrow \frac{\frac{1}{\infty}}{\frac{1}{\infty}} = \frac{\infty}{\infty} = \frac{1}{1}
$$
Afterthought
The equation you are trying to prove assumes the relation that you are having fault with.
This is assumed because it is first assumed that both f and g are differentiable functions and that  y is differentiable with respect to x. It isn't an algebraic manipulation, but rather an equality made by the construction of the functions themselves.
Since formulation of $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ assumes that $ \frac{dy}{dx}$, the relation is true by construction.
