When is $\frac{dx}{dt}=\frac{\Delta x}{\Delta t}$ a valid approximation? It is often said that when the change in e.g. $\Delta x$ is small than we can make the approximation:
$$\frac{dx}{dt}=\frac{\Delta x}{\Delta t}$$
But it is not enough to say $\Delta x$ is small because we can define units to make it large. So we need $\Delta x$ to be small compared to something else. What is this something else?
 A: I guess the OP was looking to some more easy answer. I will try to give one. You may be aware of the geometrical meaning of the derivative.
Look at this picture (courtesy of http://www.sagemath.org/calctut/tangent.html)

The idea, as you may know, is to push the point on the right toward the origin until the point are close enough so that you may ignore the curvature of the blue graph (this is a very inexact but I think rather intuitive way of seeing it) between the two points. The point on the right would be the point at $t+\Delta t$ and the point on the left would be the point at $t$. Now that was the only graph I was able to quickly find, and is a bit unfortunate that the left point is the origin. But I hope you get the idea.
Remember that the derivative, geometrically, is the value of the slope of the tangent to your function in a certain given point. In addition you are probably talking about $x$ as a function of $t$ so that $x=x(t)$. Otherwise the derivative does not make much sense. So the question to ask is not how "small" must $\Delta x$ be, but how small $\Delta t$ must be. $\Delta x(t)=x(t+\Delta t)-x(t)$ can be a lot bigger than $\Delta t$.
So to summarize, intuitively $\Delta t$ must be so small that you may neglect the curvature of your function between $t$ and $t+\Delta t$ (you may recognise here a reference to the second derivative) that is to say you must choose a $\Delta t$ so small that your function is well approximated by only the first derivative.
I hope this may help you in getting an intuitive idea of the derivative.
A: Small is an ambiguous term.
We can use $\frac{dx}{dt}$ to approximate $f(t).$
That is,
$f(t+\delta) = f(t) + \frac {dx}{dt} \delta + \epsilon.$
Where $f(t) + \frac {dx}{dt}\delta$ is our linear approximation in a neighborhood of $t$ and $\epsilon$ is the error in the approximation.
Can we bound $\epsilon$?
Turns out we can.
$\epsilon |\le \frac 12 \max(|\frac{d^2f}{dt^2}|)\delta^2|$
Where $\max(|\frac{d^2f}{dt^2}|)$ is the maximal value of the absolute value of the $2^{nd}$ derivative of $f(t)$ in the interval $(t,t+\delta)$
