Derivatives in the sense that you are using them are a much more profound concept than division by zero: they are a way to resolve an indeterminate expression usefully.
When we take dx/dt (velocity), it is true that we are putting the limit as change in time approaches zero in the denominator, but there is a crucial distinction: change in position, in the numerator, is also limited to zero.
Zero divided by zero is a class wholly outside the scope of any non-zero constant over zero; the latter is undefined (or, in some applications in physics, infinity), whereas the former is indeterminate. Getting an indeterminate solution is not the end of the line; it just means that the guy with the pencil and paper did something clumsy and removed all the usefulness from an equation. For example, take the following solvable linear equation:
3x + 5 = 20
Now, I sit down with my pencil and decide that what I really want to do is multiply both sides of the equation by zero and then solve for x. Don't ask me for my logic; I'm intentionally being clumsy.
0x = 0
x = 0/0
So what's the value of X? According to my solution, it is indeterminate. That doesn't mean that X doesn't have a value; it simply means we haven't found it, or in some cases, that we cannot find it.
On the other hand, if you misused calculus and ended up with the following:
n/dt
where n is any non-zero expression, the fraction would become a problem of division-by-zero and the mathematical system would explode like a wet turtle in a microwave oven.
Anyway, my point is that the fact that there is a zero (or a zero-limit) in the denominator alone does not mean that the concept of division-by-zero ought even enter our thoughts, or that the result is undefined.
And that's more than just clever semantics. A practical application can be found in physics: a photon has no mass, but it does have a measurable momentum. How is that possible?
Momentum is (elementarily) defined as mass times velocity:
P = m *
v
However, this is where special relativity dives in and plays a trick on us: when something reaches the speed of light, its mass is increased infinityfold. So the effective momentum of the photon is:
P = (0 *
infinity) *
v
Of course, in this case, infinity can be reduced to any constant over zero, which changes the expression to:
P = (0/0) *
v
Is that a problem? Not for the universe; photons have a definite and measurable (albeit slight) momentum.
So: indeterminate and undefined are completely different beasts.