# Definition2 of derivative

1. Sep 11, 2006

### lokofer

"Definition2 of derivative...

We have the defintion (taking the limit) for hte derivative:

$$\frac{f(x)-f(a)}{x-a}$$ for an Euclidean Space...

But what keeps us from defining another metric (on an Euclidean or other abstract space) so the derivative takes de form:

$$\frac{(df(x),f(a))}{d(x,a}$$ so "d" is a distance in the form that you can define "infinitesimal2 elements on an space and that for x=a ,d=0

Then the "abstract" definition of integral is:

$$\sum_{i} f(X_i ) d(X_{i+1},X_{i})$$

of course in the limit that the distance $$|| X_{i+1}-X_í}||\rightarrow 0$$ and ¿what happens if we had an "Infinite" dimensional space...so

- it is "numerable" (ie: R^{n})
- it's not "numerable" (function space)

2. Sep 11, 2006

### Hurkyl

Staff Emeritus

I don't see any technical problem with defining

$$f'(a) := \lim_{d(x, a) \rightarrow 0} \frac{d(f(x), f(a))}{d(x, a)}$$

on an arbitrary metric space... but it won't really behave algebraically like a derivative. I don't know if it will exist very often for something that doesn't look like a linear space. It also doesn't agree with the ordinary derivative on Euclidean space: by your definition, the derivative of -x is 1. (and not -1) (because, for the real numbers, d(x, y) = |x - y|)

Last edited: Sep 11, 2006
3. Sep 11, 2006

### MeJennifer

But the metric space has to be smooth right?

4. Sep 11, 2006

### Hurkyl

Staff Emeritus
The definition makes literal sense for any metric space. (Though I suspect such limits might not exist very often)

I'm not sure what a "smooth metric space" is -- if you've heard that, then that's probably a good criterion that ensures lots of these "derivatives" will exist.

5. Sep 11, 2006

### HallsofIvy

What is (or is not) a "smooth" metric space?

A standard definition, given in most "Calculus III" courses, for the derivative is

A function, f:X-> Y, where X and Y are metric spaces, is said to be "differentiable at a" if and only if there exist a linear function, L:X->Y, and a function $\epsilon$ such that
$$f(x)= f(a)+ L(x-a)+ \epsilon(x-a)$$
and
$$\lim_{x\rightarrow a}\epsilon(x-a)||x-a||= 0$$

What more do you want?

6. Sep 11, 2006

### Hurkyl

Staff Emeritus
Hrm. I had assumed that he wanted to talk about derivatives for something that isn't a linear space, but now that I read again, that wasn't a good assumption!