1. ## Alternate Derivative Definition.

I think this is a nice problem. It is a known fact, just nobody tries to fully prove it.

2. Originally Posted by ThePerfectHacker
It is a known fact, just nobody tries to fully prove it.
What do you mean by that?
Surely there is nothing to prove there. It just two different ways of writing the same limit. Isn’t it?

3. Originally Posted by Plato
What do you mean by that?
Surely there is nothing to prove there. It just two different ways of writing the same limit. Isn’t it?
I mean to show that they are really the same thing. It takes a little work to show that.

I just realized people use that without really justifing it.

4. Are you asking us to justify the definition of the derivative?

If so:

Let f(x) be a function that is continuous on some interval I containing x0. If we wish to find the gradient (or slope) of the line connecting f(x0) to some other point f(x1) (which is the slope of the secant line between these points) we may use the formula:
M = [f(x0) - f(x1)]/[x0 - x1]

Let's redefine x1 as some point h units from x0. Therefore, we have x1 = x0 + h, and the slope of the of the secant line becomes:
M = [f(x0) - f(x0 + h)]/[x0 - (x0 + h)] = [f(x0) - f(x0 + h)]/h

Now, we may find the slope of a line tangent to f(x0) if we allow the distance between x0 and x0 + h to go to zero (in other words, we are letting h (the distance between x0 and x1) go to zero). Doing so we obtain a new limit:
L = lim{h->0} [f(x0) - f(x0 + h)]/h

We define this limit to be the slope of the tangent line to f(x) at x0, which we further define to be the derivative of f(x): f'(x) if we let x0 be any arbitrary value of x on the domain of f(x).
f'(x) = L = lim{h->0} (f(x) - f(x + h))/h

5. Originally Posted by ecMathGeek
Are you asking us to justify the definition of the derivative?
The point is that there are two representations of the definition of ‘derivative’ .
They are the two sides of the same limit coin.

6. Originally Posted by Plato
The point is that there are two representations of the definition of ‘derivative’ .
They are the two sides of the same limit coin.
Can you elaborate on this?

7. Originally Posted by Plato
The point is that there are two representations of the definition of ‘derivative’ .
They are the two sides of the same limit coin.
Can you elaborate on this?
The definition [f(x)-f(x_0)](x-x_0) and [f(x+h)-f(x)]/h
Are equivalent.

If the first one exists then the second one exists and furthermore the same value.

If the scond one exists then the first one exists and furthermore the same value.

8. Originally Posted by ThePerfectHacker
The definition [f(x)-f(x_0)](x-x_0) and [f(x+h)-f(x)]/h
Are equivalent.

If the first one exists then the second one exists and furthermore the same value.

If the scond one exists then the first one exists and furthermore the same value.
Why wouldn't they be equivalent? One is just a linear transformation of the x-axis to the other. Limits are independent of such transformations.

-Dan

9. Originally Posted by topsquark
Why wouldn't they be equivalent?
Because I like to do these things very very carefully. Some might say too carefully. (In fact, I think we need to know the functions are continous for you to make such a statement . Thus, in general what you said is probably not true . Yes, here it is).

Maybe I will post my solution.

10. This is what would do.