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Differentiation: Magic?

  1. Jun 29, 2005 #1
    How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of [itex]\frac{0}{0}[/itex]?
     
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  3. Jun 29, 2005 #2
    [tex]\lim_{\Delta x \rightarrow 0} \Delta x = 0 \neq dx[/tex]

    we are considering the function [tex]\frac{dy}{dx}[/tex]

    and the limit of the whole thing, [tex]\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}[/tex]

    is different than the limit of it's parts.
     
  4. Jun 29, 2005 #3

    robphy

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    Try f(x)=x^2 in quetzalcoatl9's formula (the definition of the derivative).
    Expand out the numerator. Do some algebra.
    Lastly, take the limit.
    See what you get.

    Try x^3, etc... sin(x),exp(x), etc...

    In some sense what you're doing is measuring the relative rate between [tex] \Delta f[/tex] (which depends on [tex]\Delta x[/tex]) and [tex]\Delta x[/tex] itself as [tex]\Delta x[/tex] gets small.
     
  5. Jun 29, 2005 #4
    They aren't the exact same coordinate. Hence the [itex]x+\Delta{x}[/tex]

    I'm kind of confused on your question. If you could explain your thoughts more, perhaps we could help you more.
     
  6. Jun 29, 2005 #5

    Hurkyl

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    The question you need to be asking yourself is "what is a limit?" Once you have that down, it will be (more) clear how the definition of a derivative in terms of a limit works.
     
  7. Jun 29, 2005 #6

    Tom Mattson

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    To expand on Jameson's comment....

    cscott, go back to the definition of a limit, which says:

    Note that the second inequality really implies that [itex]0 \leq |f(x)-L| < \epsilon[/itex], which means that [itex]f(x)-L[/itex] can indeed equal zero. But in the first inequality, [itex]|x-c|[/itex] is forever restricted to be above zero.

    So in the definition of the derivative:

    [tex]f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex],

    we can see that it is [itex]|\Delta x - 0|[/itex] that is forever restricted to be above zero.

    edit:

    Hmmm....Why didn't LaTeX format this right?

    [itex]\lim_{x \rightarrow c}f(x)=L[/itex]
     
    Last edited: Jun 29, 2005
  8. Jun 29, 2005 #7
    I think "itex" is limited to one line... so the the x ->c can't fit under.
     
  9. Jun 30, 2005 #8
    Thanks for the new insight. I'm going to read more on limits and such.
     
  10. Jul 1, 2005 #9

    mathwonk

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    or you could take the classical way out, of descartes etc.

    i.e. when f is a polynomial, then f(x)-f(a) = [x-a][g(x)] for some factor g.

    then you are asking essentially for the value of the quotient [f(x)-f(a)]/(x-a)

    = [(x-a)g(x)]/(x-a) at x=a.

    obviously you should factor out the x-a first, then the answeer is g(a), which has nothing to do with the fact that you get 0/0 before you factor out.


    now in cases where you do not know how to factor, you plug in a sequence of numbers for x that are merely close to a, i.e. you let x = a+e for small e.

    then you get the sequence of approximations
    [(e) g(a + e)]/e = g(a+e). as e gets smaller and smaller, hopefully you are eventually able to guiess what g(a) should be from all these approximations of form g(a+e).

    if so you have "taken the limit" as x goes to a.
     
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