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Math Help - differantiation proof question..

  1. #1
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    differantiation proof question..

    if f(x) is differentiable on x_0
    prove that


    ??

    using the given i could
    say that lim [f(x+x_0) - f(x0)]/[x-x_0] exist
    Last edited by transgalactic; January 21st 2009 at 11:26 AM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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     f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

    now let  u = x+h \Rightarrow x = u-h

    So  f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h}

    Since  h \to 0, f'(u-h) \to f'(u)

    Therefore  f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}

     f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)

     2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h}

    Hence,  f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}
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  3. #3
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    how did you know to pick those variables??
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  4. #4
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    and why you didnt use the point x_0
    that they presented
    there is another definition to limit
    <br />
f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}<br />

    is there a way to use this definition like you did before
    what variables to pick??
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by transgalactic View Post
    and why you didnt use the point x_0
    that they presented
    there is another definition to limit
    <br />
f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}<br />

    is there a way to use this definition like you did before
    what variables to pick??
    Does your problem require you to use that definition of  f'(x) ?

    Also what variables were you talking about in your other post?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    and why you didnt use the point x_0
    that they presented
    there is another definition to limit
    <br />
f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}<br />

    is there a way to use this definition like you did before
    what variables to pick??
    A simple substitution will change your defition to Chiph's. Now what substitution is that?
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  7. #7
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    if ill take x-x_0 =h
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    correct
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