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Thread: heres a derivative problem for ya!

  1. #1
    Junior Member
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    heres a derivative problem for ya!

    Suppose f is differentiable. Prove that
    $\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+g(h))-f(x)}{g(h)}$ exists, where g is a nonzero function of h such that $\displaystyle \lim_{h\to 0}g(h)=0$ and find the value of $\displaystyle f'(x)$.
    I think I need to change the terms of the limit from h to g(h) going to zero, not sure how to though.
    Any thoughts?
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  2. #2
    MHF Contributor chisigma's Avatar
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    May be that a little problem exist in this case. If $\displaystyle f(*)$ is differentiable in $\displaystyle x=x_{0}$ , then...

    $\displaystyle f^{'}(x) = \lim_{h \rightarrow 0+} \frac{f(x_{0}+h)-f(x_{0})}{h}= \lim_{h \rightarrow 0-} \frac{f(x_{0}+h)-f(x_{0})}{h}$ (1)

    ... i.e. both the derivatives 'from left' and 'from wright' exist and they are equal. In the proposed 'alternative' if , for instance, is $\displaystyle g(h)=h^{2}$ , then is...

    $\displaystyle \lim_{h \rightarrow 0} \frac{f(x_{0}+g(h))-f(x_{0})}{g(h)}= \lim_{h \rightarrow 0+} \frac{f(x_{0}+h)-f(x_{0})}{h}$ (2)

    ... so that we have no information about the derivative 'from left' of $\displaystyle f(*)$ in $\displaystyle x=x_{0}$ ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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