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Math Help - derivate from first principles.

  1. #1
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    derivate from first principles.

    Need help with derivate of this function.

    f(x) = 1/x, x can not be 0

    lim (f(x+h)-f(x)) / h =
    h->0

    lim ((1/(x-h)) - (1/x)) / h =
    h->0

    lim (x-(x+h)) / (x(x+h)h) =
    h->0

    lim (-h) / (hx^2 + xh^2) =
    h->0

    Need help with this step

    -1/x^2
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  2. #2
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    e^(i*pi)'s Avatar
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    Factor out h from the denominator and cancel (it only approaches 0 so it can cancel)

    \displaystyle \lim_{h \to 0} \left(\dfrac{-h}{xh(x+h)}\right) = \lim_{h \to 0} \left(-\dfrac{1}{x(x+h)}\right)


    Since we are taking the limit of h, not x it follows that \displaystyle \lim_{h \to 0} (x+h) = x. When this is put back into the original equation you get:

    -\dfrac{1}{x(x)} = -\dfrac{1}{x^2}

    Which is what we'd expect knowing the power rule of differentiation.
    Last edited by mr fantastic; March 3rd 2011 at 12:16 PM. Reason: Suggestion for correct subforum.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by paulaa View Post
    Need help with derivate of this function.

    f(x) = 1/x, x can not be 0

    lim (f(x+h)-f(x)) / h =
    h->0

    lim ((1/(x-h)) - (1/x)) / h =
    h->0

    lim (x-(x+h)) / (x(x+h)h) =
    h->0

    lim (-h) / (hx^2 + xh^2) =
    h->0

    Need help with this step

    -1/x^2
    When you are at this step you are almost done
    \displaystyle \lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h} =  \lim_{h \to 0} \frac{\frac{x-(x+h)}{(x)(x+h)}}{h} =  \lim_{h \to 0} \frac{-h}{x(x+h)h} =  \lim_{h \to 0}\frac{-1}{x(x+h)}=-\frac{1}{x^2}

    too slow
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