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Math Help - Clarification Needed: Proof of the Product Rule

  1. #1
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    Clarification Needed: Proof of the Product Rule

    Here's the proof, as provided by text:

    \frac{d}{dx}(uv)=\lim_{h\to\0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}
    =\lim_{h\to\0}[u(x+h)\frac{v(x+h)-v(x)}{h}+v(x)\frac{u(x+h)-u(x)}{h}]
    =\lim_{h\to\0}u(x+h)*\lim_{h\to\0}\frac{v(x+h)-v(x)}{h}+v(x)*\lim_{h\to\0}\frac{u(x+h)-u(x)}{h}.

    I understand that u(v+h)v(x)-u(v+h)v(x) is introduced into the expression so that the equation can be expressed by definition as \frac{dv}{dx} and \frac{du}{dx}. Although in step 2, wouldn't the denominator, h, retain its form into step 3 as:

    \lim_{h\to\0}\frac{u(x+h)}{h}*\lim_{h\to\0}\frac{v  (x+h)-v(x)}{h}+\lim_{h\to\0}\frac{v(x)}{h}*\lim_{h\to\0}  \frac{u(x+h)-u(x)}{h} ?

    It seems to me the h was omitted from a factor for each term.
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  2. #2
    Newbie LordoftheFlies's Avatar
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    Re: Clarification Needed: Proof of the Product Rule

    Think about it: that is equivalent to saying:

    \dfrac{xy}{h}=\dfrac{x}{h}\cdot\dfrac{y}{h}=\dfrac  {xy}{h^2}

    ... which of course, is false!
    Last edited by LordoftheFlies; January 1st 2013 at 02:45 PM.
    Thanks from Lambin and Paze
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