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Math Help - proof of derivation product rule

  1. #1
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    proof of derivation product rule

    I have question regarding to product rule.

    We know that derivative of y with respect to x is:
     <br />
\frac{dy}{dx} = f'(x) = \lim_{h\rightarrow 0} \: \frac{f(x+h) - f(x)}{x+h-x} = \lim_{h\rightarrow 0} \: \frac{f(x+h) - f(x)}{h} = \frac{\Delta y}{\Delta x}

    Now I'm trying to understand the proof of product rule:
    f'(x)\cdot g'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)<br />

    to simplify notation:
    let\;u=f(x)\;and\;v=g(x).\;\;Then \;D(u\cdot v) = u' \cdot v + u \cdot v'

    Ok, now to proof:
    (u \cdot v)' = \lim_{h\rightarrow 0} \frac{\left[u(x+h) \cdot v(x+h) \right] - \left[u(x) \cdot v(x) \right]}{h}<br />

    we know that \Delta u = u(x+h) - u(x) \;,\; thus\; u(x+h) = \Delta u + u(x) so we can substitute this with respected to u and v in equation and get:
    (u \cdot v)' = \lim_{h\rightarrow 0} \frac{\left[(\Delta u+u(x)) \cdot (\Delta v + v(x)) \right] - \left[u(x) \cdot v(x) \right]}{h}

    Now to do some algebra we get:
    = \lim_{h\rightarrow 0} \left\{ \frac{\Delta u \cdot \Delta v}{h} + \frac{\Delta u}{h} \cdot v(x) + \frac{\Delta v}{h} \cdot u(x) + \frac{u(x) \cdot v(x)}{h} - \frac{u(x) \cdot v(x)}{h} \right\}

    = \lim_{h\rightarrow 0} \left\{ \frac{\Delta u \cdot \Delta v}{h} + \frac{\Delta u}{h} \cdot v(x) + \frac{\Delta v}{h} \cdot u(x) \right\}

    And we know that u' = \lim_{h\rightarrow 0} \frac{\Delta u}{h} \; , \; v' = \lim_{h\rightarrow 0} \frac{\Delta v}{h}

    so finally we substitute u' and v' to the equation we get:
    = \lim_{h\rightarrow 0}  \frac{\Delta u \cdot \Delta v}{h} + u' \cdot v + v' \cdot u

    How i can get rid of \lim_{h\rightarrow 0}  \frac{\Delta u \cdot \Delta v}{h} since this is some error in my proof.
    Last edited by tabularasa; November 17th 2008 at 04:30 AM.
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  2. #2
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    Quote Originally Posted by tabularasa View Post
    How i can get rid of \lim_{h\rightarrow 0}  \frac{\Delta u \cdot \Delta v}{h} since this is some error in my proof.
    It is not an error. Write it as \lim_{h\rightarrow 0}  h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h}. Then \lim_{h\to0}\frac{\Delta u}h = u'(x), \lim_{h\to0}\frac{\Delta v}h = v'(x), and of course \lim_{h\to0}h = 0. Therefore, by the product rule for limits, \lim_{h\rightarrow 0}  h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h} = 0.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    It is not an error. Write it as \lim_{h\rightarrow 0}  h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h}. Then \lim_{h\to0}\frac{\Delta u}h = u'(x), \lim_{h\to0}\frac{\Delta v}h = v'(x), and of course \lim_{h\to0}h = 0. Therefore, by the product rule for limits, \lim_{h\rightarrow 0}  h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h} = 0.
    Where's that h \lim_{h\to0} h\cdot\ coming from?
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  4. #4
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    Quote Originally Posted by tabularasa View Post
    Where's that h \lim_{h\to0} h\cdot\ coming from?
    Notice that he split the denominator into h^2, so there has to be another h in the numerator to counter that.
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  5. #5
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    Quote Originally Posted by FusionHK View Post
    Notice that he split the denominator into h^2, so there has to be another h in the numerator to counter that.
    Aah, too clever. i just didn't saw that one coming. Thank you!
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