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Math Help - Proof of the Product Rule

  1. #1
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    Proof of the Product Rule

    I understand the product rule and how to use it, but I've been told to prove it, and I don't really know how. Could I be helped with this?
    Thank you
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  2. #2
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    Shouldn't be too hard

    Here's the start

    f(x) = v(x)\times u(x)

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

    we have

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

    can you take it from here?
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  3. #3
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    I have understood what you have done using the calculus theorem, however I don't know what to do from were you have left it, usually the ' h's would cancel out,
    Sorry, could you give more clues
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  4. #4
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    There's only a couple of steps left to the proof, I don't want to spoil it completely!

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

    here's a little bit more

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

    Now factor out v(x+h) and u(x) in groups.
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  5. #5
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    I may have got it I think,

    \frac{v(x+h)u(x+h)-v(x+h)u(x)+v(x+h)u(x)-v(x)u(x)}{h}

    = v(x+h)\frac{u(x+h)-u(x)}{h}+u(x)\frac{v(x+h)-v(x)}{h}

    but \frac{u(x+h)-u(x)}{h} can be written as u'(x) as h tends to zero

    thus f'(x)=v(x)u'(x)+u(x)v'(x) as h tends to zero

    am I correct?
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  6. #6
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    Nice work!
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  7. #7
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    wahay, kinda guessed that,
    thank you very much
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