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Math Help - differentiation problem

  1. #1
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    differentiation problem

    Differentiate for x

    (3x^2-4x+3)sin(2x) / 9

    (6x-x^-1/2+3)2cos(2x)/9

    am i on the right track -
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  2. #2
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    Not yet, you need to employ the product rule here which is, if \displaystyle y= vu \implies y'=vu'+uv'

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  3. #3
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    Nope

    Barely touched calculus- derived instantaneous gradient and looked at x^4 becomes 4x^3, and table of derivatives. But i'm a real newbie.
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  4. #4
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    Quote Originally Posted by stophe73 View Post
    Differentiate for x

    (3x^2-4x+3)sin(2x) / 9

    This guy here is the product rule

    \displaystyle y= vu \implies y'=vu'+uv'

    used when you differentiate a product.

    In your case make \displaystyle v= \frac{(3x^2-4x+3)}{9}, u =\sin(2x) ifnd u' and v'

    so we get,

    <br />
\displaystyle \left(\frac{(3x^2-4x+3)\sin(2x)}{9}\right)' = \frac{(3x^2-4x+3)}{9}(\sin(2x))'}+\left(\frac{(3x^2-4x+3)}{9}\right)'\sin(2x)
    Last edited by pickslides; February 15th 2011 at 12:02 AM. Reason: Bad Latex
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  5. #5
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    Thats a big Help.....thanks.
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