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Math Help - Quotient rule

  1. #1
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    Quotient rule

    Hi all,

    I'm struggling to solve this equation using the quotient rule.

    y = 3cos2x
    2sin3x


    Any help appreciated
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by astrostar View Post
    Hi all,

    I'm struggling to solve this equation using the quotient rule.
    y = 3cos2x
    2sin3x


    Any help appreciated
    If you can't remember the quotient rule, you can use the product rule.

    \frac d{dx} \frac {3cos(2x)}{2sin(3x)} ~~~~=~~~~ \frac 32 \frac d{dx} \frac 1{sin(3x)} cos(2x) ~~~~=~~~~ \frac 32 \frac d{dx} [sin(3x)]^{-1}* cos(2x)

    The quotient rule is just a version of the product rule, if you use the product rule on the version of your problem that I showed you, you will end up getting your problem into the same format as the quotient rule after you simplify.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    If you mean

    you are trying to find f'(x) when f(x)=\frac{3cos(2x)}{2sin(3x)}...then here is what you do...its Low(D[hi])-hi[D(low)]/low˛...so f'(x)=\frac{2sin(3x)[3\cdot{-sin(2x)}\cdot{2}]-3cos(2x)[2\cdot{cos(3x)}\cdot{3}]}{4sin(3x)^2}...which simplifies to f'(x)=\frac{-3(3cos(2x)\cdot{cos(3x)}+2sin(2x)\cdot{sin(3x)})}{  2sin(3x)^2}
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  4. #4
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    Thanks for the replies,

    I have to find the answer using the quotient rule, my workng is,

    y = 3cos2x
    2sin3x


    let u = 3cos2x

    du = -6sinx
    dx


    let v = 2sin3x

    dv = 6cos3x
    dx

    this is where I get stuck
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Ok your doing it a weird way

    but if you want to do it that way youd say that \frac{D[\frac{u}{v}]}{dx}=\frac{v\cdot\frac{du}{dx}-u\cdot\frac{dv}{dx}}{v^2}
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