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Math Help - Quotient Rule Problem

  1. #1
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    Quotient Rule Problem

    y = \dfrac{x + 3}{x^{3} + x - 5}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][\dfrac{d}{dx}(x + 3)] - [x + 3][\dfrac{d}{dx}(x^{3} + x - 5)]}{(x^{3} + x - 5)^{2}}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x + 1]}{(x^{3} + x - 5)^{2}}
    Last edited by Jason76; October 3rd 2013 at 01:00 PM.
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  2. #2
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    Re: Quotient Rule Problem

    Quote Originally Posted by Jason76 View Post
    y = \dfrac{x + 3}{x^{3} + x - 5}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][\dfrac{d}{dx}(x + 3)] - [x + 3][\dfrac{d}{dx}(x^{3} + x - 5)]}{(x^{3} + x - 5)^{2}}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x + 1]}{(x^{3} + x - 5)^{2}}
    The derivative of x^3 is 3x^2, not 3x. (That may have been a typo.)
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  3. #3
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    Re: Quotient Rule Problem

    y = \dfrac{x + 3}{x^{3} + x - 5}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][\dfrac{d}{dx}(x + 3)] - [x + 3][\dfrac{d}{dx}(x^{3} + x - 5)]}{(x^{3} + x - 5)^{2}}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x^{2} + 1]}{(x^{3} + x - 5)^{2}}

    \dfrac{dy}{dx} = \dfrac{[x^{3} + x - 5][1] - [x + 3][3x^{2} + 1]}{(x^{3} + x - 5)^{2}}

    \dfrac{dy}{dx} = \dfrac{x^{3} +  x - 5}{(x^{3} + x - 5)^{2}} - \dfrac{3x^{3} + x  + 9x^{2} + 3}{(x^{3} + x - 5)^{2}}

    \dfrac{dy}{dx} = \dfrac{x^{3} +  x - 5}{(x^{3} + x - 5)(x^{3} + x - 5)} - \dfrac{3x^{3} + x  + 9x^{2} + 3}{(x^{3} + x - 5)(x^{3} + x - 5)}

    \dfrac{dy}{dx} = \dfrac{1}{(x^{3} + x - 5)} -  \dfrac{3x^{3} + x  + 9x^{2} + 3}{(x^{3} + x - 5)(x^{3} + x - 5)} Can this simplified further?
    Last edited by Jason76; October 3rd 2013 at 07:55 PM.
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  4. #4
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    Re: Quotient Rule Problem

     \text{From your last line, } \frac{dy}{dx}=\frac{x^3+x-5-(3x^3+x+9x^2+3)}{(x^3+x-5)^2}=\frac{-2x^3-9x^2-8}{(x^3+x-5)^2}

    I believe this is the simplified form, since the numerator has no rational root. If someone disagrees, please inform the poster and/ or me.
    Last edited by chen09; October 3rd 2013 at 09:15 PM.
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    Re: Quotient Rule Problem

    Quote Originally Posted by chen09 View Post
     \text{From your last line, } \frac{dy}{dx}=\frac{x^3+x-5-(3x^3+x+9x^2+3)}{(x^3+x-5)^2}=\frac{-2x^3-9x^2-8}{(x^3+x-5)^2}

    I believe this is the simplified form, since the numerator has no rational root. If someone disagrees, please inform the poster and/ or me.
    It can be factored out with (x + 4.6821). I did not do this using the radical method, I got it from the graph.
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  6. #6
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    Re: Quotient Rule Problem

    ok, looks good. The computer says it's right. Thanks.
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