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Math Help - Help with using the quotient rule on a derivative?

  1. #1
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    Help with using the quotient rule on a derivative?

    I don't know why, but for some reason I absolutely cannot get the answer to this derivative:

    h(x) = sqrt(x)/x^3+1

    I got as far as writing everything out and trying to simplify it:
    (1/2x^-1/2)(x^3+1)-(x^1/2)(3x^2)/(x^3+1)^2
    Simplifying:
    (1/2)x^-3/2+(1/2)x^-(1/2)-3x/x^3+1

    Somehow, I'm supposed to make this equal
    1-5x^3/2sqrt(x)(x^3+1)^2

    I think I might be making a simple algebra mistake, but I can't find it, so if someone could show me the process to get to the answer, that would help immensely
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  2. #2
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    Quote Originally Posted by Ohoneo View Post
    I don't know why, but for some reason I absolutely cannot get the answer to this derivative:

    h(x) = sqrt(x)/x^3+1

    I got as far as writing everything out and trying to simplify it:
    (1/2x^-1/2)(x^3+1)-(x^1/2)(3x^2)/(x^3+1)^2
    Simplifying:
    (1/2)x^-3/2+(1/2)x^-(1/2)-3x/x^3+1

    Somehow, I'm supposed to make this equal
    1-5x^3/2sqrt(x)(x^3+1)^2

    I think I might be making a simple algebra mistake, but I can't find it, so if someone could show me the process to get to the answer, that would help immensely

    h'(x) = \frac{(x^3+1)\frac{1}{2\sqrt{x}} -  \sqrt{x}(3x^2)}{(x^3+1)^2}

    multiply by \frac{2\sqrt{x}}{2\sqrt{x}} ...

    h'(x) = \frac{(x^3+1) -   2x(3x^2)}{2\sqrt{x}(x^3+1)^2}

    h'(x) = \frac{1 - 5x^3}{2\sqrt{x}(x^3+1)^2}
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  3. #3
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    Ah, thank you. I was getting too ahead of myself and forgetting to rationalize. Thanks
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