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Math Help - Find the Derivative

  1. #1
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    Find the Derivative

    y=(9+4x-8 sqrt x)/x

    Using the quotient rule I get

    [x(4-4x^-1/2)-(9+4x-8x^1/2)]x^2

    [4x-(4x^1/2)-9-4x+(8x^1/2)]x^2

    [(8x^1/2)-9]x^2

    My final answer is (4sqrtx-9)/x^2

    However the boosk final answer is

    -9/x^2 + 4/xsqrtx

    It has been a year since I took trig and I know my algebra is horrible but how did they get 4/xsqrtx from 4sqrtx/x^2

    Thanks for the help.
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  2. #2
    Junior Member beebe's Avatar
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    Re: Find the Derivative

    Quote Originally Posted by oriont View Post
    y=(9+4x-8 sqrt x)/x

    Using the quotient rule I get

    [x(4-4x^-1/2)-(9+4x-8x^1/2)]x^2

    [4x-(4x^1/2)-9-4x+(8x^1/2)]x^2

    [(8x^1/2)-9]x^2

    My final answer is (4sqrtx-9)/x^2

    However the boosk final answer is

    -9/x^2 + 4/xsqrtx

    It has been a year since I took trig and I know my algebra is horrible but how did they get 4/xsqrtx from 4sqrtx/x^2

    Thanks for the help.
    \frac{4\sqrt{x}}{x^2}=(4x^{1/2})(x^{-2})=4x^{1/2-2}=4x^{-3/2}=\frac{4}{x\sqrt{x}}

    You have the same answer as the book.
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Find the Derivative

    As an alternative, you can use logarithmic differentiation.

    y=\frac{9+4x-8\sqrt{x}}{x}

     \frac{1}{y}\frac{dy}{dx}=\frac{4+\frac{8}{2\sqrt{x  }}}{xy}-\frac{1}{x}

    \therefore \frac{dy}{dx}=\cdots = \frac{4\sqrt{x}-9}{x^2}
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  4. #4
    MHF Contributor
    skeeter's Avatar
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    Re: Find the Derivative

    y = \frac{9+4x-8\sqrt{x}}{x} = \frac{9}{x} + \frac{4x}{x} - \frac{8\sqrt{x}}{x} = 9x^{-1} + 4 - 8x^{-1/2}

    y' =  -9x^{-2} + 4x^{-3/2} = -\frac{9}{x^2} + \frac{4}{x^{3/2}} = -\frac{9}{x^2} + \frac{4}{x\sqrt{x}}
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  5. #5
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    Re: Find the Derivative

    Thanks, appreciate the answer.
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