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Math Help - Derivative Problem

  1. #1
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    Derivative Problem

    Find the derivate of the following algebraic function:
    f(x) = (x^3 + 3x + 2) / (x^2 - 1)

    I think I am supposed to use the quotient rule, but I can't seem to get the algebra to work. My calculator says the derivative at, say, 1 is 300,000, and I get far less than that. I'm wondering if I am entering it incorrectly in my calculator or just not solving it right as it doesn't seem very simplified when I get through with it.
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  2. #2
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    Quote Originally Posted by seuzy13 View Post
    Find the derivate of the following algebraic function:
    f(x) = (x^3 + 3x + 2) / (x^2 - 1)

    I think I am supposed to use the quotient rule, but I can't seem to get the algebra to work. My calculator says the derivative at, say, 1 is 300,000, and I get far less than that. I'm wondering if I am entering it incorrectly in my calculator or just not solving it right as it doesn't seem very simplified when I get through with it.
    f(x)=\frac{x^3+3x+2}{x^2-1}

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

    You can't evaluate the derivative at 1 because the denominator is 0 when x=1
    Last edited by artvandalay11; September 23rd 2009 at 06:22 PM.
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  3. #3
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    You mean you can't simplify it any further than that? I'm not necessarily concerned with the derivative at 1, I was just checking it as an example to see if it I was working it out right. Appears I should have chosen a different number. Anyway, I'm trying to find the derivate with respect to x not the derivative at 1.
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  4. #4
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    Sure you can simplify more, I was just showing you what direct use of the quotient rule gives you

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

    =\frac{3x^4+3x^2-3x^2-3-(2x^4+6x^2+4x)}{(x^2-1)^2}

    =\frac{x^4-6x^2-4x-3}{(x^2-1)^2}
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  5. #5
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    I see now. I was making a simple multiplication mistake. It's funny how you can make the same silly mistake no matter how many times you try to do the problem. The answer still seems complicated, but I guess that's as far as it goes.
    Thank you!
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