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Math Help - find the derivative

  1. #1
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    find the derivative

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

    I used quotient rule and chain rule.
    The solution keep getting is wrong. I think my algebra is off.: y'= \frac{12x(x^2+1)^2}{(x^2-1)^4}
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  2. #2
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    y'=3{(x^2+1)/(x^2-1)}^2 * {2x(x^2-1)-2x(x^2+1)}/{x^2-1}^2=...
    }
    =3{(x^2+1)/(x^2-1)}^2 * {2x{x^2-1-x^2-1}/{(x^2-1)^2}=

    =3{(x^2+1)/(x^2-1)}^2 * {2x{-2}/{(x^2-1)^2}=

    =3{(x^2+1)/(x^2-1)}^2 * {-4x/{(x^2-1)^2}=

    =-12x/(x^2-1)}^4
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  3. #3
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    Quote Originally Posted by superduper1 View Post
    y=(\frac{x^2+1}{x^2-1})^3

    I used quotient rule and chain rule.
    The solution keep getting is wrong. I think my algebra is off.: y'= \frac{12x(x^2+1)^2}{(x^2-1)^4}
    show your work so that someone may see your mistake.
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  4. #4
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    I get y = \frac{6x(x^2-1)^3(x^2+1)^2-6x(x^2-1)^2(x^2-1)^3}{(x^2-1)^6}

    take out common factors to do some additional cancellations.
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  5. #5
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    I think the mistake is somewhere after here:

    y'=3(\frac{x^2+1}{x^2-1})^2 * \frac{2x[x^2+1-x^2+1]}{(x^2-1)^2}
    y'=3(\frac{x^2+1}{x^2-1})^2 * \frac{2x(2)}{(x^2+1)^2}

    What is this when simplified? I think this is throwing me off. The 3 in front and the ^2 in the back
    y'=3(\frac{x^2+1}{x^2-1})^2
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  6. #6
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    \displaystyle y=\left(\frac{x^2+1}{x^2-1}\right)^3

    \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2}

    \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{2x[x^2-1 - (x^2+1)]}{(x^2-1)^2}

    \displaystyle \frac{dy}{dx} = 3\left(\frac{x^2+1}{x^2-1}\right)^2 \cdot \frac{-4x}{(x^2-1)^2}
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