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Math Help - Finding Second Derivatives

  1. #1
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    Finding Second Derivatives

    This is a continuation of a problem which I have the derivative of. It is -2x^3+6x^2-6x+2/(x^3-3x^2+2x)^2

    The question is to find the second derivative but I keep getting caught up in the simplifying areas around it after taking the derivative of it. Can anyone help me out by even posting a couple of main steps and the final answer part?

    I'm not sure if I'm doing it right either.

    Thanks again.
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  2. #2
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    Simplify first

    Hello Shadoowned
    Quote Originally Posted by Shadoowned View Post
    This is a continuation of a problem which I have the derivative of. It is -2x^3+6x^2-6x+2/(x^3-3x^2+2x)^2

    The question is to find the second derivative but I keep getting caught up in the simplifying areas around it after taking the derivative of it. Can anyone help me out by even posting a couple of main steps and the final answer part?

    I'm not sure if I'm doing it right either.

    Thanks again.
    Have you tried factorising the expression first?

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

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

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

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

    Have another go, starting from here.

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello ShadoownedHave you tried factorising the expression first?

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

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

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

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

    Have another go, starting from here.

    Grandad
    Alright so would I just multiply the -2 on the top through it and then take the derivative of this?
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  4. #4
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    I think I may be lost. I'm trying to get this done before class today which is in an hour too.
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