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Math Help - Second derivative of implicit differentiation

  1. #1
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    Second derivative of implicit differentiation

    If x^2 + x + y^3 + y =1 then find y". Your answer must not contain y'. Do not simplify at all.

    So I found the first derivative because I dont understand why the answer cannot contain y' if you have to find y" :S but i found the first derivative and got:

    2x + 1/ 3y^2 = dy/dt

    then my second deriv was y"= 6y(y')(2x-1) - (2)(3y^2)/ (3y^2)^2

    I got 1.5/3 and I dont understand why, any help would be greatly appreciated
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  2. #2
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    It "can't contain y' " because that was a condition set by the problem!

    x^2 + x + y^3 + y =1
    so 2x+ 1+ 3y^2y'+ y'= 0

    Your first derivative is wrong because you forgot the y' needed for the derivative of y^3 with respect to x.

    I have no idea in the world why you think the answer should be a number!

    From 2x+ 1+ (3y^2+ 1)y'= 0 I would differentiate with respect to x again to get
    2+ (6y)y'+ (3y^2+ 1)y''= 0 so that y''= -\frac{(2+ 6y)y'}{3y^2+ 1}

    Now, simply because of the requirement that y' not appear in that formula, go back to 2x+1+ (3y^2+ 1)y'= 0 to get y'= -\frac{2+ 1}{3y^2+ 1} and use that to replace y' in that formula.
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  3. #3
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    OHHH
    wow, how could I not see that?

    ookay thank you very much!
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