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Posted by on Saturday, March 30, 2013 at 12:47pm.

y' and y'' for x^2- xy + y^2 = 3

y'= (y-2x)/(2y-x)

y'' = [(y'-2)(2y-x)] - [(2yy' - x)(y- 2x)]/ ((2x-y)^2)

final answer for second derivative is:

[2y^2- 5xy + 2x^2 - 2y^3 + 8xy^2 - 8x^2y/ ((2x-y)^3)] - [(4y + 2x + xy - 2x^2)/((2y-x)^2)]

is this right. thank you

  • Did I do this right?-Calculus - , Saturday, March 30, 2013 at 2:14pm

    I'll let you verify whether our answers agree, but I think you may find it easier to do it like this

    y'= (y-2x)/(2y-x)
    (2y-x)y' = y-2x
    2yy' -xy' = y-2x
    2y'^2 + 2yy" - y' - xy" = y'-2
    y"(2y-x) = y'-2-2y'^2+y'

    y" = 2(y'^2-2y'+2)/(x-2y)
    = 2((2x-y)^2 - (2x-y)(x-2y) + 1)/(x-2y)^3

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