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
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