Did I do this right?Calculus
posted by Iliana on .
y' and y'' for x^2 xy + y^2 = 3
y'= (y2x)/(2yx)
y'' = [(y'2)(2yx)]  [(2yy'  x)(y 2x)]/ ((2xy)^2)
final answer for second derivative is:
[2y^2 5xy + 2x^2  2y^3 + 8xy^2  8x^2y/ ((2xy)^3)]  [(4y + 2x + xy  2x^2)/((2yx)^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'= (y2x)/(2yx)
(2yx)y' = y2x
2yy' xy' = y2x
2y'^2 + 2yy"  y'  xy" = y'2
y"(2yx) = y'22y'^2+y'
y" = 2(y'^22y'+2)/(x2y)
= 2((2xy)^2  (2xy)(x2y) + 1)/(x2y)^3