3x^4-4xy^2+7=2y
12x^3 - 4y^2 - 8xyy' = 2y'
(8xy+2)y' = 12x^3-4y^2
y' = (6x^3-3y^2)/(4xy+1)
y" =
(18x^2-6yy')(4xy+1) - (6x^3-3y^2)(4y+4xy')
-------------------------------------------------------
(4xy+1)
Now it's just algebra. Substitute the value of y' in the above expression and then massage it as you will.
find the first and second derivatives for 3x^4-4xy^2+7=2y
3 answers
oops. The denominator is (4xy+1)^2
See if you can get what wolframalpha.com gets:
http://www.wolframalpha.com/input/?i=%28%2818x^2-6y%28%286x^3-3y^2%29%2F%284xy%2B1%29%29%29%284xy%2B1%29+-+%286x^3-3y^2%29%284y%2B4x%28%286x^3-3y^2%29%2F%284xy%2B1%29%29%29+%29%2F%284xy%2B1%29+^2
http://www.wolframalpha.com/input/?i=%28%2818x^2-6y%28%286x^3-3y^2%29%2F%284xy%2B1%29%29%29%284xy%2B1%29+-+%286x^3-3y^2%29%284y%2B4x%28%286x^3-3y^2%29%2F%284xy%2B1%29%29%29+%29%2F%284xy%2B1%29+^2