Question

Find
y''
by implicit differentiation.
3x3 + 4y3 = 1

Answer 1

3x^3 + 4y^3 = 1

9x^2 + 12y^2 y' = 0
y' = -3x^2/4y^2

now, using the quotient rule,

y'' = (-24xy^2 + 3x^2(8yy'))/ 16y^4
= (-3xy^2 + 3x^2yy')/2y^4
= -3x (y - xy')/2y^3

= -3x(y - x*(-3x^2/4y^2))/2y^3
= -3x(4y^3 + 3x^3)/8y^5

or, you can use implicit differentiation twice:

9x^2 + 12y^2 y' = 0
18x + 24y (y')^2 + 12y^2 y'' = 0
3x + 4y y'^2 + 2y^2 y'' = 0
2y^2 y'' = -(3x + 4y (y')^2)
y'' = -(3x + 4y (y')^2)/2y^2
= -(3x + 4y * 9x^4/16y^4)/2y^2
= -3x(4y^3 + 3x^3)/8y^5