Differentiate both sides with respect to x.
4 x^3 + 4y^3 * dy/dx = 0
y'(x) = dy/dx = -x^3/y^3
Now do it again.
y''(x) = d^2y/dx^2 = -3x^2/y^3
-x^3*(-3y^4)*dy/dx
= -3x^2/y^3 +3x^3*y^4*(-x^3/y^3)
= -3x^2/y^3 -3x^6*y
find y'' if x^4+y^4=16
1 answer