Question

Assuming that the equation determines a differentialable function f such that y = f(x), find y'.

2x^3 + x^2y + y^3 = 1

Answer 1

Uhm, x^2y? That's got me stumped.

Answer 2

\[6x^2+2xy+x^2y'+3y^2y'=0\rightarrow y'=\ -\frac{ 6x^2+2xy }{ x^2+3y^2 }\]

Answer 3

I believe it's supposed to be done through implicit differentiation.