Question

How do you implicitly differentiate `-1=-y^3x-2xy+3x^2y`?

Answer 1

`dy/dx = (y^3+2y-6xy)/(6x^2-3xy^2-2x)`

Explanation:

Differentiate both sides of the equation with respect to `x`:

`0 = -3xy^2y'-y^3 -2y -2xy'+6xy+6x^2y'`

`0 = y'(-3xy^2-2x +6x^2) -y^3 -2y +6xy`

`y' = (y^3+2y-6xy)/(6x^2-3xy^2-2x)`