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

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Answer 1 · 2017-02-17T15:09:06

$\frac{d y}{d x} = \frac{y^{3} + 2 y - 6 x y}{6 x^{2} - 3 x y^{2} - 2 x}$ $dy/dx = (y^3+2y-6xy)/(6x^2-3xy^2-2x)$

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

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

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

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

How do you implicitly differentiate -1=-y^3x-2xy+3x^2y −1=−y3x−2xy+3x2y-1=-y^3x-2xy+3x^2y ?