How do you use implicit differentiation to find dy/dx given $x^3+y^3=6xy-1$ ?

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Answer 1 · 2016-11-04T20:27:50

$dy/dx = (2y-x^2)/(y^2-2x)$

$x^3+y^3=6xy-1$

$3x^2+3y^2 dy/dx=6y+6x dy/dx$

$3y^2 dy/dx-6x dy/dx=6y-3x^2$

$dy/dx (3y^2-6x)=6y-3x^2$

$dy/dx=(6y-3x^2)/(3y^2-6x)$

$dy/dx = (2y-x^2)/(y^2-2x)$

How do you use implicit differentiation to find dy/dx given x^3+y^3=6xy-1x^3+y^3=6xy-1?