How do you use implicit differentiation to find dy/dx given $2x-y+y^2=0$ ?

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Answer 1 · 2016-11-13T02:47:22

$dy/dx=(-2)/(2y-1)$

Given-

$2x-y+y^2=0$

$2-dy/dx+2y*dy/dx=0$

$-dy/dx+2y*dy/dx=-2$

$dy/dx(-1+2y)=-2$

$dy/dx(2y-1)=-2$

$dy/dx=(-2)/(2y-1)$

How do you use implicit differentiation to find dy/dx given 2x-y+y^2=02x-y+y^2=0?