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

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

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Answer 1 · 2016-11-13T02:47:22

$\frac{d y}{d x} = \frac{- 2}{2 y - 1}$ $dy/dx=(-2)/(2y-1)$

Explanation:

Given-

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

$2 - \frac{d y}{d x} + 2 y \cdot \frac{d y}{d x} = 0$ $2-dy/dx+2y*dy/dx=0$

$- \frac{d y}{d x} + 2 y \cdot \frac{d y}{d x} = - 2$ $-dy/dx+2y*dy/dx=-2$

$\frac{d y}{d x} (- 1 + 2 y) = - 2$ $dy/dx(-1+2y)=-2$

$\frac{d y}{d x} (2 y - 1) = - 2$ $dy/dx(2y-1)=-2$

$\frac{d y}{d x} = \frac{- 2}{2 y - 1}$ $dy/dx=(-2)/(2y-1)$

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

1 Answer

Explanation:

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How do you use implicit differentiation to find dy/dx given 2x-y+y^2=02x−y+y2=02x-y+y^2=0?

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