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

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

Given-

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To find ( \frac{dy}{dx} ) using implicit differentiation for the equation ( 2x - y + y^2 = 0 ):

- Differentiate both sides of the equation with respect to ( x ).
- For terms involving ( y ), apply the chain rule.
- Solve the resulting equation for ( \frac{dy}{dx} ).

Differentiating both sides of the equation with respect to ( x ):

[ \frac{d}{dx}(2x - y + y^2) = \frac{d}{dx}(0) ]

Apply the chain rule for terms involving ( y ):

[ \frac{d}{dx}(2x) - \frac{d}{dx}(y) + \frac{d}{dx}(y^2) = 0 ]

[ 2 - \frac{dy}{dx} + 2y\frac{dy}{dx} = 0 ]

Solve for ( \frac{dy}{dx} ):

[ 2 - \frac{dy}{dx} + 2y\frac{dy}{dx} = 0 ] [ 2 + 2y\frac{dy}{dx} = \frac{dy}{dx} ] [ 2 = \frac{dy}{dx}(1 - 2y) ] [ \frac{2}{1 - 2y} = \frac{dy}{dx} ]

So, ( \frac{dy}{dx} = \frac{2}{1 - 2y} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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