# How do you implicitly differentiate #11=-xy^2-(xy)/(2x-y)#?

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To implicitly differentiate the given equation, you need to differentiate each term with respect to ( x ) and apply the chain rule wherever necessary.

Given the equation: ( 11 = -xy^2 - \frac{xy}{2x - y} )

Differentiating both sides with respect to ( x ), you get:

( \frac{d}{dx}[11] = \frac{d}{dx}[-xy^2] - \frac{d}{dx}\left[\frac{xy}{2x - y}\right] )

Simplify each term:

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

Now, apply the chain rule and product rule where necessary:

( 0 = -y^2 - x(2y \frac{dy}{dx}) - \left[\frac{(2x - y)(y + x\frac{dy}{dx}) - xy(2 - \frac{dy}{dx})}{(2x - y)^2}\right] )

Simplify and solve for ( \frac{dy}{dx} ):

( 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(2xy - y^2 + xy + x^2\frac{dy}{dx}) - (2xy - xy^2)}{(2x - y)^2}\right] )

( 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(2xy - y^2 + xy + x^2\frac{dy}{dx}) - 2xy + xy^2}{(2x - y)^2}\right] )

( 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(2xy - y^2 + xy + x^2\frac{dy}{dx}) - 2xy + xy^2}{(2x - y)^2}\right] )

( 0 = -y^2 - 2xy\frac{dy}{dx} - \left[\frac{(x^2\frac{dy}{dx} - y^2)}{(2x - y)^2}\right] )

( 0 = -y^2 - 2xy\frac{dy}{dx} - \frac{x^2\frac{dy}{dx} - y^2}{(2x - y)^2} )

Now, solve for ( \frac{dy}{dx} ).

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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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