# How do you differentiate #sqrt(e^(2x-2y))-xy^2=6#?

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To differentiate (\sqrt{e^{2x-2y}} - xy^2 = 6), differentiate both sides of the equation with respect to (x), treating (y) as a function of (x) using the chain rule and the product rule. The implicit derivative is given by:

[ \frac{d}{dx} \left(\sqrt{e^{2x-2y}} - xy^2\right) = \frac{d}{dx} (6) ]

[ \frac{1}{2\sqrt{e^{2x-2y}}} \frac{d}{dx} (e^{2x-2y}) - (x \frac{d}{dx}(y^2) + y^2) ]

Using the chain rule for the first term and the product rule for the second term:

[ \frac{1}{2\sqrt{e^{2x-2y}}} \cdot e^{2x-2y} (2 - 2y') - (y^2 + x \cdot 2y y') ]

Simplify and solve for (y'), the derivative of (y) with respect to (x):

[ \frac{e^{2x-2y}}{\sqrt{e^{2x-2y}}} - (2y - 2y') - (y^2 + 2xyy') ]

[ y' = \frac{e^{2x-2y}}{\sqrt{e^{2x-2y}}} - 2y + 2y' - y^2 - 2xyy' ]

[ y' - 2y' + 2xyy' = \frac{e^{2x-2y}}{\sqrt{e^{2x-2y}}} - 2y - y^2 ]

[ y'(1 - 2 + 2xy) = \frac{e^{2x-2y}}{\sqrt{e^{2x-2y}}} - 2y - y^2 ]

[ y' = \frac{\frac{e^{2x-2y}}{\sqrt{e^{2x-2y}}} - 2y - y^2}{1 - 2 + 2xy} ]

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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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