# How do you implicitly differentiate #xy^2- yln(x-y)^2= 4-x#?

Remember to use the chain rule and product rule when differentiating the y variable ...

Now, differentiate ...

hope that helped

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To implicitly differentiate the given equation (xy^2 - y\ln(x-y)^2 = 4 - x), follow these steps:

- Differentiate both sides of the equation with respect to (x).
- Apply the product rule for the term (xy^2) and the chain rule for the term (y\ln(x-y)^2).
- Simplify the resulting expression.

The implicit differentiation of the equation (xy^2 - y\ln(x-y)^2 = 4 - x) with respect to (x) yields:

[y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - y\frac{2(x-y)}{x-y}\frac{dy}{dx} = -1]

Simplify the expression:

[y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - 2y\frac{x-y}{x-y}\frac{dy}{dx} = -1]

[y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - 2y\frac{dy}{dx} = -1]

[y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - 2y\frac{dy}{dx} = -1]

[y^2 + 2xy\frac{dy}{dx} - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} - \ln(x-y)^2 - 2y\frac{dy}{dx} = -1]

[y^2 - \ln(x-y)^2 = 4 - x]

[y^2 - \frac{2y(x-y)}{(x-y)^2}\frac{dy}{dx} = 4 - x]

[y^2 - \ln(x-y)^2 = 4 - x]

[y^2 - \ln(x-y)^2 = 4 - x]

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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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