# How do you implicitly differentiate # y^2+(y-x)^2-y/x^2-3y#?

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To implicitly differentiate ( y^2+(y-x)^2-\frac{y}{x^2}-3y ) with respect to ( x ), you would apply the chain rule and product rule as needed. The steps are as follows:

- Differentiate each term with respect to ( x ).
- Use the chain rule for terms involving ( y ) to differentiate the outer function and then multiply by the derivative of the inner function.
- Simplify the result.

The derivative of the given expression is:

[ \frac{d}{dx} \left( y^2 + (y-x)^2 - \frac{y}{x^2} - 3y \right) = 2y\frac{dy}{dx} + 2(y-x)(\frac{dy}{dx}-1) - \frac{x^2y' - 2xy}{x^4} - 3 ]

[ = 2y\frac{dy}{dx} + 2(y-x)\frac{dy}{dx} - 2(y-x) - \frac{x^2y' - 2xy}{x^4} - 3 ]

[ = 2y\frac{dy}{dx} + 2y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y + 2x - \frac{x^2y' - 2xy}{x^4} - 3 ]

[ = 4y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y - \frac{x^2y' - 2xy}{x^4} + 2x - 3 ]

[ = 4y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y - \frac{y'}{x^2} + \frac{2y}{x^3} + 2x - 3 ]

[ = (4y - 2x)\frac{dy}{dx} - 2y + \frac{2y}{x^3} + 2x - 3 - \frac{y'}{x^2} ]

So, the implicit derivative is ( (4y - 2x)\frac{dy}{dx} - 2y + \frac{2y}{x^3} + 2x - 3 - \frac{y'}{x^2} ).

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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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