# How do you implicitly differentiate # y^2-xy-6=0#?

You can do it like this:

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To implicitly differentiate the equation ( y^2 - xy - 6 = 0 ), follow these steps:

- Differentiate each term of the equation with respect to ( x ).
- Apply the chain rule whenever differentiating ( y ) terms.
- Solve for ( \frac{{dy}}{{dx}} ) to express it explicitly in terms of ( x ) and ( y ).

Differentiating each term:

[ \frac{{d}}{{dx}}(y^2) - \frac{{d}}{{dx}}(xy) - \frac{{d}}{{dx}}(6) = 0 ]

Applying the chain rule for ( y^2 ) and ( xy ):

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

Rearranging terms and solving for ( \frac{{dy}}{{dx}} ):

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

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To implicitly differentiate ( y^2 - xy - 6 = 0 ), follow these steps:

- Differentiate each term with respect to ( x ).
- Apply the chain rule whenever you differentiate ( y ) terms.
- Treat ( y ) as a function of ( x ) and apply the product rule when differentiating ( xy ).

Differentiating ( y^2 ): [ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} ]

Differentiating ( xy ): [ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y ]

The derivative of a constant term ((-6)) is zero.

So, the implicit differentiation of ( y^2 - xy - 6 = 0 ) gives:

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

You can then solve this equation for ( \frac{dy}{dx} ) if necessary.

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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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