# How do you find the slope of the line tangent to #x^2-y^2=3# at (2,1), (2,-1), (sqrt3,0)?

Differentiate each term with respect to x.

Solve for

Evaluate #m = dy(x,y)/dx at the points.

Differentiate each term with respect to x:

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To find the slope of the line tangent to the curve x^2 - y^2 = 3 at a given point, we can use the derivative of the equation. Taking the derivative of x^2 - y^2 = 3 with respect to x, we get 2x - 2yy' = 0. Solving for y', we have y' = x/y.

At the point (2,1), the slope of the tangent line is 2/1 = 2. At the point (2,-1), the slope of the tangent line is 2/-1 = -2. At the point (√3,0), the slope of the tangent line is (√3)/0, which is undefined.

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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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