# How do you find an equation for the tangent line to #x^4=y^2+x^2# at #(2, sqrt12)#?

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Differentiating,

The equation to the tangent at P is

y-sqrt13=7/sqrt3(x-2), giving

graph{(x^2-sqrt(x^2+y^2))(7x-sqrt3 y-8.2)=0 [-7, 7, -35, 35]}

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To find the equation of the tangent line to the curve (x^4 = y^2 + x^2) at the point ((2, \sqrt{12})), you can follow these steps:

- Differentiate both sides of the equation (x^4 = y^2 + x^2) implicitly with respect to (x) to find the derivative (\frac{dy}{dx}).
- Substitute the coordinates of the given point ((2, \sqrt{12})) into the derivative to find the slope of the tangent line.
- Use the point-slope form of the equation of a line, (y - y_1 = m(x - x_1)), where (m) is the slope and ((x_1, y_1)) is the given point, to find the equation of the tangent line.

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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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