# What is the slope of the tangent line of # (x^2-y^2)/(sqrt(4y-x)+xy) =C #, where C is an arbitrary constant, at #(1,1)#?

Please see below.

Differentiating implicitly, we have:

graph{(x^2-y^2)/(sqrt(4y-x)+xy) =0 [-6.496, 7.55, -1.824, 5.2]}

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To find the slope of the tangent line at the point (1, 1) for the equation

[ \frac{x^2 - y^2}{\sqrt{4y - x} + xy} = C ]

we first differentiate implicitly with respect to ( x ), then substitute the point (1, 1) into the derivative to find the slope.

Differentiating both sides of the equation implicitly gives us:

[ \frac{d}{dx} \left( \frac{x^2 - y^2}{\sqrt{4y - x} + xy} \right) = \frac{d}{dx} C ]

Applying the quotient rule and chain rule on the left side:

[ \frac{(2x - 2yy')(\sqrt{4y - x} + xy) - (x^2 - y^2) \left( \frac{1}{2\sqrt{4y - x}} - y - x \right)}{(\sqrt{4y - x} + xy)^2} = 0 ]

Now, substitute ( x = 1 ) and ( y = 1 ) into the equation above to find ( y' ), the derivative of ( y ) with respect to ( x ) at the point (1, 1). After simplifying, we can find the slope of the tangent line at that point.

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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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