# What is the slope of the tangent line of # (xy-y^2)(1+x) =C #, where C is an arbitrary constant, at #(-3,1)#?

As# P(-3, 1) is on the graph,

C = ((-3)(1)-1^2)(1-3)=0#.

Differentiating,

I have used a parallel line, in proximity of the tangent, to keep off the

gap, at the point of contact. In this graphics method, P appears as a

gap, for the exact equation of the tangent. For the interested reader,

this graph is also included. In this graph, the pixels at P do not glow.

graph{((xy-y^2)(1+x)-8)(3x-5y+13.7)=0 [-9.79, 9.785, -4.895, 4.895]}

graph{((xy-y^2)(1+x)-8)(3x-5y+14)=0 [-9.79, 9.785, -4.895, 4.895]}

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To find the slope of the tangent line at the point (-3, 1) for the curve ( (xy - y^2)(1+x) = C ), we need to differentiate the equation implicitly with respect to x and then evaluate it at the given point (-3, 1).

Differentiating implicitly, we get:

[ (xy - y^2)(1+x) = C ]

[ (y + xy' - 2yy')(1 + x) + (xy - y^2)(1) = 0 ]

[ (y + xy' - 2yy' + xy - y^2) + x(y + xy' - 2yy') = 0 ]

[ xy' - 2yy' + xy - y^2 + xy + x^2y' - 2xyy' = 0 ]

[ y'(x + x^2 - 2y - 2xy) = y^2 - xy ]

[ y' = \frac{y^2 - xy}{x^2 - 2xy + x} ]

Now, substitute (-3, 1) into the expression for ( y' ) to find the slope of the tangent line at that point.

[ y' = \frac{(1)^2 - (-3)(1)}{(-3)^2 - 2(-3)(1) - 3} ]

[ y' = \frac{1 + 3}{9 + 6 - 3} ]

[ y' = \frac{4}{12} ]

[ y' = \frac{1}{3} ]

So, the slope of the tangent line at the point (-3, 1) is ( \frac{1}{3} ).

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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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