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

Answer 1

Isaiah Allen

#4/7#

#xy-y^2+x =C#

#(xy)' -(y^2)'+(x)' =0#

#(y + xy') -(2y y')+1 =0#

# y' (x -2y) =-(1 + y)#

# y' =(1 + y)/(2y-x)#

#y'_{(-1,3)} = (1 + 3)/(2(3) - (-1)) = 4/7#

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Answer 2

Cameron Alexander

To find the slope of the tangent line at a point (-3, 1) on the curve xy - y^2 + x = C, differentiate the equation implicitly with respect to x, then plug in the coordinates of the point and solve for the derivative. The slope of the tangent line is the value of this derivative at that point. Differentiating the equation implicitly with respect to x yields:

dy/dx = (2y - x) / (2x - y)

Substitute the coordinates (-3, 1) into the derivative:

dy/dx = (21 - (-3)) / (2(-3) - 1) = (2 + 3) / (-6 - 1) = 5 / -7

So, the slope of the tangent line at (-3, 1) is -5/7.

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Answer from HIX Tutor

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.