What is the slope of the tangent line of #(1-x)(3-4y^2)-lny = C #, where C is an arbitrary constant, at #(1,1)#?

Answer 1

#m=1#

#f(x,y)=(1-x)(3-4y^2)-lny-C#
#dy/dx = - f_x/f_y#
# = -((-1)(3-4y^2))/((1-x)(-8y)-1/y)#
At #(1,1)#, we get
#dy/dx = -((-1)(-1))/((0)(-8)-1) = 1#
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Answer 2

To find the slope of the tangent line at the point (1,1) for the curve (1-x)(3-4y^2) - ln(y) = C:

  1. Differentiate implicitly with respect to x.
  2. Find the derivative of each term using the product rule and the chain rule.
  3. Plug in the coordinates (1,1) to find the slope of the tangent line.
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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.

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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