If y is a differentiable function of x, then the slope of the curve of xy^2 - 2y + 4y^3 = 6 at the point where y=1 is

(show work! )

a -1/18
b -1/26
c 5/18
d -11/18
e 0

  a.  -1/18

Step-by-step explanation:

Differentiating implicitly, you have ...

  y^2 +2xyy' -2y' +12y^2y' = 0

Solving for y', we get ...

  y'(2xy -2 +12y^2) = -y^2

  y' = -y^2/(2xy -2 +12y^2)

To make use of this, we need to know the value of x at y=1. Filling in y=1 into the given equation, we have ...

  x -2 +4 = 6

  x = 4 . . . . . . . . subtract 2

So, at the point (x, y) = (4, 1), the slope is ...

  y' = -1/(8 -2 +12)

  y' = -1/18

_____

The attached graph shows that the line with slope -1/18 appears to be tangent to the curve at (4, 1).

0.02040816326