calculus

Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .
a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1)
b. Write an equation of each horizontal tangent line to the curve.
c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x – and y – coordinates of point P.

2y^3 + (6x^2)(y) - 12x^2 + 6y = 1
then your dy/dx is not correct

I get:
8y^2 dy/dx + 6x^2 dy/dx + 12xy - 24x + 6dy/dx = 0
dy/dx(8y^2 + 6x^2 + 6) = 24x - 12xy
dy/dx = 12x(2 - y)/(2(4y^2 + 3x^2 + 3) )
= 6x(2-y)/(4y^2 + 3x^2 + 3)

posted by Reiny
2. This is what I found in the paper though.

posted by Taeyeon
3. \$4000 is invested at 9% compounded quarterly. In how many years will the account have grown to \$14,500? Round your answer to the nearest tenth of a year

posted by Anonymous

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