Posted by Taeyeon on .
Consider the curve defined by 2y^3+6X^2(y) 12x^2 +6y=1 .
a. Show that dy/dx= (4x2xy)/(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.

calculus 
Reiny,
If I read your equation as
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(2y)/(4y^2 + 3x^2 + 3)
Please check before I proceed 
calculus 
Taeyeon,
This is what I found in the paper though.

calculus 
Anonymous,
$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