calculus
posted by Tom .
considering that x is the independent variable in this equation:
y + y^3 + 3 = e^y^2 + 3^x * cos(3y)  x
Evaluate dy/dx
I get to:
dy/dx (1+3y) = e^y^2 * 3y^2 * dy/dx + 3^x * ln3 * (sin(3y)) * 3 *dy/dx 1
Will it give me?
dy/dx(1+3y) =
dy/dx((e^y^2 * 3y^2) +
(3^x * ln3 * (sin(3y)) * 3 ) 1
dy/dx(1+3y) =
dy/dx((e^y^2 * 3y^2)  (9^xsin(3y)ln3))
1
1 / ((e^y^2 * 3y^2)  (9^xsin(3y)ln3)  (1+3y))

last one is supposed to be
dy/dx = 1 / ((e^y^2 * 3y^2)  (9^xsin(3y)ln3)  (1+3y))