calculus
posted by Sasha .
what is the equation of the tangent line to the curve x^3 + 2y^2 + 3xy = 6 at the point (1,1)?

3x^2 + 4y dy/dx + 3xdy/dx + 3y = 0
dy/dx(4y+3x) = 3x^2  3y
dy/dx = (3x^2  3y)/(4y+3x)
at (1,1) dy/dx = (33)/(4+3) = 6/7
y1 = (6/7)(x1
times 7
7y  7 = 6(x1)
7y  7 = 6x + 6
6x + 7y = 13