Calculus
posted by Anonymous on .
Find the equation of the tangent line to the curve x^(3)+3xy+y^(3)=1 at the point (1,0).

Implicit derivative:
3x^2 + 3x dy/dx + 3y + 3y^2 dy/dx = 0
dy/dx(3x + 3y^2) = 3x^2  3y
dy/dx = (x^2  y)/(x + y^2)
at (1,0)
dy/dx = 1  0)/(1 + 0) = 1
let y = x + b
at (1,0)
0 = 1 + b
b = 1
tangent equation at (1,0) is
y = x + 1