Find an equation for the tangent line to the curve x2y + xy3 = 2 at the point (1, 1).
x^2 y + x y^3 = 2
x^2 dy/dx +2 x y +3 x y^2 dy/dx +y^3 = 0
dy/dx (x^2+3xy^2) = -2xy-y^3
at (1,1)
dy/dx(1+3) =-2-1
or
dy/dx = slope = m = -3/4
so
y = -(3/4) x + b , what is b, put in (1,1)
1 = -(3/4)(1) + b
b = 1 3/4 = 7/4
so y = -(3/4)x + 7/4
or
4 y = -3 x + 7