Question

If the equation of the tangent line to the curve:

xy^3+xy=6, (3,1) find m and b

Answer 1

x(3 y^2 dy/dx) + y^3 + x dy/dx + y = 0

3 (3 dy/dx) + 1 + 3 dy/dx = 0
12 dy/dx = -1
dy/dx = -1/12 = m

then y = m x + b
1 = -(1/12) (3) + b
1 = -1/4 + b
b = 5/4
check my arithmetic