Question

find the equation of the tangent line for (xy)^2+(xy)-x=11 at the point (-11,0)

Answer 1

2(xy)(x dy/dx + y) + x dy/dx + y - 1 = 0
plug in (-11,0)
2(-11)(0)(-11dy/dx + 0) + (-11)dy/dx + 0 - 1 = 0
-11 dy/dx = 1
dy/dx = -1/11

so now you have the slope -1/11 and the point(-11,0),

equation of tangent:
0 = -1/11(-11) + b
0 = 1+b --> b= -1

y = (-1/11)x - 1