Use implicit differentiation to find the equation of the tangent line to the curve xy^3+2xy=9 at the point (31). The equation of this tangent line can be written in the form y=mx+b where m is

Answer 1

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

dy (3xy^2+2x) = -dx (y^3+2y)

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

I assume you mean the point (3,1)
x = 3
y = 1
so
m = dy/dx = -(1+2)/(9+6) = - 3/15 = -1/5
put in point
1 = m * 3 + b
1 = -3/5 + b
b = 8/5
y = -x/5 + 8/5
5y = 8-x