xy^3 + 2y - 2x = 1
y^3 + 3xy^2 y' + 2y' - 2 = 0
y' = (2-y^3)/(3xy^2+2)
At (1,1), y' = 2/5
Now you have a point: (1,1) and a slope: 2/5
I'm sure you know the point-slope form of a line.
Find the tangent line to the curve xy3+2y−2x=1 through the point (x,y)=(1,1).
(Enter ∗ for multiplication: type 2*x for 2x. Enter / for division: type 1/2 for 12.)
Tangent Line y=
2 answers
y´=1/5*x-4/5