Find an equation of the tangent plane to the surface given by z = f (x,y )=𝑥^2 y + x𝑦^3 at the point (x,y)=(2,1).

Answer 1

∂z/∂x = 2xy + y^3
∂z/∂y = x^2 + 3xy^2
z(2,1) = 4+2 = 6
So the tangent plane at (1,2) is
z-6 = 5(x-2) + 10(y-1)