A differentiable function f(x,y) has the property that f(2,2)=2 and fx(2,2)=1 and fy(2,2)=2. Find the equation of the tangent plane at the point on the surface z=f(x,y) where x=2, y=2.

Answer 1

for f(x,y,z) = 0,

the plane is

fx(2,2)(x-2) + fy(2,2)(y-2) + fz(2,2)(z-2) = 0

1(x-2) + 2(y-2) - 1(z-2) = 0
x + 2y - z = 4