z = √(x^2+y^2)+xy
∂z/∂x = x/√(x^2+y^2) + y
∂^2z/∂x^2 = y^2/(x^2+y^2)^(3/2)
∂^2z/∂x∂y = 1 - xy/(x^2+y^2)^(3/2)
Since z is symmetric in x and y, the y derivatives are the same, just using y instead of x.
As usual, check my algebra.
Z=square root of(x^2+y^2) + xy
Find the partial derivatives:
x, xx, y, yy, xy.
Thanks.
1 answer