Final answer:
The second partial derivatives of v = (xy)/(x-y) are vxx = 1/(x-y) - (y-x)/(x-y)^2, vxy = -1/(x-y) + (y-x)/(x-y)^2, vyx = 0, and vyy = 0.
Step-by-step explanation:
To find the second partial derivatives of v = (xy)/(x-y), we need to differentiate it twice with respect to each variable. Let's start with the first partial derivatives:
vx = y/(x-y) - xy/(x-y)2 = y/(x-y) - x/(x-y) = (y-x)/(x-y)
vy = x/(x-y) - xy/(x-y)2 = x/(x-y) - y/(x-y) = (x-y)/(x-y) = 1
Now, let's differentiate the first partial derivatives with respect to x and y respectively:
vxx = d(vx)/dx = d((y-x)/(x-y))/dx = 1/(x-y) - (y-x)/(x-y)2
vxy = d(vx)/dy = d((y-x)/(x-y))/dy = -1/(x-y) + (y-x)/(x-y)2
vyx = d(vy)/dx = d(1)/dx = 0
vyy = d(vy)/dy = d(1)/dy = 0
Therefore, the second partial derivatives of v = (xy)/(x-y) are:
vxx = 1/(x-y) - (y-x)/(x-y)2
vxy = -1/(x-y) + (y-x)/(x-y)2
vyx = 0
vyy = 0