Find all partial derivatives?
v = (xy)/(x-y)
vxx=
vxy=
vyx=
vyy=
vx = -y^2/(x-y)^2
vy = x^2/(x-y)^2
vxx = 2y^2/(x-y)^3
vyy = 2x^2/(x-y)^3
vxy = vyx = -2xy/(x-y)^3
v=xy/x-y
To find all partial derivatives of the function v = (xy)/(x-y), we have:
vxx = d²v/dx²
vxy = d²v/dxdy
vyx = d²v/dydx
vyy = d²v/dy²
Let's find each of these partial derivatives step-by-step.
Step 1: Find the first partial derivative with respect to x.
To do this, we treat y as a constant and differentiate the function with respect to x.
v_x = d(v)/dx = [(x-y)*d(xy)/dx - (xy)*d(x-y)/dx]/(x-y)²
= [(x-y)*(y) - (xy)*(-1)]/(x-y)²
= (xy - y² + xy)/(x-y)²
= (2xy - y²)/(x-y)²
Step 2: Find the first partial derivative with respect to y.
To do this, we treat x as a constant and differentiate the function with respect to y.
v_y = d(v)/dy = [(x-y)*d(xy)/dy - (xy)*d(x-y)/dy]/(x-y)²
= [(x-y)*(x) - (xy)*(1)]/(x-y)²
= (x² - xy)/(x-y)²
Step 3: Find the second partial derivative with respect to x.
To do this, we differentiate the previously obtained v_x with respect to x.
vxx = d(v_x)/dx = d[(2xy - y²)/(x-y)²]/dx
Differentiating each term separately, we get:
vxx = d(2xy - y²)/dx - d(x-y)²/dx
Using chain rule and product rule, we have:
vxx = 2y - 2xy/x-y + 2(x-y)/x-y
= 2y - 2xy/x-y + 2
Step 4: Find the second partial derivative with respect to y.
To do this, we differentiate the previously obtained v_y with respect to y.
vyy = d(v_y)/dy = d[(x² - xy)/(x-y)²]/dy
Differentiating each term separately, we get:
vyy = d(x² - xy)/dy - d(x-y)²/dy
Using chain rule and product rule, we have:
vyy = -x + (x-y)(2)/x-y
= -x + 2
Step 5: Find the mixed partial derivatives.
To find the mixed partial derivatives, we differentiate v_x with respect to y and v_y with respect to x.
vxy = d(v_x)/dy = d[(2xy - y²)/(x-y)²]/dy
Differentiating each term separately, we get:
vxy = 2x - 2xy/x-y
vyx = d(v_y)/dx = d[(x² - xy)/(x-y)²]/dx
Differentiating each term separately, we get:
vyx = y - (x-y)(2)/x-y
= y - 2
Therefore, the partial derivatives of the function v = (xy)/(x-y) are:
vxx = 2y - 2xy/(x-y) + 2
vxy = 2x - 2xy/(x-y)
vyx = y - 2
vyy = -x + 2
To find the partial derivatives of the given function `v = (xy)/(x-y)`, we will use the quotient rule and the product rule.
First, let's find `vxx` (the second partial derivative of `v` with respect to `x`):
Start by finding the first partial derivative of `v` with respect to `x` (v_x):
v_x = (y * (x - y) - x * (xy))/((x - y)^2)
Now, differentiate v_x with respect to x to get vxx:
vxx = [(y * (x - y) - x * (xy)) * (x - y)^2 -
(2 * (y * (x - y) - x * (xy)) * (x - y) * (1))]/((x - y)^4)
Simplify this expression to get vxx.
Next, let's find vxy (the second partial derivative of v with respect to x and y):
To find vxy, start by finding the first partial derivative of v with respect to y (v_y):
v_y = (x * (x - y) - y * (xy))/((x - y)^2)
Now, differentiate v_y with respect to x to get vxy:
vxy = [(x * (x - y) - y * (xy)) * (x - y)^2 -
(2 * (x * (x - y) - y * (xy)) * (x - y) * (1))]/((x - y)^4)
Simplify this expression to get vxy.
Similarly, we can find vyy (the second partial derivative of v with respect to y):
Start by finding the first partial derivative of v with respect to y (v_y):
v_y = (x * (x - y) - y * (xy))/((x - y)^2)
Now, differentiate v_y with respect to y to get vyy:
vyy = [(x * (x - y) - y * (xy)) * (x - y)^2 -
(2 * (x * (x - y) - y * (xy)) * (x - y) * (-1))]/((x - y)^4)
Simplify this expression to get vyy.
Finally, the derivative vyx (the second partial derivative of v with respect to y and x) is the same as vxy due to the symmetry of mixed partial derivatives. So, the value of vyx is the same as vxy.