Verify Clairaut's theorem for u=ln(x^(2)-y^(2)).
∂u/∂x = 2x/(x^2-y^2)
∂u/∂y = -2y/(x^2-y^2)
∂2u/∂x∂y = ∂2u/∂y∂x = -2(x^2+y^2)/(x^2-y^2)^2
To verify Clairaut's theorem for a given function, we need to check if the partial derivatives with respect to x and y are equal.
Let's start by finding the partial derivatives of u with respect to x and y.
1. Partial Derivative with respect to x (du/dx):
To find du/dx, we differentiate u with respect to x while considering y as a constant.
Using the chain rule, we have:
du/dx = (∂u/∂x) + (∂u/∂y) * (dy/dx)
Since u = ln(x^2 - y^2), we have:
∂u/∂x = 2x / (x^2 - y^2) (using the chain rule)
Notice that we need (∂u/∂y) and (dy/dx) for the second term, which we will calculate shortly.
2. Partial Derivative with respect to y (du/dy):
To find du/dy, we differentiate u with respect to y while considering x as a constant.
Using the chain rule, we have:
du/dy = (∂u/∂x) * (dx/dy) + (∂u/∂y)
Again, we will need (∂u/∂x) and (dx/dy) for the first term, which we will calculate shortly.
Calculating (∂u/∂y):
∂u/∂y = -2y / (x^2 - y^2) (using the chain rule)
Calculating (dx/dy):
To find (dx/dy), we differentiate x with respect to y. Since x is not a function of y, dx/dy will be zero.
Therefore, (dx/dy) = 0.
Now, let's substitute the calculated values back into the partial derivatives:
du/dx = 2x / (x^2 - y^2)
du/dy = -2y / (x^2 - y^2)
According to Clairaut's theorem, if du/dx = du/dy, then the theorem is verified.
Let's check if the two partial derivatives are equal:
2x / (x^2 - y^2) = -2y / (x^2 - y^2)
Canceling out the common denominator (x^2 - y^2), we are left with:
2x = -2y
Dividing both sides by 2, we get:
x = -y
So, we have verified Clairaut's theorem for the function u = ln(x^2 - y^2), where x = -y.