[tex]f(x,y)=\log(x-y)[/tex]
has first-order partial derivatives
[tex]\dfrac{\partial f}{\partial x}=\dfrac1{x-y}[/tex]
[tex]\dfrac{\partial f}{\partial y}=-\dfrac1{x-y}[/tex]
Then the second-order partial derivatives are
[tex]\dfrac{\partial^2f}{\partial x^2}=-\dfrac1{(x-y)^2}[/tex]
[tex]\dfrac{\partial^2f}{\partial x\partial y}=\dfrac1{(x-y)^2}[/tex]
[tex]\dfrac{\partial^2f}{\partial y\partial x}=\dfrac1{(x-y)^2}[/tex]
[tex]\dfrac{\partial^2f}{\partial y^2}=-\dfrac1{(x-y)^2}[/tex]
We want to get the second partial derivatives of the function f(x, y) = log(x - y).
[tex]\frac{d^2f(x,y)}{dx^2} = \frac{1}{(x - y)^2*ln(10)} \\\\\frac{df(x,y)}{dxdy} = -\frac{1}{(x - y)^2*ln(10)} \\\\\frac{d^2f(x,y)}{dydx} = -\frac{1}{(x - y)^2*ln(10)} \\\\\frac{d^2f(x,y)}{dy^2} = \frac{1}{(x - y)^2*ln(10)}[/tex]
Remember that:
if g(x) = log(x)
[tex]\frac{dg(x)}{dx} = \frac{1}{x*ln(10)}[/tex]
Now we can use this rule to derive our function, we will get the first partial derivatives.
[tex]\frac{df(x,y)}{dx} = \frac{1}{(x - y)*ln(10)} \\\\\frac{df(x,y)}{dy} = -\frac{1}{(x - y)*ln(10)}[/tex]
Now we need to derive these again to get:
[tex]\frac{d^2f(x,y)}{dx^2} = \frac{1}{(x - y)^2*ln(10)} \\\\\frac{df(x,y)}{dxdy} = -\frac{1}{(x - y)^2*ln(10)} \\\\\frac{d^2f(x,y)}{dydx} = -\frac{1}{(x - y)^2*ln(10)} \\\\\frac{d^2f(x,y)}{dy^2} = \frac{1}{(x - y)^2*ln(10)}[/tex]
So we can see that the mixed partial derivatives are equal, meaning that the theorem is true in this case.
If you want to learn more, you can read:
https://brainly.com/question/17330601
