Partial derivative. Explanation of equal the function

[Guest]

Hello, please help wit the explanation of this:

There is function:
[tex]z\:=\:y\:ln\left(x^2-y^2\right)\:[/tex]

Need to explain that:
[tex]\frac{\partial \:^2z}{\partial x\partial y}\:=\:\frac{\partial ^2z}{\partial y\partial x}[/tex]

[shyamjayakannan]

LHS [tex]=\frac{\partial^2z}{\partial x\partial y}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial}{\partial x}\left[\frac{\partial}{\partial y}\left\{y\ln\left(x^2-y^2\right)\right\}\right]=\frac{\partial}{\partial x}\left\{\ln\left(x^2-y^2\right)-\frac{2y^2}{x^2-y^2}\right\}=\frac{2x}{x^2-y^2}+\frac{4y^2x}{\left(x^2-y^2\right)^2}[/tex]

RHS [tex]=\frac{\partial^2z}{\partial y\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial y}\left[\frac{\partial}{\partial x}\left\{y\ln\left(x^2-y^2\right)\right\}\right]=\frac{\partial}{\partial y}\left(\frac{2xy}{x^2-y^2}\right)=\frac{2x}{x^2-y^2}+\frac{4y^2x}{\left(x^2-y^2\right)^2}[/tex]