[m]dz=\frac{ ∂z }{ ∂x }dx+\frac{ ∂z }{ ∂y }dy[/m]
[m]\frac{ ∂z }{ ∂x }=z`_{x}=(arctg\frac{y}{x}+arctg\frac{x}{y})`_{x}[/m]
Применяем формулу: [m] (arctg u)`=\frac{1}{1+u^2}\cdot u`[/m]
[m]\frac{ ∂z }{ ∂x }=\frac{1}{1+(\frac{y}{x})^2}\cdot(\frac{y}{x})`_{x}+\frac{1}{1+(\frac{x}{y})^2}\cdot(\frac{x}{y})`_{x}[/m]
[m]\frac{ ∂z }{ ∂x }=\frac{1}{1+(\frac{y}{x})^2}\cdot y\cdot (-\frac{1}{x^2})+\frac{1}{1+(\frac{x}{y})^2}\cdot(\frac{1}{y})[/m]
[m]\frac{ ∂z }{ ∂x }=\frac{x^2}{x^2+y^2}\cdot (-\frac{y}{x^2})+\frac{y^2}{x^2+y^2}\cdot(\frac{1}{y})[/m]
[m]\frac{ ∂z }{ ∂x }=-\frac{y}{x^2+y^2}+\frac{y}{x^2+y^2}=0[/m]
[m]\frac{ ∂z }{ ∂y }dy=z`_{y}=(arctg\frac{y}{x}+arctg\frac{x}{y})`_{y}[/m]
Применяем формулу: [m] (arctg u)`=\frac{1}{1+u^2}\cdot u`[/m]
[m]\frac{ ∂z }{ ∂y }=\frac{1}{1+(\frac{y}{x})^2}\cdot(\frac{y}{x})`_{y}+\frac{1}{1+(\frac{x}{y})^2}\cdot(\frac{x}{y})`_{y}[/m]
[m]\frac{ ∂z }{ ∂y }=\frac{1}{1+(\frac{y}{x})^2}\cdot(\frac{1}{x})+\frac{1}{1+(\frac{x}{y})^2}\cdot x \cdot (-\frac{1}{y^2})[/m]
[m]\frac{ ∂z }{ ∂y }=\frac{x^2}{x^2+y^2}\cdot (\frac{1}{x})+\frac{y^2}{x^2+y^2}\cdot(-\frac{x}{y^2})[/m]
[m]\frac{ ∂z }{ ∂y }=\frac{x}{x^2+y^2}-\frac{x}{x^2+y^2}=0[/m]
[m]dz=0dx+0dy[/m]