z =f (x; y)
Показать, что
[m]\frac{ ∂z }{ ∂ x}=(sinx+siny+sin(2x-y))`_{x}=(sinx)`_{x}+(siny)`_{x}+sin(2x-y))`_{x}=cosx+0+cos(2x-y)\cdot (2x-y)`_{x}=cosx+2cos(2x-y)[/m]
[m]\frac{ ∂z }{ ∂ y}=(sinx+siny+sin(2x-y))`_{y}=(sinx)`_{y}+(siny)`_{y}+sin(2x-y))`_{y}=0+cosy+cos(2x-y)\cdot (2x-y)`_{y}=cosy-cos(2x-y)[/m]
[m]F=\frac{ ∂z }{ ∂ x}-\frac{ ∂z }{ ∂ y}-cosx+cosy-3cos(2x-y)[/m]
[m]F=\underbrace{cosx+2cos(2x-y)}_{\frac{ ∂z }{ ∂ x}}-\underbrace{(cosy-cos(2x-y))}_{\frac{ ∂z }{ ∂ y}}-cosx+cosy-3cos(2x-y)[/m]
[m]F=0[/m]- верно