z=arctg(5x-3xy+y^2)
[m]\frac{ ∂ z}{ ∂ x}=(arctg(5x–3xy+y^2))`_{x}=\frac{1}{1+(5x–3xy+y^2)^2)}\cdot (5x-3xy+y^2)`_{x}=\frac{3-3y}{1+(5x–3xy+y^2)^2)}[/m]
[m]\frac{ ∂ z}{ ∂ y}=(arctg(5x–3xy+y^2))`_{y}=\frac{1}{1+(5x–3xy+y^2)^2)}\cdot (5x-3xy+y^2)`_{y}=\frac{(-3x+2y)}{1+(5x–3xy+y^2)^2)}[/m]
[m]dz=\frac{3-3y}{1+(5x–3xy+y^2)^2)} dx+\frac{(-3x+2y)}{1+(5x–3xy+y^2)^2)}dy[/m] - о т в е т.
2)
Находим
[m]\frac{ ∂^2z}{ ∂ x^2}=(\frac{(3-3y)}{1+(5x–3xy+y^2)^2)})`_{x}[/m]=...
[m]\frac{ ∂^2z}{ ∂ x ∂y }=(\frac{(3-3y)}{1+(5x–3xy+y^2)^2)})`_{y}[/m]=...
[m]\frac{ ∂^2z}{ ∂ y^2}=(\frac{(-3x+2y)}{1+(5x–3xy+y^2)^2)})`_{y}[/m]=...
cчитаете...
и подставляете в формулу:
[m]d^2z=\frac{ ∂^2 z}{ ∂ x^2}(dx)^2+2\frac{ ∂^2z}{ ∂ x ∂y }dxdy+\frac{ ∂^2 z}{ ∂ y^2}(dy)^2[/m]