Задача 64871 Кто сможет?...
Условие
Решение
1)
[m]\frac{ ∂z }{ ∂x }=(\frac{tg(x-y)}{x})`_{x}=\frac{(tg(x-y))`_{x}\cdot x-tg(x-y)\cdot (x)`_{x}}{x^2}=\frac{\frac{1}{cos^2(x-y)}\cdot (x-y)`_{x}\cdot x-tg(x-y)}{x^2}=\frac{\frac{1}{cos^2(x-y)}\cdot 1 \cdot x-tg(x-y)}{x^2}[/m]
[m]\frac{ ∂z }{ ∂y }=(\frac{tg(x-y)}{x})`_{y}=\frac{1}{x}\cdot (tg(x-y))`_{y}=\frac{1}{x}\cdot \frac{1}{cos^2(x-y)}\cdot (x-y)`_{y}=\frac{1}{x}\cdot \frac{1}{cos^2(x-y)}\cdot (-1)=-\frac{1}{x\cdot cos^2(x-y)}[/m]
2)
[m]\frac{ ∂^2z }{ ∂x ∂y }=(\frac{ ∂z }{ ∂x })`_{y}=(\frac{\frac{1}{cos^2(x-y)}\cdot 1 \cdot x-tg(x-y)}{x^2})`_{y}=\frac{1}{x^2}\cdot (x\cdot (-2cos^{-3}(x-y)\cdot (x-y)`_{y}-\frac{1}{cos^2(x-y)}\cdot (x-y)`_{y}=\frac{1}{x^2}\cdot (\frac{2x}{cos^3(x-y)}+\frac{1}{cos^2(x-y)})[/m]
[m]\frac{ ∂^2z }{ ∂y^2 }=(\frac{ ∂z }{ ∂y })`_{y}=(-\frac{1}{x\cdot cos^2(x-y)})`_{y}=-\frac{1}{x}\cdot (-2cos^{-3}(x-y)\cdot (x-y)`_{y}=-\frac{2}{x\cdot cos^3(x-y)}[/m]
Тогда
[m]\frac{ ∂^2z }{ ∂x ∂y }+\frac{ ∂^2z }{ ∂y^2 }+\frac{1}{x}\cdot\frac{ ∂z }{ ∂y }= \frac{1}{x^2}\cdot (\frac{2x}{cos^3(x-y)}+\frac{1}{cos^2(x-y)})-\frac{2}{x\cdot cos^3(x-y)}+\frac{1}{x}\cdot(-\frac{1}{x\cdot cos^2(x-y)})=0[/m]- верно