Задача 63768 2.Найти полные дифференциалы функций ...
Условие
Решение
Так как [m]tg\frac{x+y}{x-y}=\frac{sin\frac{x+y}{x-y}}{cos\frac{x+y}{x-y}}[/m]
По формуле логарифма частного:
[m]ln\frac{sin\frac{x+y}{x-y}}{cos\frac{x+y}{x-y}}=ln sin\frac{x+y}{x-y}- ln cos\frac{x+y}{x-y}[/m]
Значит
[m]z=ln sin\frac{x+y}{x-y}- ln cos\frac{x+y}{x-y}[/m]
[m]dz=\frac{ ∂z }{ ∂x }dx+\frac{ ∂z }{ ∂y }dy[/m]
[m]\frac{ ∂z }{ ∂x }=z`_{x}=(ln sin\frac{x+y}{x-y}- ln cos\frac{x+y}{x-y})`_{x}=(ln sin\frac{x+y}{x-y})`_{x}-( ln cos\frac{x+y}{x-y})`_{x}[/m]
Применяем формулу [m] (lnu)`=\frac{1}{u}\cdot u`[/m]
[m]=\frac{1}{sin\frac{x+y}{x-y}}\cdot (sin\frac{x+y}{x-y})`_{x}-\frac{1}{cos\frac{x+y}{x-y}}\cdot (cos\frac{x+y}{x-y})`_{x}=[/m]
[m]=\frac{1}{sin\frac{x+y}{x-y}}\cdot cos\frac{x+y}{x-y}\cdot (\frac{x+y}{x-y})`_{x}-\frac{1}{cos\frac{x+y}{x-y}}\cdot (-sin\frac{x+y}{x-y})\cdot (\frac{x+y}{x-y})`_{x}=[/m]
[m]=\frac{1}{sin\frac{x+y}{x-y}}\cdot cos\frac{x+y}{x-y}\cdot \frac{(x+y)`_{x}\cdot (x-y)-(x+y)\cdot(x-y)`_{x}}{(x-y)^2}-\frac{1}{cos\frac{x+y}{x-y}}\cdot (-sin\frac{x+y}{x-y})\cdot\frac{(x+y)`_{x} \cdot (x-y)-(x+y)\cdot(x-y)`_{x}}{(x-y)^2}=[/m]
[m]=\frac{cos\frac{x+y}{x-y}}{sin\frac{x+y}{x-y}}\cdot \frac{(x-y)-(x+y)}{(x-y)^2}+\frac{sin\frac{x+y}{x-y}}{cos\frac{x+y}{x-y}}\cdot\frac{(x-y)-(x+y)}{(x-y)^2}=[/m]
[m]=ctg\frac{x+y}{x-y}\cdot \frac{x-y-x-y}{(x-y)^2}+tg\frac{x+y}{x-y}\cdot\frac{x-y-x-y}{(x-y)^2}=[/m]
[m]=-ctg\frac{x+y}{x-y}\cdot \frac{2y}{(x-y)^2}-tg\frac{x+y}{x-y}\cdot\frac{2y}{(x-y)^2}=\frac{2y}{(x-y)^2}\cdot(-ctg\frac{x+y}{x-y}-tg\frac{x+y}{x-y}) [/m]
[m]\frac{ ∂z }{ ∂y }=z`_{y}=(ln sin\frac{x+y}{x-y}- ln cos\frac{x+y}{x-y})`_{y}=[/m]
Применяем формулу [m] (lnu)`=\frac{1}{u}\cdot u`[/m]
[m]=\frac{1}{sin\frac{x+y}{x-y}}\cdot (sin\frac{x+y}{x-y})`_{y}-\frac{1}{cos\frac{x+y}{x-y}}\cdot (cos\frac{x+y}{x-y})`_{y}=[/m]
[m]=\frac{1}{sin\frac{x+y}{x-y}}\cdot cos\frac{x+y}{x-y}\cdot (\frac{x+y}{x-y})`_{y}-\frac{1}{cos\frac{x+y}{x-y}}\cdot (-sin\frac{x+y}{x-y})\cdot (\frac{x+y}{x-y})`_{y}=[/m]
[m]=\frac{1}{sin\frac{x+y}{x-y}}\cdot cos\frac{x+y}{x-y}\cdot\frac{(x+y)`_{y}\cdot (x-y)-(x+y)\cdot(x-y)`_{y}}{(x-y)^2}-\frac{1}{cos\frac{x+y}{x-y}}\cdot (-sin\frac{x+y}{x-y})\cdot \frac{(x+y)`_{y}\cdot (x-y)-(x+y)\cdot(x-y)`_{y}}{(x-y)^2}=[/m]
[m]=\frac{cos\frac{x+y}{x-y}}{sin\frac{x+y}{x-y}}\cdot \frac{1\cdot (x-y)-(x+y)(-1)}{(x-y)^2}+\frac{sin\frac{x+y}{x-y}}{cos\frac{x+y}{x-y}}\cdot\frac{1\cdot (x-y)-(x+y)\cdot (-1))}{(x-y)^2}=[/m]
[m]=ctg\frac{x+y}{x-y}\cdot \frac{x-y+x+y}{(x-y)^2}+tg\frac{x+y}{x-y}\cdot\frac{x-y+x+y}{(x-y)^2}=[/m]
[m]=ctg\frac{x+y}{x-y}\cdot \frac{2x}{(x-y)^2}+tg\frac{x+y}{x-y}\cdot\frac{2x}{(x-y)^2}=\frac{2x}{(x-y)^2}\cdot (ctg\frac{x+y}{x-y}+tg\frac{x+y}{x-y})[/m]
О т в е т.
[m]dz=(ctg\frac{x+y}{x-y}+tg\frac{x+y}{x-y})\cdot\frac{1}{(x-y)^2}(-2ydx+ 2xdy)[/m]