Задача 58902 Помогите пожалуйста!...
Условие
Решение
[m]z`_{x}=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot (\frac{x-y^2}{x^2-y})`_{x}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{1\cdot (x^2-y)-(x-y^2)\cdot (2x)}{(x^2-y)^2}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{x^2-y-2x^2+2xy^2}{(x^2-y)^2}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{2xy^2-y-x^2}{(x^2-y)^2}=[/m]
[m]z`_{y}=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot (\frac{x-y^2}{x^2-y})`_{y}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{(x-y^2)`_{y}\cdot (x^2-y)-(x-y^2)\cdot (x^2-y)`_{y}}{(x^2-y)^2}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{(-2y)\cdot (x^2-y)-(x-y^2)\cdot (-1)}{(x^2-y)^2}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{-2x^2y+2y^2+x+y^2)\cdot (-1)}{(x^2-y)^2}=[/m]
[m]=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{3y^2+x-2x^2y}{(x^2-y)^2}=[/m]
О т в е т.
[m]dz=2\cdot cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{2xy^2-y-x^2}{(x^2-y)^2}dx+2cos\frac{x-y^2}{x^2-y}\cdot (-sin\frac{x-y^2}{x^2-y})\cdot\frac{3y^2+x-2x^2y}{(x^2-y)^2}dy[/m]
[m]dz=-2\cdot cos\frac{x-y^2}{x^2-y}\cdot sin\frac{x-y^2}{x^2-y}\cdot \frac{1}{(x^2-y)^2}\cdot( (2xy^2-y-x^2)dx+(3y^2+x-2x^2y)dy)[/m]