Задача 71111 z = yf(cos(x-y)) доказать что dx/dx +...
Условие
Решение
[m]=y\cdot φ `_{x}(cos(x-y))\cdot (-sin(x-y))\cdot (x-y)`_{x}=y\cdot φ `_{x}(cos(x-y))\cdot (-sin(x-y))\cdot 1=y\cdot φ `_{x}(cos(x-y))\cdot (-sin(x-y))[/m]
[m]\frac{ ∂z }{ ∂y }=(y\cdot φ (cos(x-y))`_{y}=y`\cdot φ (cos(x-y))+y \cdot φ `_{y}(cos(x-y))\cdot ((cos(x-y))`_{y} =[/m]
[m]= φ (cos(x-y))+y \cdot φ `_{y}(cos(x-y))\cdot (-sin(x-y))\cdot (x-y)`_{y}=φ (cos(x-y))+y \cdot φ `_{y}(cos(x-y))\cdot (-sin(x-y))\cdot (-1) =[/m]
[m]=φ (cos(x-y))+y \cdot φ `_{y}(cos(x-y))\cdot (sin(x-y))[/m]
Тогда
[m]\frac{ ∂z }{ ∂x }+\frac{ ∂z }{ ∂y }=y\cdot φ `_{x}(cos(x-y))\cdot (-sin(x-y))+φ (cos(x-y))+y \cdot φ `_{y}(cos(x-y))\cdot (sin(x-y))=[/m]
[m]=y\cdot sin(x-y)\cdot (φ `_{y}(cos(x-y))- φ `_{x}(cos(x-y)))[/m]
Это не равно [m]\frac{x}{y}[/m]