Задача 60833 Найти частные производные функций по...
Условие
Решение
a)
[m]\frac{ ∂ z}{ ∂ x}=(\frac{x}{3y})`_{x}-(\frac{3y}{x})`_{x}[/m](y= const)=[m]\frac{1}{3y}\cdot (x)`-3y\cdot (\frac{1}{x})`=\frac{1}{3y}-3y\cdot (-\frac{1}{x^2})=\frac{1}{3y}+\frac{3y}{x^2}[/m]
[m]\frac{ ∂ z}{ ∂ y}=(\frac{x}{3y})`_{y}-(\frac{3y}{x})`_{y}[/m](x= const)=[m]\frac{x}{3}\cdot (\frac{1}{y})`-\frac{3}{x}\cdot (\frac{1}{y})`=\frac{x}{3}-\frac{3}{x}\cdot (-\frac{1}{y^2})=\frac{x}{3}+\frac{3}{xy^2})[/m]
б)
[m]\frac{ ∂ z}{ ∂ x}=-\frac{1}{\sqrt{1-(\frac{x}{2y})^2}}\cdot (\frac{x}{2y})`_{x}=-\frac{1}{\sqrt{1-(\frac{x}{2y})^2}}\cdot (\frac{1}{2y})\cdot (x)`=-\frac{1}{\sqrt{1-(\frac{x}{2y})^2}}\cdot (\frac{1}{2y})\cdot 1[/m]
[m]\frac{ ∂ z}{ ∂ y}=-\frac{1}{\sqrt{1-(\frac{x}{2y})^2}}\cdot (\frac{x}{2y})`_{y}=-\frac{1}{\sqrt{1-(\frac{x}{2y})^2}}\cdot \frac{x}{2}\cdot (\frac{1}{y})`=-\frac{1}{\sqrt{1-(\frac{x}{2y})^2}}\cdot \frac{x}{2}(-\frac{1}{y^2})[/m]
в)
[m]\frac{ ∂ z}{ ∂ x}=\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot (\frac{2y}{4+xy})`_{x}=\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot (2y)\cdot (\frac{1}{4+xy})`_{x} =\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot (2y)\cdot (-\frac{1}{(4+xy)^2})\cdot (4+xy)`_{x} =\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot (2y)\cdot (-\frac{1}{(4+xy)^2})\cdot (0+y) [/m]
[m]\frac{ ∂ z}{ ∂ y}=\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot (\frac{2y}{4+xy})`_{y}=\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot \frac{(2y)`_{y}\cdot (4+xy)+(2y)\cdot (4+xy)`_{y}}{(4+xy)^2}=\frac{1}{1+(\frac{2y}{4+xy})^2}\cdot \frac{(2\cdot (4+xy)+(2y)\cdot (0+x)}{(4+xy)^2}[/m]