Задача 80774 Решите примеры ...
Условие
Решение
[m]\frac{dz}{dx} = 0,5y^{-1} + (-\frac{1}{2})x^{-3/2}y = \frac{1}{2y} - \frac{y}{2x^{3/2}}[/m]
[m]\frac{dz}{dy} = 0,5x \cdot (-1)y^{-2} + x^{-1/2} = -\frac{x}{2y^2} + \frac{1}{\sqrt{x}}[/m]
б) [m]z = arcsin \sqrt{\frac{x}{y}}[/m]
[m]\frac{dz}{dx} = \frac{1}{\sqrt{1 - x/y}} \cdot \frac{1}{2\sqrt{x/y}} \cdot \frac{1}{y} = \frac{\sqrt{y}}{\sqrt{y-x}} \cdot \frac{\sqrt{y}}{2\sqrt{x}} \cdot \frac{1}{y} = \frac{1}{2\sqrt{x}\sqrt{y-x}}[/m]
[m]\frac{dz}{dy} = \frac{1}{\sqrt{1 - x/y}} \cdot \frac{1}{2\sqrt{x/y}} \cdot (-\frac{x}{y^2}) = -\frac{\sqrt{y}}{\sqrt{y-x}} \cdot \frac{\sqrt{y}}{2\sqrt{x}} \cdot \frac{x}{y^2} = -\frac{\sqrt{x}}{2y\sqrt{y-x}}[/m]
в) [m]z = arcctg \frac{2x}{xy+1} = arctg \frac{xy+1}{2x}[/m]
[m]\frac{dz}{dx} = \frac{1}{ 1 + (\frac{xy+1}{2x})^2} \cdot \frac{y \cdot 2x - (xy+1) \cdot 2}{4x^2} = \frac{4x^2}{4x^2 + (xy+1)^2} \cdot \frac{2xy - 2(xy+1)}{4x^2} = \frac{-2}{4x^2 + (xy+1)^2}[/m]
[m]\frac{dz}{dy} = \frac{1}{ 1 + (\frac{xy+1}{2x})^2} \cdot \frac{x}{2x} = \frac{4x^2}{4x^2 + (xy+1)^2} \cdot \frac{1}{2} = \frac{2x^2}{4x^2 + (xy+1)^2}[/m]