Задача 79063 Решите задачу...
Условие
Решение
Нужно найти производную сложной функции.
Выпишем сначала производные каждой из функций:
[m](arctg\ z)' = \frac{1}{1+z^2}[/m]
[m](\sqrt{1-x})' = \frac{1}{2\sqrt{1-x}} \cdot (-1) = -\frac{1}{2\sqrt{1-x}}[/m]
[m](1-\sqrt{x})' = -\frac{1}{2\sqrt{x}}[/m]
[m](\frac{\sqrt{1-x}}{1-\sqrt{x}})' = \frac{(\sqrt{1-x})'(1-\sqrt{x}) - \sqrt{1-x}(1-\sqrt{x})'}{(1-\sqrt{x})^2} =[/m]
[m]= [-\frac{1}{2\sqrt{1-x}} \cdot (1-\sqrt{x}) - \sqrt{1-x} \cdot (-\frac{1}{2\sqrt{x}})] : [(1-\sqrt{x})^2] =[/m]
[m]= [-\frac{1-\sqrt{x}}{2\sqrt{1-x}} + \frac{\sqrt{1-x} }{2\sqrt{x}}] : [(1-\sqrt{x})^2] = \frac{\sqrt{1-x} }{2\sqrt{x}(1-\sqrt{x})^2} - \frac{1-\sqrt{x}}{2\sqrt{1-x}(1-\sqrt{x})^2} =[/m]
[m]= \frac{\sqrt{1-x} \cdot \sqrt{1-x} - (1-\sqrt{x})\sqrt{x}}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})^2}= \frac{1-x - \sqrt{x}+x}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})^2} = \frac{1 - \sqrt{x}}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})^2} = \frac{1}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})}[/m]
Производная сложной функции:
[m]y'(z(x)) = y'(z) \cdot z'(x)[/m]
[m]y' = 1 : \bigg [1 + \bigg (\frac{\sqrt{1-x}}{1-\sqrt{x}} \bigg )^2 \bigg ] \cdot \frac{1}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})}=[/m]
[m]= 1 : [1 + \frac{1-x}{(1-\sqrt{x})^2}] \cdot \frac{1}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})} = 1 : \frac{(1-\sqrt{x})^2 + 1 - x}{(1-\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})} =[/m]
[m]= \frac{(1-\sqrt{x})^2}{1-2\sqrt{x} + x + 1 - x}\cdot \frac{1}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})} = \frac{(1-\sqrt{x})^2}{2-2\sqrt{x}}\cdot \frac{1}{2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})} =[/m]
[m]\frac{(1-\sqrt{x})^2}{2(1-\sqrt{x}) \cdot 2\sqrt{x}\sqrt{1-x}(1-\sqrt{x})} = \frac{1}{4\sqrt{x}\sqrt{1-x}}[/m]
Ответ: [m]y' = \frac{1}{4\sqrt{x}\sqrt{1-x}}[/m]