Question

Найти производную. кто тут самый умный, пусть решит если сможет.

Answer 1

$(arctg \frac{ \sqrt{1-x}}{1- \sqrt{x}})'=\frac{1}{1+(\frac{ \sqrt{1-x}}{1- \sqrt{x}})^2}\cdot(\frac{ \sqrt{1-x}}{1- \sqrt{x}})'=$

$=\frac{1}{1+\frac{1-x}{(1- \sqrt{x})^2}}\cdot\frac{ (\sqrt{1-x})'(1- \sqrt{x})-\sqrt{1-x}(1- \sqrt{x})'}{(1- \sqrt{x})^2}=$

$=\frac{(1- \sqrt{x})^2}{(1- \sqrt{x})^2+1-x}\cdot\frac{\frac{1}{2\sqrt{1-x}}(1-x)'(1- \sqrt{x})-\sqrt{1-x}(-\frac{1}{2 \sqrt{x}})}{(1- \sqrt{x})^2}=$

$=\frac{-\frac{1- \sqrt{x}}{2\sqrt{1-x}}+\frac{\sqrt{1-x}}{2 \sqrt{x}}}{1-2\sqrt{x}+x+1-x}=$

$=\frac{(\sqrt{1-x})^2-\sqrt{x}(1-\sqrt{x})}{2\sqrt{x}\sqrt{1-x}(2-2\sqrt{x})}=\frac{1-x-\sqrt{x}+x}{4\sqrt{x}\sqrt{1-x}(1-\sqrt{x})}=\frac{1-\sqrt{x}}{4\sqrt{x}\sqrt{1-x}(1-\sqrt{x})}=$

$=\frac{1}{4\sqrt{x}\sqrt{1-x}}$

Answer 2

$(arctg\frac{\sqrt{1-x}}{1-\sqrt{x}})' = \frac{1}{\frac{1-x}{(1-\sqrt{x})^2}+1}*\frac{-\frac{(-1)(1-\sqrt{x})}{2\sqrt{1-x}}+\frac{\sqrt{1-x}}{2\sqrt{x}}}{(1-\sqrt{x})^2}=$$\frac{(1-\sqrt{x})^2}{1-x+1-2\sqrt{x}+x}*(\frac{\sqrt{1-x}}{2\sqrt{x}}+\frac{1-\sqrt{x}}{2\sqrt{1-x}}):(1-\sqrt{x})$$\frac{(1-\sqrt{x})^2}{2(1-\sqrt{x})}*\frac{(1-x)+\sqrt{x}-x}{2\sqrt{x(1-x})(1-\sqrt{x})}}}}=$$\frac{(1-\sqrt{x})^2(1-2x+\sqrt{x})}{2(1-\sqrt{x})2\sqrt{x(1-x)}(1-\sqrt{x})}=\frac{1-2x+\sqrt{x}}{4\sqrt{x(1-x)}}$