## Task Condition

Find the derivative.

$$
y=\operatorname{arctg} \frac{\sqrt{1-x}}{1-\sqrt{x}}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\sqrt{1-x}}{1-\sqrt{x}}\right)^{\prime}=\frac{1}{1+\left(\frac{\sqrt{1-x}}{1-\sqrt{x}}\right)^{2}} \cdot\left(\frac{\sqrt{1-x}}{1-\sqrt{x}}\right)^{\prime}= \\
& =\frac{(1-\sqrt{x})^{2}}{(1-\sqrt{x})^{2}+1-x} \cdot\left(\frac{(\sqrt{1-x})^{\prime} \cdot(1-\sqrt{x})-(\sqrt{1-x}) \cdot(1-\sqrt{x})^{\prime}}{(1-\sqrt{x})^{2}}\right)= \\
& =\frac{\frac{1}{2 \sqrt{1-x}} \cdot(-1) \cdot(1-\sqrt{x})-(\sqrt{1-x}) \cdot\left(-\frac{1}{2 \sqrt{x}}\right)}{1-2 \sqrt{x}+x+1-x}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{-\frac{1-\sqrt{x}}{2 \sqrt{1-x}}+\frac{\sqrt{1-x}}{2 \sqrt{x}}}{2-2 \sqrt{x}}=\frac{-\frac{\sqrt{x}(1-\sqrt{x})}{2 \sqrt{x} \sqrt{1-x}}+\frac{\sqrt{1-x} \sqrt{1-x}}{2 \sqrt{x} \sqrt{1-x}}}{2(1-\sqrt{x})}= \\
& =\frac{-\sqrt{x}+x+1-x}{4 \sqrt{x} \sqrt{1-x}(1-\sqrt{x})}=\frac{1-\sqrt{x}}{4 \sqrt{x} \sqrt{1-x}(1-\sqrt{x})}=\frac{1}{4 \sqrt{x(1-x)}}
\end{aligned}
$$

Kuznetsov Differentiation 10-25