## Task Condition

Find the derivative.

$$
y=\operatorname{arctg} \frac{\sqrt{2 \operatorname{tg} x}}{1-\operatorname{tg} x}
$$
## Solution

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

Problem Kuznetsov Differentiation 15-16