## Problem Statement

Find the derivative $y_{x}^{\prime}$.

$$
\left\{\begin{array}{l}
x=\sqrt{t-t^{2}}-\operatorname{arctg} \sqrt{\frac{1-t}{t}} \\
y=\sqrt{t}-\sqrt{1-t} \cdot \arcsin \sqrt{t}
\end{array}\right.
$$
## Solution

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

$$
\begin{aligned}
& =\frac{1-2 t}{2 \sqrt{t-t^{2}}}+\frac{\sqrt{t}}{2 t \sqrt{1-t}}=\frac{1-2 t}{2 \sqrt{t} \sqrt{1-t}}+\frac{1}{2 \sqrt{t} \sqrt{1-t}}=\frac{2(1-t)}{2 \sqrt{t} \sqrt{1-t}}=\sqrt{\frac{1-t}{t}} \\
& y_{t}^{\prime}=(\sqrt{t}-\sqrt{1-t} \cdot \arcsin \sqrt{t})^{\prime}= \\
& =\frac{1}{2 \sqrt{t}}-\left(\frac{1}{2 \sqrt{1-t}} \cdot(-1) \cdot \arcsin \sqrt{t}+\sqrt{1-t} \cdot \frac{1}{\sqrt{1-(\sqrt{t})^{2}}} \cdot \frac{1}{2 \sqrt{t}}\right)= \\
& =\frac{1}{2 \sqrt{t}}+\frac{\arcsin \sqrt{t}}{2 \sqrt{1-t}}-\frac{1}{2 \sqrt{t}}=\frac{\arcsin \sqrt{t}}{2 \sqrt{1-t}}
\end{aligned}
$$

We obtain:

$$
y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{\arcsin \sqrt{t}}{2 \sqrt{1-t}}\right) /\left(\sqrt{\frac{1-t}{t}}\right)=\frac{\sqrt{t} \cdot \arcsin \sqrt{t}}{2(1-t)}
$$

## Problem Kuznetsov Differentiation 16-26