## Problem Statement

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

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

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

$=\frac{\sqrt{1-t^{2}}-t \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)}{1-t^{2}} \cdot \arcsin t+\frac{t}{\sqrt{1-t^{2}}} \cdot \frac{1}{\sqrt{1-t^{2}}}+\frac{1}{\sqrt{1-t^{2}}} \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=$

$=\frac{\sqrt{1-t^{2}}+\frac{t^{2}}{\sqrt{1-t^{2}}}}{1-t^{2}} \cdot \arcsin t+\frac{t}{1-t^{2}}-\frac{t}{1-t^{2}}=\frac{1-t^{2}+t^{2}}{\left(1-t^{2}\right) \sqrt{1-t^{2}}} \cdot \arcsin t=$

$=\frac{\arcsin t}{\left(1-t^{2}\right) \sqrt{1-t^{2}}}$

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

We obtain:

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

## Problem Kuznetsov Differentiation 16-30