## Condition of the problem

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

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

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

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

We obtain:

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

## Problem Kuznetsov Differentiation $16-18$