## Problem Statement

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

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

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

$$
\begin{aligned}
& =\frac{1-t^{2}+t \cdot t}{\sqrt{\left(1-t^{2}\right)^{3}}}=\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}}} \\
& y_{t}^{\prime}=\left(\ln \frac{1+\sqrt{1-t^{2}}}{t}\right)^{\prime}=\frac{t}{1+\sqrt{1-t^{2}}} \cdot \frac{\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t) \cdot t-\left(1+\sqrt{1-t^{2}}\right) \cdot 1}{t^{2}}= \\
& =\frac{1}{1+\sqrt{1-t^{2}}} \cdot \frac{-t^{2}-\sqrt{1-t^{2}}-\left(1-t^{2}\right)}{t \sqrt{1-t^{2}}}=\frac{1}{1+\sqrt{1-t^{2}}} \cdot \frac{-\left(\sqrt{1-t^{2}}+1\right)}{t \sqrt{1-t^{2}}}= \\
& =-\frac{1}{t \sqrt{1-t^{2}}}
\end{aligned}
$$

We obtain:

$$
y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{1}{t \sqrt{1-t^{2}}}\right) /\left(\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}}}\right)=-\frac{\sqrt{\left(1-t^{2}\right)^{3}}}{t \sqrt{1-t^{2}}}=-\frac{1-t^{2}}{t}=\frac{t^{2}-1}{t}
$$

## Problem Kuznetsov Differentiation 16-14