## Problem Statement

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

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

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

$=\frac{-1-t-1+t}{1-t^{2}}=-\frac{2}{1-t^{2}}$

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

We obtain:

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

$=\frac{t \cdot \sqrt{1-t^{2}}}{2}$

## Problem Kuznetsov Differentiation 16-16