## Problem Statement

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

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

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

We obtain:

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

## Problem Kuznetsov Differentiation 16-10