## Problem Statement

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

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

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

We obtain:

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

## Problem Kuznetsov Differentiation 16-24