## Task Condition

Find the derivative.

$y=\frac{1}{\sqrt{8}} \arcsin \frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{8}} \arcsin \frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}\right)^{\prime}= \\
& =\frac{1}{\sqrt{8}} \cdot \frac{1}{\sqrt{1-\left(\frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}\right)^{2}}} \cdot\left(\frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}\right)^{\prime}= \\
& =\frac{1}{\sqrt{8}} \cdot \frac{1+3 \operatorname{ch} x}{\sqrt{(1+3 \operatorname{ch} x)^{2}-(3+\operatorname{ch} x)^{2}}} \cdot \frac{\operatorname{sh} x \cdot 3 \operatorname{ch} x-(3+\operatorname{ch} x) \cdot 3 \operatorname{sh} x}{(1+3 \operatorname{ch} x)^{2}}= \\
& =\frac{1}{\sqrt{8}} \cdot \frac{1}{\sqrt{1+6 \operatorname{ch} x+9 \operatorname{ch}^{2} x-9-6 \operatorname{ch} x-\operatorname{ch}^{2} x}} \cdot \frac{3 \operatorname{sh} x \cdot \operatorname{ch} x-9 \operatorname{sh} x-3 \operatorname{sh} x \cdot \operatorname{ch} x}{1+3 \operatorname{ch} x}= \\
& =\frac{1}{\sqrt{8}} \cdot \frac{1}{\sqrt{8 \operatorname{ch}^{2} x-8}} \cdot \frac{-9 \operatorname{sh} x}{1+3 \operatorname{ch} x}=\frac{1}{8 \sqrt{\operatorname{sh}^{2} x}} \cdot \frac{-9 \operatorname{sh} x}{1+3 \operatorname{ch} x}=-\frac{9 \operatorname{sh} x}{8(1+3 \operatorname{ch} x)}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 11-18