## Task Condition

Find the derivative.

$y=-\frac{1}{4} \arcsin \frac{5+3 \operatorname{ch} x}{3+5 \operatorname{ch} x}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(-\frac{1}{4} \arcsin \frac{5+3 \operatorname{ch} x}{3+5 \operatorname{ch} x}\right)^{\prime}= \\
& =-\frac{1}{4} \cdot \frac{1}{\sqrt{1-\left(\frac{5+3 \operatorname{ch} x}{3+5 \operatorname{ch} x}\right)^{2}}} \cdot \frac{3 \operatorname{sh} x \cdot(3+5 \operatorname{ch} x)-(5+3 \operatorname{ch} x) \cdot 5 \operatorname{sh} x}{(3+5 \operatorname{ch} x)^{2}}= \\
& =-\frac{1}{4} \cdot \frac{3+5 \operatorname{ch} x}{\sqrt{(3+5 \operatorname{ch} x)^{2}-(5+3 \operatorname{ch} x)^{2}}} \cdot \frac{9 \operatorname{sh} x+15 \cdot \operatorname{sh} x \cdot \operatorname{ch} x-25 \operatorname{sh} x-15 \operatorname{ch} x \cdot \operatorname{sh} x}{(3+5 \operatorname{ch} x)^{2}}= \\
& =-\frac{1}{4} \cdot \frac{1}{\sqrt{9+30 \operatorname{ch} x+25 \operatorname{ch}^{2} x-25-30 \operatorname{ch} x-9 \operatorname{ch}^{2} x}} \cdot \frac{9 \operatorname{sh} x-25 \operatorname{sh} x}{3+5 \operatorname{ch} x}= \\
& =-\frac{1}{4} \cdot \frac{1}{\sqrt{16 \operatorname{ch}^{2} x-16}} \cdot \frac{-16 \operatorname{sh} x}{3+5 \operatorname{ch} x}=\frac{1}{4 \sqrt{\operatorname{ch}^{2} x-1}} \cdot \frac{4 \operatorname{sh} x}{3+5 \operatorname{ch} x}= \\
& =\frac{1}{\sqrt{\operatorname{sh}^{2} x}} \cdot \frac{\operatorname{sh} x}{3+5 \operatorname{ch} x}=\frac{1}{\operatorname{sh} x} \cdot \frac{\operatorname{sh} x}{3+5 \operatorname{ch} x}=\frac{1}{3+5 \operatorname{ch} x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 11-21