## problem statement

Find the derivative.

$$
y=\frac{1}{6} \ln \frac{1-\operatorname{sh} 2 x}{2+\operatorname{sh} 2 x}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{6} \ln \frac{1-\operatorname{sh} 2 x}{2+\operatorname{sh} 2 x}\right)^{\prime}= \\
& =\frac{1}{6} \cdot \frac{2+\operatorname{sh} 2 x}{1-\operatorname{sh} 2 x} \cdot \frac{-\operatorname{ch} 2 x \cdot 2 \cdot(2+\operatorname{sh} 2 x)-(1-\operatorname{sh} 2 x) \cdot \operatorname{ch} 2 x \cdot 2}{(2+\operatorname{sh} 2 x)^{2}}= \\
& =\frac{1}{6} \cdot \frac{1}{1-\operatorname{sh} 2 x} \cdot \frac{-4 \operatorname{ch} 2 x-2 \operatorname{ch} 2 x \cdot \operatorname{sh} 2 x-2 \operatorname{ch} 2 x+2 \operatorname{sh} 2 x \cdot \operatorname{ch} 2 x}{2+\operatorname{sh} 2 x}= \\
& =\frac{1}{6} \cdot \frac{1}{1-\operatorname{sh} 2 x} \cdot \frac{-6 \operatorname{ch} 2 x}{2+\operatorname{sh} 2 x}=-\frac{\operatorname{ch} 2 x}{2+\operatorname{sh} 2 x-2 \operatorname{sh} 2 x-\operatorname{sh}^{2} 2 x}= \\
& =\frac{\operatorname{ch} 2 x}{\operatorname{sh}^{2} 2 x+\operatorname{sh} 2 x-2}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 11-10