## Problem Statement

Find the derivative.

$$
y=\frac{1}{4} \ln \left|\operatorname{th} \frac{x}{2}\right|-\frac{1}{4} \cdot \ln \frac{3+\operatorname{ch} x}{\operatorname{sh} x}
$$
## Solution

$y^{\prime}=\left(\frac{1}{4} \ln \left|\operatorname{th} \frac{x}{2}\right|-\frac{1}{4} \cdot \ln \frac{3+\operatorname{ch} x}{\operatorname{sh} x}\right)^{\prime}=$

$=\frac{1}{4} \cdot \frac{1}{\operatorname{th} \frac{x}{2}} \cdot \frac{1}{\operatorname{ch}^{2} \frac{x}{2}} \cdot \frac{1}{2}-\frac{1}{4} \cdot \frac{\operatorname{sh} x}{3+\operatorname{ch} x} \cdot \frac{\operatorname{sh} x \cdot \operatorname{sh} x-(3+\operatorname{ch} x) \cdot \operatorname{ch} x}{\operatorname{sh}^{2} x}=$

$=\frac{1}{4} \cdot \frac{1}{2 \operatorname{sh} \frac{x}{2} \cdot \operatorname{ch}^{2} \frac{x}{2}}-\frac{\operatorname{sh}^{2} x-3 \operatorname{ch} x-\operatorname{ch}^{2} x}{4 \cdot(3+\operatorname{ch} x) \cdot \operatorname{sh} x}=$

$=\frac{1}{4} \cdot \frac{1}{\operatorname{sh} x}-\frac{-1-3 \operatorname{ch} x}{4 \cdot(3+\operatorname{ch} x) \cdot \operatorname{sh} x}=\frac{1}{4 \operatorname{sh} x}+\frac{1}{4 \operatorname{sh} x}=\frac{1}{2 \operatorname{sh} x}$

Kuznetsov Differentiation 11-20