Condition of the problem

Find the differential $d y$.

$y=\operatorname{arctg}(\operatorname{sh} x)+(\operatorname{sh} x) \ln (\operatorname{ch} x)$
## Solution

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

## Problem Kuznetsov Differentiation 4-7