Condition of the problem

Find the derivative.

$y=\frac{1}{2} \cdot \ln \left(\epsilon^{2 x}+1\right)-2 \operatorname{arctg} e^{x}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot \ln \left(\epsilon^{2 x}+1\right)-2 \operatorname{arctg} e^{x}\right)^{\prime}=\frac{1}{2} \cdot \frac{1}{\varepsilon^{2 x}+1} \cdot 2 \epsilon^{2 x}-2 \cdot \frac{1}{1+\epsilon^{2 x}} \cdot e^{x}= \\
& =\frac{\epsilon^{2 x}}{1+\epsilon^{2 x}}-\frac{2 e^{x}}{1+\epsilon^{2 x}}=\frac{\epsilon^{2 x}-\epsilon^{x}}{1+\epsilon^{2 x}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 7-7