## Task Condition

Find the derivative.

$y=\ln \left(e^{x}+\sqrt{e^{2 x}-1}\right)+\arcsin e^{-x}$
## Solution

$y^{\prime}=\left(\ln \left(e^{x}+\sqrt{e^{2 x}-1}\right)+\arcsin e^{-x}\right)^{\prime}=$

$=\frac{1}{e^{x}+\sqrt{e^{2 x}-1}} \cdot\left(e^{x}+\frac{1}{2 \sqrt{e^{2 x}-1}} \cdot 2 e^{2 x}\right)+\frac{1}{\sqrt{1-e^{-2 x}}} \cdot\left(-e^{-x}\right)=$

$$
\begin{aligned}
& =\frac{e^{x}}{e^{x}+\sqrt{e^{2 x}-1}} \cdot \frac{\sqrt{e^{2 x}-1}+e^{x}}{\sqrt{e^{2 x}-1}}-\frac{1}{e^{x} \cdot \sqrt{1-e^{-2 x}}}=\frac{e^{x}}{\sqrt{e^{2 x}-1}}-\frac{1}{\sqrt{e^{2 x}-1}}= \\
& =\frac{e^{x}-1}{\sqrt{\left(e^{x}-1\right) \cdot\left(e^{x}+1\right)}}=\sqrt{\frac{e^{x}-1}{e^{x}+1}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation $7-17$

This problem may have a different condition (possibly due to different editions or errors). See below.