## Task Condition

Find the derivative.

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

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

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

$$
\begin{aligned}
& =\frac{1}{\arcsin \sqrt{1-e^{2 x}}} \cdot \frac{1}{\sqrt{1-1+e^{2 x}}} \cdot \frac{-e^{2 x}}{\sqrt{1-e^{2 x}}}= \\
& =\frac{1}{\arcsin \sqrt{1-e^{2 x}}} \cdot \frac{1}{\sqrt{e^{2 x}}} \cdot \frac{-e^{2 x}}{\sqrt{1-e^{2 x}}}=-\frac{e^{x}}{\sqrt{1-e^{2 x}} \cdot \arcsin \sqrt{1-e^{2 x}}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 8-21