## Task Condition

Find the derivative.

$y=\ln \arccos \sqrt{1-e^{4 x}}$
Solution

$$
\begin{aligned}
& y^{\prime}=\left(\ln \arccos \sqrt{1-e^{4 x}}\right)^{\prime}=\frac{1}{\arccos \sqrt{1-e^{4 x}}} \cdot \frac{-1}{\sqrt{1-\left(\sqrt{1-e^{4 x}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{1-e^{4 x}}} \cdot\left(-4 e^{4 x}\right)= \\
& =\frac{1}{\arccos \sqrt{1-e^{4 x}}} \cdot \frac{1}{\sqrt{e^{4 x}}} \cdot \frac{2 e^{4 x}}{\sqrt{1-e^{4 x}}}=\frac{2 e^{2 x}}{\sqrt{1-e^{4 x}} \cdot \arccos \sqrt{1-e^{4 x}}}
\end{aligned}
$$

Kuznetsov Differentiation 8-22