## Task Condition

Find the derivative.
$y=\frac{e^{2 x}(2-\sin 2 x-\cos 2 x)}{8}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{e^{2 x}(2-\sin 2 x-\cos 2 x)}{8}\right)^{\prime}= \\
& =\frac{1}{8} \cdot\left(2 e^{2 x}(2-\sin 2 x-\cos 2 x)+e^{2 x}(-\cos 2 x \cdot 2+\sin 2 x \cdot 2)\right)= \\
& =\frac{1}{4} \cdot e^{2 x} \cdot(2-\sin 2 x-\cos 2 x-\cos 2 x+\sin 2 x)= \\
& =\frac{1}{2} \cdot e^{2 x} \cdot(1-\cos 2 x)=\frac{1}{2} \cdot e^{2 x} \cdot\left(1-\left(1-2 \sin ^{2} x\right)\right)=e^{2 x} \cdot \sin ^{2} x
\end{aligned}
$$

## Problem Kuznetsov Differentiation 7-2