## Task Condition

Find the derivative.

$$
y=\frac{\sin (\cos 3) \cdot \cos ^{2} 2 x}{4 \sin 4 x}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{\sin (\cos 3) \cdot \cos ^{2} 2 x}{4 \sin 4 x}\right)^{\prime}=\sin (\cos 3) \cdot\left(\frac{\cos ^{2} 2 x}{8 \sin 2 x \cdot \cos 2 x}\right)^{\prime}= \\
& =\sin (\cos 3) \cdot\left(\frac{\cos 2 x}{8 \sin 2 x}\right)^{\prime}=\frac{\sin (\cos 3)}{8} \cdot(\operatorname{ctg} 2 x)^{\prime}= \\
& =\frac{\sin (\cos 3)}{8} \cdot \frac{-1}{\sin ^{2} 2 x} \cdot 2=-\frac{\sin (\cos 3)}{4 \sin ^{2} 2 x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 9-6