Condition of the problem

Find the derivative.

$y=\cos \ln 2-\frac{1}{3} \cdot \frac{\cos ^{2} 3 x}{\sin 6 x}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\cos \ln 2-\frac{1}{3} \cdot \frac{\cos ^{2} 3 x}{\sin 6 x}\right)^{\prime}=0-\frac{1}{3} \cdot\left(\frac{\cos ^{2} 3 x}{\sin 6 x}\right)^{\prime}= \\
& =-\frac{1}{3} \cdot\left(\frac{\cos ^{2} 3 x}{2 \sin 3 x \cdot \cos 3 x}\right)^{\prime}=-\frac{1}{6} \cdot\left(\frac{\cos 3 x}{\sin 3 x}\right)^{\prime}=
\end{aligned}
$$

$=-\frac{1}{6} \cdot(\operatorname{ctg} 3 x)^{\prime}=-\frac{1}{6} \cdot \frac{-1}{\sin ^{2} 3 x} \cdot 3=\frac{1}{2 \sin ^{2} 3 x}$

## Problem Kuznetsov Differentiation 9-2