## Task Condition

Find the derivative.

$y=\operatorname{ctg}(\cos 2)+\frac{1}{6} \cdot \frac{\sin ^{2} 6 x}{\cos 12 x}$
## Solution

$y^{\prime}=\left(\operatorname{ctg}(\cos 2)+\frac{1}{6} \cdot \frac{\sin ^{2} 6 x}{\cos 12 x}\right)^{\prime}=0+\frac{1}{6} \cdot\left(\frac{\sin ^{-2} 6 x}{\cos 12 x}\right)^{\prime}=$

$=\frac{1}{6} \cdot \frac{\left(\sin ^{2} 6 x\right)^{\prime} \cdot \cos 12 x-\sin ^{2} 6 x \cdot(\cos 12 x)^{\prime}}{\cos ^{2} 12 x}=$

$=\frac{1}{6} \cdot \frac{2 \sin 6 x \cdot(\sin 6 x)^{\prime} \cdot \cos 12 x-\sin ^{2} 6 x \cdot(-\sin 12 x) \cdot 12}{\cos ^{2} 12 x}=$

$$
\begin{aligned}
& =\frac{1}{6} \cdot \frac{2 \sin 6 x \cdot \cos 6 x \cdot 6 \cdot \cos 12 x+12 \sin ^{2} 6 x \cdot \sin 12 x}{\cos ^{2} 12 x}= \\
& =\frac{1}{6} \cdot \frac{6 \sin 12 x \cdot \cos 12 x+12 \sin ^{2} 6 x \cdot \sin 12 x}{\cos ^{2} 12 x}= \\
& =\frac{\sin 12 x \cdot\left(\cos 12 x+2 \sin ^{2} 6 x\right)}{\cos ^{2} 12 x}=\frac{\sin 12 x \cdot\left(1-2 \sin ^{2} 6 x+2 \sin ^{2} 6 x\right)}{\cos ^{2} 12 x}= \\
& =\frac{\sin 12 x}{\cos ^{2} 12 x}=\frac{\operatorname{tg} 12 x}{\cos 12 x}
\end{aligned}
$$

Problem Kuznetsov Differentiation 9-9