Condition of the problem

Find the derivative.

$$
y=\frac{\cos \ln 7 \cdot \sin ^{2} 7 x}{7 \cos 14 x}
$$
## Solution

$y^{\prime}=\left(\frac{\cos \ln 7 \cdot \sin ^{2} 7 x}{7 \cos 14 x}\right)^{\prime}=\frac{\cos \ln 7}{7} \cdot\left(\frac{\sin ^{2} 7 x}{\cos 14 x}\right)^{\prime}=$

$=\frac{\cos \ln 7}{7} \cdot \frac{2 \sin 7 x \cdot \cos 7 x \cdot 7 \cdot \cos 14 x-\sin ^{2} 7 x \cdot(-\sin 14 x) \cdot 14}{\cos ^{2} 14 x}=$

$=\cos \ln 7 \cdot \frac{\sin 14 x \cdot \cos 14 x+2 \sin ^{2} 7 x \cdot \sin 14 x}{\cos ^{2} 14 x}=$

$=\cos \ln 7 \cdot \frac{\sin 14 x \cdot\left(\cos 14 x+2 \sin ^{2} 7 x\right)}{\cos ^{2} 14 x}=$

$=\cos \ln 7 \cdot \frac{\sin 14 x \cdot\left(1-2 \sin ^{2} 7 x+2 \sin ^{2} 7 x\right)}{\sin 14 x \quad \begin{array}{c}\cos ^{2} 14 x \\ \cos \ln 7 \cdot \operatorname{tg} 14 x\end{array}}=$

$=\cos \ln 7 \cdot \frac{\sin 14 x}{\cos ^{2} 14 x}=\frac{\cos \ln 7 \cdot \operatorname{tg} 14 x}{\cos 14 x}$

## Problem Kuznetsov Differentiation 9-7