## Task Condition

Find the derivative.

$$
y=\frac{\operatorname{tg}(\ln 2) \cdot \sin ^{2} 19 x}{19 \cos 38 x}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{\tan(\ln 2) \cdot \sin ^{2} 19 x}{19 \cos 38 x}\right)^{\prime}=\frac{\tan(\ln 2)}{19} \cdot\left(\frac{\sin ^{2} 19 x}{\cos 38 x}\right)^{\prime}= \\
& =\frac{\tan(\ln 2)}{19} \cdot\left(\frac{\sin ^{2} 19 x}{\cos 38 x}\right)^{\prime}= \\
& =\frac{\tan(\ln 2)}{19} \cdot \frac{2 \sin 19 x \cdot \cos 19 x \cdot 19 \cdot \cos 38 x-\sin ^{2} 19 x \cdot(-\sin 38 x) \cdot 38}{\cos ^{2} 38 x}= \\
& =\tan(\ln 2) \cdot \frac{\sin 38 x \cdot \cos 38 x+2 \sin ^{2} 19 x \cdot \sin 38 x}{\cos ^{2} 38 x}= \\
& =\tan(\ln 2) \cdot \frac{\sin 38 x \cdot\left(\cos 38 x+2 \sin ^{2} 19 x\right)}{\cos ^{2} 38 x}= \\
& =\tan(\ln 2) \cdot \frac{\sin 38 x \cdot\left(1-2 \sin ^{2} 19 x+2 \sin ^{2} 19 x\right)}{\cos ^{2} 38 x}= \\
& =\tan(\ln 2) \cdot \frac{\sin 38 x}{\cos ^{2} 38 x}=\frac{\tan(\ln 2)^{2} \cdot \tan 38 x}{\cos 38 x}
\end{aligned}
$$

Problem Kuznetsov Differentiation 9-19