## Task Condition

Find the derivative.

$y=\log _{4} \log _{2} \operatorname{tg} x$
## Solution

$y^{\prime}=\left(\log _{4} \log _{2} \operatorname{tg} x\right)^{\prime}=\frac{1}{\log _{2} \operatorname{tg} x \cdot \ln 4} \cdot\left(\log _{2} \operatorname{tg} x\right)^{\prime}=$

$=\frac{1}{\log _{2} \operatorname{tg} x \cdot \ln 4} \cdot \frac{1}{\operatorname{tg} x \cdot \ln 2} \cdot \frac{1}{\cos ^{2} x}=$

$$
\begin{aligned}
& =\frac{1}{2 \sin x \cdot \cos x \cdot \ln ^{2} 2 \cdot \log _{2} \operatorname{tg} x}= \\
& =\frac{1}{\sin 2 x \cdot \ln ^{2} 2 \cdot \frac{\ln \operatorname{tg} x}{\ln 2}}=\frac{1}{\sin 2 x \cdot \ln 2 \cdot \ln \operatorname{tg} x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 8-15