## Task Condition

Find the derivative.

$$
y=\sqrt[7]{\tan(\cos 2)}+\frac{\sin ^{2} 27 x}{27 \cos 54 x}
$$
## Solution

$y^{\prime}=(\sqrt[7]{\operatorname{tg}(\cos 2)})^{\prime}+\left(\frac{\sin ^{2} 27 x}{27 \cos 54 x}\right)^{\prime}=0+\frac{1}{27}\left(\frac{\sin ^{2} 27 x}{\cos 54 x}\right)^{\prime}=\frac{1}{27}\left(\frac{\left(\sin ^{2} 27 x\right)^{\prime} \cos 54 x-\sin ^{2} 27 x(\cos 54 x)^{\prime}}{\cos 54 x}\right)=$

$=\frac{1}{27}\left(\frac{2 \sin 27 x \cdot \cos 27 x \cdot 27 \cdot \cos 54 x-\sin ^{2} 27 x \cdot(-\sin 54 x) \cdot 54}{\cos ^{2} 54 x}\right)=$

$=\frac{1}{27}\left(\frac{\sin 54 x \cdot 27 \cdot \cos 54 x+\sin ^{2} 27 x \cdot \sin 54 x \cdot 54}{\cos ^{2} 54 x}\right)=$
$=\frac{1}{27}\left(\frac{27 \sin 54 x \cdot\left(\cos 54 x+54 \sin ^{2} 27 x\right)}{\cos ^{2} 54 x}\right)=\frac{\sin 54 x \cdot\left(\cos ^{2} 27 x-\sin ^{2} 27 x+2 \sin ^{2} 27 x\right)}{\cos ^{2} 54 x}=$

$=\frac{\sin 54 x \cdot\left(\cos ^{2} 27 x+\sin ^{2} 27 x\right)}{\cos ^{2} 54 x}=\frac{\sin 54 x}{\cos ^{2} 54 x}=\sin 54 x \cdot \sec ^{2} 54 x=\operatorname{tg} 54 x \cdot \sec 54 x$

## Problem Kuznetsov Differentiation 9-27