## Task Condition

Find the derivative.

$$
y=\sin \sqrt{3}+\frac{1}{3} \cdot \frac{\sin ^{2} 3 x}{\cos 6 x}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\sin \sqrt{3}+\frac{1}{3} \cdot \frac{\sin ^{2} 3 x}{\cos 6 x}\right)^{\prime}=\frac{1}{3} \cdot\left(\frac{\sin ^{2} 3 x}{\cos 6 x}\right)^{\prime}= \\
& =\frac{1}{3} \cdot \frac{2 \sin 3 x \cdot \cos 3 x \cdot 3 \cdot \cos 6 x-\sin ^{2} 3 x \cdot(-\sin 6 x \cdot 6)}{\cos ^{2} 6 x}= \\
& =\frac{\sin 6 x \cdot \cos 6 x+2 \sin ^{2} 3 x \cdot \sin 6 x}{\cos ^{2} 6 x}=\frac{\sin 6 x \cdot\left(\cos 6 x+2 \sin ^{2} 3 x\right)}{\cos ^{2} 6 x}= \\
& =\frac{\sin 6 x \cdot\left(1-2 \sin ^{2} 3 x+2 \sin ^{2} 3 x\right)}{\cos ^{2} 6 x}=\frac{\sin 6 x}{\cos ^{2} 6 x}=\frac{\tan 6 x}{\cos 6 x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 9-1

Find the derivative.

$y=\operatorname{arctg} \frac{\tan x-\cot x}{\sqrt{2}}$

Solution

$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\tan x-\cot x}{\sqrt{2}}\right)^{\prime}=\frac{1}{1+\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)^{2}} \cdot\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)^{\prime}= \\
& =\frac{2}{2+\tan^{2} x-2 \tan x \cdot \cot x+\cot^{2} x} \cdot \frac{1}{\sqrt{2}} \cdot\left(\frac{1}{\cos ^{2} x}-\frac{-1}{\sin ^{2} x}\right)= \\
& =\frac{\sqrt{2}}{2+\tan^{2} x-2+\cot^{2} x} \cdot\left(\frac{\sin ^{2} x}{\cos ^{2} x \cdot \sin ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x}\right)= \\
& =\frac{\sqrt{2}}{\tan^{2} x+\cot^{2} x} \cdot \frac{\sin ^{2} x+\cos ^{2} x}{\cos ^{2} x \cdot \sin ^{2} x}= \\
& =\frac{\sqrt{2}}{\frac{\sin ^{2} x}{\cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x}} \cdot \frac{1}{\cos ^{2} x \cdot \sin ^{2} x}=\frac{\sqrt{2}}{\sin ^{4} x+\cos ^{4} x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 10-1