## Task Condition

Find the derivative.

$$
y=\frac{\cos \left(\tan \frac{1}{3}\right) \cdot \sin ^{2} 15 x}{15 \cos 30 x}
$$
## Solution

$y^{\prime}=\left(\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right) \cdot \sin ^{2} 15 x}{15 \cos 30 x}\right)^{\prime}=$

$=\frac{\cos \left(\operatorname{tg}_{\frac{1}{3}}\right)}{15} \cdot \frac{2 \sin 15 x \cdot \cos 15 x \cdot 15 \cdot \cos 30 x-\sin ^{2} 15 x \cdot(-\sin 30 x) \cdot 30}{\cos ^{2} 30 x}=$

$=\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right)}{15} \cdot \frac{15 \sin 30 x \cdot \cos 30 x+30 \sin ^{2} 15 x \cdot \sin 30 x}{\cos ^{2} 30 x}=$

$=\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right)}{15} \cdot \frac{15 \sin 30 x \cdot\left(\cos 30 x+2 \sin ^{2} 15 x\right)}{\cos ^{2} 30 x}=$

$=\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right)}{15} \cdot \frac{15 \sin 30 x \cdot\left(1-2 \sin ^{2} 15 x+2 \sin ^{2} 15 x\right)}{\cos ^{2} 30 x}=$

$=\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right)}{15} \cdot \frac{15 \sin 30 x}{\cos ^{2} 30 x}=\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right) \cdot \operatorname{tg} 30 x}{\cos 30 x}$

## Problem Kuznetsov Differentiation 9-15