## Condition of the problem

Find the derivative.
$y=\operatorname{ctg}\left(\sin \frac{1}{13}\right)-\frac{1}{48} \cdot \frac{\cos ^{2} 24 x}{\sin 48 x}$
## Solution

$y^{\prime}=\left(\operatorname{ctg}\left(\sin \frac{1}{13}\right)-\frac{1}{48} \cdot \frac{\cos ^{2} 24 x}{\sin 48 x}\right)^{\prime}=0-\frac{1}{48} \cdot\left(\frac{\cos ^{2} 24 x}{\sin 48 x}\right)^{\prime}=$

$=-\frac{1}{48} \cdot \frac{\left(\cos ^{2} 24 x\right)^{\prime} \cdot \sin 48 x-\cos ^{2} 24 x \cdot(\sin 48 x)^{\prime}}{\sin ^{2} 48 x}=$

$=-\frac{1}{48} \cdot \frac{2 \cdot \cos 24 x \cdot(-\sin 24 x) \cdot 24 \cdot \sin 48 x-\cos ^{2} 24 x \cdot \cos 48 x \cdot 48}{\sin ^{2} 48 x}=$

$=\frac{\cos 24 x \cdot \sin 24 x \cdot \sin 48 x+\cos ^{2} 24 x \cdot \cos 48 x}{\sin ^{2} 48 x}=$

$=\frac{2 \cos ^{2} 24 x \cdot \sin ^{2} 24 x+\cos ^{2} 24 x \cdot \cos 48 x}{4 \sin ^{2} 24 x \cdot \cos ^{2} 24 x}=$

$=\frac{2 \sin ^{2} 24 x+\cos 48 x}{4 \sin ^{2} 24 x}=\frac{2 \sin ^{2} 24 x+1-2 \sin ^{2} 24 x}{4 \sin ^{2} 24 x}=$

$=\frac{1}{4 \sin ^{2} 24 x}$

## Problem Kuznetsov Differentiation 9-24