## Problem Statement

Find the derivative.

$y=\frac{\ln (\operatorname{ctg} x+\operatorname{ctg} \alpha)}{\sin \alpha}$
## Solution

$y^{\prime}=\left(\frac{\ln (\operatorname{ctg} x+\operatorname{ctg} \alpha)}{\sin \alpha}\right)^{\prime}=\frac{1}{\sin \alpha \cdot(\operatorname{ctg} x+\operatorname{ctg} \alpha)} \cdot \frac{-1}{\sin ^{2} x}=$
$=-\frac{1}{\sin \alpha \cdot \cos x+\cos \alpha \cdot \sin x}=-\frac{1}{\sin (\alpha+x)}$

## Problem Kuznetsov Differentiation 15-26