## Problem Statement

Find the derivative.

$y=x \cdot \cos \alpha+\sin \alpha \cdot \ln \sin (x-\alpha)$
## Solution

$$
\begin{aligned}
& y^{\prime}=(x \cdot \cos \alpha+\sin \alpha \cdot \ln \sin (x-\alpha))^{\prime}=\cos \alpha+\sin \alpha \cdot \frac{1}{\sin (x-\alpha)} \cdot \cos (x-\alpha)= \\
& =\frac{1}{\sin (x-\alpha)} \cdot(\cos \alpha \cdot \sin (x-\alpha)+\sin \alpha \cdot \cos (x-\alpha))=\frac{1}{\sin (x-\alpha)} \cdot \sin (\alpha+(x-\alpha))= \\
& =\frac{\sin x}{\sin (x-\alpha)}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-2