## Task Condition

Find the derivative.

$y=\ln \left(\sin \frac{2 x+4}{x+1}\right)$
## Solution

$y^{\prime}=\left(\ln \left(\sin \frac{2 x+4}{x+1}\right)\right)^{\prime}=\frac{1}{\sin \frac{2 x+4}{x+1}} \cdot\left(\sin \frac{2 x+4}{x+1}\right)^{\prime}=$

$=\frac{1}{\sin \frac{2 x+4}{x+1}} \cdot \cos \frac{2 x+4}{x+1} \cdot\left(\frac{2 x+4}{x+1}\right)^{\prime}=$

$=\operatorname{ctg} \frac{2 x+4}{x+1} \cdot \frac{2(x+1)-(2 x+4) \cdot 1}{(x+1)^{2}}=$

$=\operatorname{ctg} \frac{2 x+4}{x+1} \cdot \frac{2 x+2-2 x-4}{(x+1)^{2}}=-\frac{2}{(x+1)^{2}} \cdot \operatorname{ctg} \frac{2 x+4}{x+1}$