## Task Condition

Find the derivative.

$$
y=\ln \cos \frac{2 x+3}{x+1}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\ln \cos \frac{2 x+3}{x+1}\right)^{\prime}=\frac{1}{\cos \frac{2 x+3}{x+1}} \cdot\left(\cos \frac{2 x+3}{x+1}\right)^{\prime}= \\
& =\frac{1}{\cos \frac{2 x+3}{x+1}} \cdot\left(-\sin \frac{2 x+3}{x+1}\right) \cdot\left(\frac{2 x+3}{x+1}\right)^{\prime}= \\
& =-\tan \frac{2 x+3}{x+1} \cdot \frac{2 \cdot(x+1)-(2 x+3) \cdot 1}{(x+1)^{2}}= \\
& =-\tan \frac{2 x+3}{x+1} \cdot \frac{2 x+2-2 x-3}{(x+1)^{2}}=\frac{\tan \frac{2 x+3}{x+1}}{(x+1)^{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 7-17(2)