## Task Condition

Find the derivative.

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

$y^{\prime}=\left(\ln \cos \frac{2 x+3}{2 x+1}\right)^{\prime}=\frac{1}{\cos \frac{2 x+3}{2 x+1}} \cdot\left(\cos \frac{2 x+3}{2 x+1}\right)^{\prime}=$

$=\frac{1}{\cos \frac{2 x+3}{2 x+1}} \cdot\left(-\sin \frac{2 x+3}{2 x+1}\right) \cdot\left(\frac{2 x+3}{2 x+1}\right)^{\prime}=$

$=-\operatorname{tg} \frac{2 x+3}{2 x+1} \cdot\left(1+\frac{2}{2 x+1}\right)^{\prime}=-\operatorname{tg} \frac{2 x+3}{2 x+1} \cdot\left(-\frac{2}{(2 x+1)^{2}} \cdot 2\right)=$
$=\operatorname{tg} \frac{2 x+3}{2 x+1} \cdot \frac{4}{(2 x+1)^{2}}=\frac{4 \operatorname{tg} \frac{2 x+3}{2 x+1}}{(2 x+1)^{2}}$

## Problem Kuznetsov Differentiation 8-17