[tex]y=\ln\mathrm{tg}(2x+1)[/tex]
[tex]y'=\dfrac{1}{\mathrm{tg}(2x+1)} \cdot\left(\mathrm{tg}(2x+1)\right)'=\dfrac{1}{\mathrm{tg}(2x+1)} \cdot\dfrac{1}{\cos^2(2x+1)} \cdot(2x+1)'=[/tex]
[tex]=\dfrac{1}{\dfrac{\sin(2x+1)}{\cos(2x+1)}\cdot \cos^2(2x+1)} \cdot 2=\dfrac{2}{\sin(2x+1)\cos(2x+1)}=[/tex]
[tex]=\dfrac{2\cdot2}{2\sin(2x+1)\cos(2x+1)}=\dfrac{4}{\sin(4x+2)}[/tex]