[m]f`(\frac{π}{6})=\frac{2\cdot sin\frac{π}{6}-(2\cdot\frac{π}{6} -3)\cdot (cos\frac{π}{6})}{sin^2\frac{π}{6}}[/m]
[m]f`(\frac{π}{6})=\frac{2\cdot \frac{1}{2}-(\frac{π}{3} -3)\cdot (\frac{\sqrt{3}}{2})}{(\frac{1}{2})^2}[/m]
[m]f`(\frac{π}{6})=4-\frac{2\sqrt{3}}{3}(π-9)[/m]