If f(x) = ||2sin x - 1| - 2cot x| , then the value of f ^ '...

In the neighbourhood of $x = \dfrac{\pi}{3}$, we know,

\left(2sin x - 1\right) > 0$<br />$⇒ f\left(x\right) = \left|\left(2sin x - 1\right) - 2cot x\right|$<br />Now, for the neighbourhood of$x = \dfrac{\pi}{3}$<br />$2sin x - 1 - 2cot x = 2\left(\dfrac{\sqrt{3}}{2}\right) - 1 - \dfrac{2}{\left(\sqrt{3}\right)} = \dfrac{3 - 2}{\sqrt{3}} - 1$<br />$= \dfrac{1}{\sqrt{3}} - 1 < 0$<br />$∴ f\left(x\right) = \left(2cot x\right) - \left(2sin x - 1\right)$<br />Now,${f}^{'}\left(x\right) = 2\left(- cose{c}^{2}x\right) - 2cos x$<br />${f}^{'}\left(\dfrac{\pi}{3}\right) = 2{\left(- \dfrac{2}{\sqrt{3}}\right)}^{2} - 2\left(\dfrac{1}{2}\right)$<br />$= 2\dfrac{4}{3} - 1 = \dfrac{5}{3}$