If $f\left(x\right)=\left|\left|2\sin x-1\right|-2\cot x\right|$, then the value of $f^{\text{'}}\left(\frac{\pi }{3}\right)$ is equal to

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

$\left(2\sin x-1\right)>0$
$\Rightarrow f\left(x\right)=\left|\left(2\sin x-1\right)-2\cot x\right|$
Now, for the neighbourhood of $x=\frac{\pi }{3}$
$2\sin x-1-2\cot x=2\left(\frac{\sqrt{3}}{2}\right)-1-\frac{2}{\left(\sqrt{3}\right)}=\frac{3-2}{\sqrt{3}}-1$
$=\frac{1}{\sqrt{3}}-1<0$
$\therefore f\left(x\right)=\left(2\cot x\right)-\left(2\sin x-1\right)$
Now, $f^{\text{'}}\left(x\right)=2\left(-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^{2}x\right)-2\cos x$
$f^{\text{'}}\left(\frac{\pi }{3}\right)=2{\left(-\frac{2}{\sqrt{3}}\right)}^2-2\left(\frac{1}{2}\right)$
$=2\frac{4}{3}-1=\frac{5}{3}$