Choose the correct alternative in the following:

$\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{\sqrt{1-\left[(\cos 2 \mathrm{x})^{\frac{1}{2}}\right]^{2}}} \cdot \frac{1}{2 \sqrt{\cos 2 \mathrm{x}}} \cdot(-\sin 2 \mathrm{x}) \cdot 2$

$=\frac{1}{\sqrt{1-\cos 2 x}} \cdot \frac{1}{\sqrt{\cos 2 x}} \cdot(\sin 2 x)$

Putting $x=\pi / 6$, we get

$=\frac{1}{\sqrt{1-\left(\cos \frac{2 \pi}{6}\right)}} \cdot \frac{1}{\sqrt{\cos \frac{2 \pi}{6}}} \cdot\left(\sin \frac{2 \pi}{6}\right)$

$=\frac{1}{\sqrt{1-\cos \frac{\pi}{3}}} \cdot \frac{1}{\sqrt{\cos \frac{\pi}{3}}} \cdot\left(\sin \frac{\pi}{3}\right)$

$=\frac{1}{\sqrt{1-\frac{1}{2}} \cdot \frac{1}{\sqrt{\frac{1}{2}}} \cdot\left(\frac{\sqrt{3}}{2}\right)}$

Simplifying above we get

$=\sqrt{2} \cdot \sqrt{2} \cdot\left(\frac{\sqrt{3}}{2}\right)$

$=\sqrt{2} \cdot \sqrt{2} \cdot\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}$

$\therefore f^{\prime}(x)=\sqrt{3}=(3)^{1 / 2}$