$\frac{d}{dx}\left( \tan^{-1}\sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right)$ is equal to

If $y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$,where $0 \leq x < \frac{\pi}{2}$,then find the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{6}$.

The derivative of $\operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is

$\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left( \frac{\int_{x^3}^{(\pi / 2)^3} (\sin (2 t^{1 / 3}) + \cos (t^{1 / 3})) dt}{(x - \frac{\pi}{2})^2} \right)$ is equal to:

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

Find the derivative of ${\sec ^{ - 1}}\left( {\frac{1}{{2{x^2} - 1}}} \right)$ with respect to $\sqrt {1 + 3x} $ at $x = - \frac{1}{3}$.