If $f(x) = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$, then $f' \left( \frac{1}{2} \right) =$

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$,then $\frac{d y}{d x}=$

Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} x$ for $-1 < x < 1$ 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:

If $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$, then $\frac{d y}{d x}=$

If $y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)$,where $x \in \left(\frac{\pi}{2}, \pi\right)$,then the value of $\frac{dy}{dx}$ at $x = \frac{5\pi}{6}$ is: