For $-\frac{\pi}{2} < x < \frac{3 \pi}{2}$,the value of $\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$ is equal to

Differentiation of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right)$ is

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:

Differentiation of $\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ is $....$

$\frac{d}{dx} \left\{ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right\} = $

If $y = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$ for $0 < |x| < 1$,then $\frac{dy}{dx} = $