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

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$,then $\frac{dy}{dx} = $

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

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

If $y = \sin^{-1}(\sqrt{1 - x^2})$,then $dy/dx = $

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