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

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

$\frac{d}{dx} \left( \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}} \right) = $

If $y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right)$,where $x^2 \le 1$. Then find $\frac{dy}{dx}$.

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} = $

Let $f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$,where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is