If $y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$,then $\frac{dy}{dx} = \ldots$

If $y = \tan^{-1} \left\{ \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right\}$, where $|x| < 1$, then $\frac{dy}{dx}$ is equal to

If $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$,then $\frac{dy}{dx} = $

If $y = \tan^{-1} \left[ \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} \right]$,then $\frac{dy}{dx} = $

$\begin{aligned} & \text{If } y = \tan^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\} \\ & + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\} \text{, then } \frac{dy}{dx} = \end{aligned}$

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{x}(3 - x)}{1 - 3x} \right) \right] =$