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

If $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ up to $10$ terms, then $f'(0) = $

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$, then the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

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} = $

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

The derivative of ${\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ with respect to $\frac{x}{2}$,where $x \in \left( {0, \frac{\pi }{2}} \right)$,is