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

If $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ up to $n$ terms,then $\frac{dy}{dx}$ is equal to

If $f(x)=e^x$,$g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$,then $\frac{h^{\prime}(x)}{h(x)}$ is

If $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$,then $\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} = $

The derivative of $\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]$ with respect to $\tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]$ is