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

If $y = \sin^{-1} \left( \frac{25 - x^2}{25 + x^2} \right)$, then find the value of $y'(1)$.

If $f$ is differentiable in $(0, 6)$ and $f'(4) = 5$,then $\lim_{x \to 2} \frac{f(4) - f(x^2)}{2 - x} = $

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 $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$,then $\frac{dy}{dx}$ is equal to

Differentiate the function with respect to $x$: $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$,where $0 < x < \frac{\pi}{2}$.