If $y = \sin^{-1}\left(\frac{\log x^2}{1+(\log x)^2}\right)$,then $\left(\frac{dy}{dx}\right)_{x=1} = $

If $y = \cot^{-1}(\cos 2x)^{1/2}$,then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is:

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

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 $0 < x < \frac{\pi}{2}$.

If $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} \right)$,then $\frac{dy}{dx}$ equals

If $f:R \to R$ is a differentiable function and $f(2) = 6$,then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{2t \, dt}{x - 2}$ is