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

The derivative of $\log_8(\log_5 x)$ w. r. t. $x$ is

If $y^{\frac{1}{m}} + y^{\frac{-1}{m}} = 2x$ , then $(x^2 - 1)y_1^{\ 2} =$

If $y = \tan^{-1}\sqrt{\dfrac{1 + \sin 2x}{1 - \sin 2x}}$, then $\dfrac{\text{d}y}{\text{d}x}$ at $x = \dfrac{\pi}{6}$ is