The slope of the normal to the curve $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

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

If the parametric equations of a curve are given by $x = \cos \theta + \log \tan \frac{\theta}{2}$ and $y = \sin \theta$,then the points for which $\frac{dy}{dx} = 0$ are given by

If $x = \frac{3at}{1 + t^3}$ and $y = \frac{3at^2}{1 + t^3}$,then $\frac{dy}{dx} = $

Let $x(t) = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y(t) = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$,$t \in (0, \frac{\pi}{2})$. Then $\frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$ at $t = \frac{\pi}{4}$ is equal to:

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to