If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

If $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$ where $\theta \in [0, 2 \pi]$,then find the value of $\frac{d^{2} y}{dx^{2}}$ at $\theta = \pi$.

If $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$,then $\frac{dy}{dx} = $

If $x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$,where $\theta$ is a parameter,then $\frac{dy}{dx}$ at $(1,1)$ is equal to

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 $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{d y}{d x}$:
$x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$