The rate of change of $\sqrt{x^2+16}$ with respect to $\frac{x}{x-1}$ at $x=5$ is

If $x = at^2$ and $y = 2at$,then find $\frac{d^2 x}{dy^2}$.

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:

If $x = a (t - 1/t)$ and $y = a (t + 1/t)$,where $t$ is the parameter,then $dy/dx = ?$

If $x = a \sin 2\theta (1 + \cos 2\theta )$ and $y = b \cos 2\theta (1 - \cos 2\theta )$,then $\frac{dy}{dx} = $

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