Derivative of $\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$ with respect to $\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ is

If $\cos x = \frac{1}{\sqrt{1 + t^2}}$ and $\sin y = \frac{t}{\sqrt{1 + t^2}}$,then $\frac{dy}{dx} = $

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

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 $x = \frac{1 + t}{t^3}$ and $y = \frac{3}{2t^2} + \frac{2}{t}$,then $x \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx}$ is equal to (where $t$ is a real parameter).

Let $y = t^{10} + 1$ and $x = t^8 + 1,$ then $\frac{d^2y}{dx^2}$ is