If $\sin x = \frac{2t}{1+t^2}$ and $\tan y = \frac{2t}{1-t^2}$,then $\frac{dy}{dx}$ is equal to

If $x = a \cos \theta$ and $y = a \sin \theta$,then $\frac{d^2 y}{d x^2} =$ . . . . . . . (where $a \neq 0$ and $\theta \neq k \pi, k \in Z$)

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$,then $\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=$

If $x = a \sec^{2} \theta$ and $y = a \tan^{2} \theta$,then find $\frac{d^{2} y}{d x^{2}}$.

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$

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: