If $x=\sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y=\sqrt{2^{\sec ^{-1} t}}, |t| \geq 1$,then $\frac{d y}{d x}=$

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$,then the value of $\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}}$ is

If $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$,then $t^6\left(\frac{d u}{d t}\right)^2 \div \left(e^{2 t}(t-2)^2\right)=$

The equation of the normal to the curve $x = \theta + \sin \theta, y = 1 + \cos \theta$ at $\theta = \frac{\pi}{2}$ is

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then ${\left( {\frac{{dy}}{{dx}}} \right)^2}$ equals

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