If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=4t$ and $y=\frac{4}{t}$.

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

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

Differentiate $\sin ^2 x$ with respect to $\cos ^2 x$.

If $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$,then $\frac{dy}{dx}$ is equal to