If $x = 2 \cos^3 \theta$ and $y = 3 \sin^2 \theta$,then $\frac{dy}{dx} = $

If the equation of the normal to the curve $x=\sqrt{t}$,$y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ equals

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $(x^2 + 4) \left(\frac{dy}{dx}\right)^2 =$

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$.

If $\cos (f(x))=\frac{1-x^2}{1+x^2}$ and $\tan (g(x))=\frac{3 x-x^3}{1-3 x^2}$,then $\frac{d f}{d g}=$