If $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$,then $\frac{dy}{dx} = $ . . . . . .

The derivative of $\frac{1-x^2}{1+x^2}$ with respect to $\frac{2x}{1+x^2}$ at $x=2$ is

The slope of the normal to the curve $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

If $x = a(\cos \theta + \theta \sin \theta)$, $y = f(\theta)$, $f(2\pi) = 0$, $\frac{dy}{dx} = \frac{\tan \theta}{\theta}$, $\theta \neq 0$ and $\theta \neq (2n+1)\frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right) = $

If $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$,then $y^{3} \frac{d^{2} y}{d x^{2}} + x^{2} + y^{2} = $

If $x=\sec \theta-\cos \theta$,$y=\sec ^{10} \theta-\cos ^{10} \theta$ and $(x^2+4)(\frac{dy}{dx})^2=k(y^2+4)$,then the value of $k$ is