What is the slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $t = 2$?

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

If $x = a \cos \theta$ and $y = b \sin \theta$,then $\frac{d^3y}{dx^3}$ is equal to

If $x = \sin t$ and $y = \sin pt$,then the value of $(1 - x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x} + p^2 y =$

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) = $

$x=\cos \theta, y=\sin 5 \theta \Rightarrow (1-x^2) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)