If $x = \log p$ and $y = \frac{1}{p}$, then $\frac{dy}{dx} = $

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 $\theta$ is the angle made by the normal drawn to the curve $x=e^{t} \cos t, y=e^{t} \sin t$ at the point $(1,0)$,with the $X$-axis,then $\theta=$

Find the slope of the normal to the curve $x = 1 - a \sin \theta$,$y = b \cos^2 \theta$ at $\theta = \pi / 2$.

If $x = at^2$ and $y = 2at$, then $\frac{d^2y}{dx^2} = \dots$

If $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$ where $\theta \in [0, 2 \pi]$,then find the value of $\frac{d^{2} y}{dx^{2}}$ at $\theta = \pi$.