For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then find $\frac{dy}{dx}$.

If $x = a \sec^{2} \theta$ and $y = a \tan^{2} \theta$,then find $\frac{d^{2} y}{d x^{2}}$.

$A$ curve is given by the equations $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$. The points for which $\frac{d^2y}{dx^2} = 0$ are given by:

Consider the curve represented parametrically by the equations $x = t^3 - 4t^2 - 3t$ and $y = 2t^2 + 3t - 5$,where $t \in \mathbb{R}$. If $H$ denotes the number of points on the curve where the tangent is horizontal and $V$ denotes the number of points where the tangent is vertical,then:

If $x = e^t \sin t$ and $y = e^t \cos t$,where $t$ is a parameter,then $\frac{d^2y}{dx^2}$ at $(1, 1)$ is equal to