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 $x = at^2$ and $y = 2at$,then find $\frac{d^2 x}{dy^2}$.

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

If $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$,then $t^6\left(\frac{d u}{d t}\right)^2 \div \left(e^{2 t}(t-2)^2\right)=$

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then:

$A$ 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: