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

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

If $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$ for $|t| \ge 1$,then $\frac{dy}{dx}$ is equal to:

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$.

Find the equations of the tangent and normal to the given curve at the indicated point: $x = \cos t, y = \sin t$ at $t = \frac{\pi}{4}$.

For $x \neq -1, y \neq -1$, if $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$, then $\frac{dx}{dy} =$