The derivative of $\sin^2 x$ with respect to $\cos^2 x$ is

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \cos \theta$ and $y = b \cos \theta$.

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{d^2y}{dx^2}$ is equal to

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.

If $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$ where $t \neq -1$ is a parameter,then find $\frac{dy}{dx}$.

The function $y(x)$ represented by $x=\sin t$,$y=a e^{t \sqrt{2}}+b e^{-t \sqrt{2}}$,$t \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ satisfies the equation $(1-x^2) y^{\prime \prime}-x y^{\prime}=k y$,then the value of $k$ is