If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = \sin t, y = \cos 2t$.

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. Its acceleration at time $t$ is:

Let $t \in (0, 1)$ and $\alpha \in (0, \frac{\pi}{4})$. If $x = \text{cosec}^{-1} \left( \frac{1 + t^2}{2t} \right)$, $y = \cot^{-1} \left( \frac{\sqrt{1 - t^2}}{t} \right)$ and $\frac{dy}{dx} = f(t)$, then the value of $f(\tan \alpha)$ is

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 $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$,then $\frac{dy}{dx} = $

If $x=t-\sin t, y=1-\cos t$ and $\frac{d^2 y}{d x^2}=-1$ at $t=K, K>0$, then $\lim_{t \rightarrow K} \frac{y}{x}=$