If $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$,then $\frac{dy}{dx} = $ . . . . . .

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

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

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

If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

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