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:

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

The derivative of $\sin x$ with respect to $\log x$ is

The slope of the tangent to the curve $x = 3t^2 + 1, y = t^3 - 1$ at $x = 1$ is

If $x=2 \cos t-\cos 2 t$ and $y=2 \sin t-\sin 2 t$,then the value of $\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}$ is (in $/2$)

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