The slope of the normal to the curve $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

The equation of the normal to the curve $x = \theta + \sin \theta, y = 1 + \cos \theta$ at $\theta = \frac{\pi}{2}$ is

The derivative of $f(\tan x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$,where $f^{\prime}(1) = 2$ and $g^{\prime}(\sqrt{2}) = 4$,is:

Let $x(t) = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y(t) = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$,$t \in (0, \frac{\pi}{2})$. Then $\frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$ at $t = \frac{\pi}{4}$ is equal to:

The equation of the tangent to the curve $x = \frac{t - 1}{t + 1}, y = \frac{t + 1}{t - 1}$ at $t = 2$ is:

If $y=4 \cos ^3(t)$ and $x=4 \sin ^3(t)$,then $\frac{d y}{d x}$ is equal to