If $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$,then $\frac{d u}{d v}$ at $x=0$ is

If $y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$,where $0 \leq x < \frac{\pi}{2}$,then find the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{6}$.

Let $0 < x < \pi$ and $y(x)$ be given by $(1+\sin x)y^3 - (\cos x)y^2 + 2(1+\sin x)y - 2\cos x = 0$. The derivative of $y$ with respect to $\tan \frac{x}{2}$ at $x = \frac{\pi}{2}$ is:

If $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ up to $10$ terms, then $f'(0) = $

If $a > b > 0$ and $x$ is acute,then $\frac{d}{dx} \left[ \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right) \right] = $

$y = \operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right)$,then $\frac{dy}{dx} = $