If $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$,then $\left(\frac{d y}{d x}\right)$ at $x=0$ is

The derivative of $\sin ^{-1}\left(3 x-4 x^3\right)$ with respect to $x$ for $\frac{1}{2} < x < 1$ is:

If $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$, then $\frac{d y}{d x}=$

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:

The value of $\frac{d}{d x}\left[\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

If $y = \tan^{-1}\left( \frac{x}{1 + \sqrt{1 - x^2}} \right)$,then $\frac{dy}{dx} = $