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

Let $y=f(x)=\sin ^3\left(\frac{\pi}{3}\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^3+5 x^2+1\right)^{\frac{3}{2}}\right)\right)$. Then,at $x =1$,

For $-\frac{\pi}{2} < x < \frac{3 \pi}{2}$, the value of $\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$ is equal to

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 $y = \frac{a^{\cos^{-1}x}}{1 + a^{\cos^{-1}x}}$ and $z = a^{\cos^{-1}x}$,then $\frac{dy}{dz} = $

If $f(x)=\cos ^{-1}\left[\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right]$,then $f^{\prime}(0.5)$ is equal to