If $y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), \quad 0 \leqslant x < \frac{\pi}{2}$, then $y^{\prime}\left(\frac{\pi}{6}\right)=$

Simplify the expression

$y = \tan^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}$

using half-angle identities. Observe that

$\frac{1+\sin x}{1-\sin x} = \frac{1+\cos\left(\frac{\pi}{2}-x\right)}{1-\cos\left(\frac{\pi}{2}-x\right)} = \frac{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} = \cot^2\left(\frac{\pi}{4}-\frac{x}{2}\right).$

Since $0 \leq x < \frac{\pi}{2}$, $\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)$ is positive, so the square root yields
$\sqrt{\cot^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} = \cot\left(\frac{\pi}{4}-\frac{x}{2}\right).$

Apply the identity $\cot\theta = \tan\left(\frac{\pi}{2}-\theta\right)$ to obtain

$y = \tan^{-1}\left[\tan\left(\frac{\pi}{2} - \left(\frac{\pi}{4}-\frac{x}{2}\right)\right)\right] = \tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right].$

For $0\leq x < \frac{\pi}{2}$, $\frac{\pi}{4} \leq \frac{\pi}{4}+\frac{x}{2} < \frac{\pi}{2}$, and since $\tan^{-1}(\tan\theta) = \theta$ in this interval
$y = \frac{\pi}{4} + \frac{x}{2}.$

Differentiate to find
$y' = \frac{1}{2}.$

Evaluate at $x = \frac{\pi}{6}$:
$y'\left(\frac{\pi}{6}\right) = \frac{1}{2}.$

$\boxed{\frac{1}{2}}$