If $y = \tan^{-1}\left[\dfrac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}}\right]$ , then $\dfrac{dy}{dx} =$

Let $x = \sin \theta$, which implies $\theta = \sin^{-1} x$.

Substituting this into the given expression, we get:

$y = \tan^{-1}\left[\dfrac{\sin \theta - \sqrt{1 - \sin^2 \theta}}{\sin \theta + \sqrt{1 - \sin^2 \theta}}\right]$

$y = \tan^{-1}\left[\dfrac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}\right]$

Dividing the numerator and the denominator by $\cos \theta$, we obtain:

$y = \tan^{-1}\left[\dfrac{\tan \theta - 1}{\tan \theta + 1}\right]$

Using the trigonometric identity $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ with $A = \theta$ and $B = \dfrac{\pi}{4}$:

$y = \tan^{-1}\left[\tan\left(\theta - \dfrac{\pi}{4}\right)\right]$

$y = \theta - \dfrac{\pi}{4}$

Substituting back $\theta = \sin^{-1} x$:

$y = \sin^{-1} x - \dfrac{\pi}{4}$

Differentiating with respect to $x$:

$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}} - 0$

$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}$

Answer: $\dfrac{1}{\sqrt{1 - x^2}}$