Differentiate $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ with respect to $x$. You should simplify your answer as a single fraction in its simplest form.

1. Apply the chain rule to differentiate the inverse tangent function.

Let $y = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$.
By the chain rule, $\frac{dy}{dx} = \frac{1}{1 + \left(\frac{x}{\sqrt{1-x^2}}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{\sqrt{1-x^2}}\right)$.

2. Differentiate the inner fractional expression using the quotient rule.

$\frac{d}{dx}\left(\frac{x}{\sqrt{1-x^2}}\right) = \frac{(1)\sqrt{1-x^2} - x \cdot \frac{d}{dx}\left(\sqrt{1-x^2}\right)}{\left(\sqrt{1-x^2}\right)^2}$
$= \frac{\sqrt{1-x^2} - x \left[\frac{1}{2}(1-x^2)^{-1/2}(-2x)\right]}{1-x^2}$

3. Simplify the derivative of the inner expression.

$= \frac{\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}}{1-x^2}$
$= \frac{\frac{1-x^2 + x^2}{\sqrt{1-x^2}}}{1-x^2}$
$= \frac{1}{(1-x^2)^{3/2}}$

4. Simplify the outer chain rule factor and multiply to get the final single fraction.

First, simplify the denominator of the outer derivative:
$1 + \left(\frac{x}{\sqrt{1-x^2}}\right)^2 = 1 + \frac{x^2}{1-x^2} = \frac{1-x^2+x^2}{1-x^2} = \frac{1}{1-x^2}$
Substitute back into the chain rule expression:
$\frac{dy}{dx} = \frac{1}{\left(\frac{1}{1-x^2}\right)} \cdot \frac{1}{(1-x^2)^{3/2}}$
$= (1-x^2) \cdot \frac{1}{(1-x^2)^{3/2}}$
$= \frac{1}{(1-x^2)^{1/2}}$
$= \frac{1}{\sqrt{1-x^2}}$