Differentiate each of the following expressions with respect to $x$.

1. Apply the derivative formula for the inverse tangent function using the chain rule.

Let $y = \tan^{-1} \sqrt{2 + x^3}$.
Using the chain rule,
$\frac{dy}{dx} = \frac{1}{1 + (\sqrt{2 + x^3})^2} \cdot \frac{d}{dx} \left( \sqrt{2 + x^3} \right)$

2. Differentiate the inner square root expression.

$\frac{d}{dx} \left( (2 + x^3)^{\frac{1}{2}} \right) = \frac{1}{2}(2 + x^3)^{-\frac{1}{2}} \cdot (3x^2)$
$= \frac{3x^2}{2\sqrt{2 + x^3}}$

3. Combine and simplify to obtain the final derivative.

$\frac{dy}{dx} = \frac{1}{1 + 2 + x^3} \cdot \frac{3x^2}{2\sqrt{2 + x^3}}$
$= \frac{1}{3 + x^3} \cdot \frac{3x^2}{2\sqrt{2 + x^3}}$
$= \frac{3x^2}{2(3 + x^3)\sqrt{2 + x^3}}$

1. Use logarithm properties to simplify the expression before differentiating.

$\begin{aligned} \ln \left( \frac{x}{\sqrt{1 - x}} \right) &= \ln x - \ln(\sqrt{1 - x}) \\ &= \ln x - \ln((1 - x)^{\frac{1}{2}}) \\ &= \ln x - \frac{1}{2} \ln(1 - x) \end{aligned}$

2. Differentiate the simplified expression term by term with respect to $x$.

$\begin{aligned} \frac{d}{dx} \left[ \ln x - \frac{1}{2} \ln(1 - x) \right] &= \frac{1}{x} - \frac{1}{2} \left( \frac{1}{1 - x} \right) \cdot \frac{d}{dx}(1 - x) \\ &= \frac{1}{x} - \frac{1}{2} \left( \frac{1}{1 - x} \right) \cdot (-1) \\ &= \frac{1}{x} + \frac{1}{2(1 - x)} \end{aligned}$

3. Combine the terms into a single algebraic fraction.

$\begin{aligned} \frac{1}{x} + \frac{1}{2(1 - x)} &= \frac{2(1 - x) + x}{2x(1 - x)} \\ &= \frac{2 - 2x + x}{2x(1 - x)} \\ &= \frac{2 - x}{2x(1 - x)} \end{aligned}$