Differentiate the following function with respect to $x$:
$\sqrt{3x+2} + \frac{1}{\sqrt{2x^2+4}}$

Let $y = \sqrt{3x+2} + \frac{1}{\sqrt{2x^2+4}} = (3x+2)^{\frac{1}{2}} + (2x^2+4)^{-\frac{1}{2}}$.
Applying the chain rule,we differentiate each term with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[(3x+2)^{\frac{1}{2}}] + \frac{d}{dx}[(2x^2+4)^{-\frac{1}{2}}]$
$= \frac{1}{2}(3x+2)^{-\frac{1}{2}} \cdot \frac{d}{dx}(3x+2) + \left(-\frac{1}{2}\right)(2x^2+4)^{-\frac{3}{2}} \cdot \frac{d}{dx}(2x^2+4)$
$= \frac{1}{2}(3x+2)^{-\frac{1}{2}} \cdot (3) - \frac{1}{2}(2x^2+4)^{-\frac{3}{2}} \cdot (4x)$
$= \frac{3}{2\sqrt{3x+2}} - \frac{2x}{(2x^2+4)^{\frac{3}{2}}}$