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

$\frac{3}{2\sqrt{3x+2}} - \frac{2x}{(2x^2+4)^{3/2}}$

The correct answer is Option (3) → $\frac{3}{2\sqrt{3x+2}} - \frac{2x}{(2x^2+4)^{3/2}}$ ##

Let $y = \sqrt{3x+2} + \frac{1}{\sqrt{2x^2+4}} = (3x+2)^{\frac{1}{2}} + (2x^2+4)^{-\frac{1}{2}}$

Note that this function is defined at all real numbers $x > -\frac{2}{3}$. Therefore

$\frac{dy}{dx} = \frac{1}{2}(3x+2)^{\frac{1}{2}-1} \cdot \frac{d}{dx}(3x+2) + \left(-\frac{1}{2}\right)(2x^2+4)^{-\frac{1}{2}-1} \cdot \frac{d}{dx}(2x^2+4)$

$= \frac{1}{2}(3x+2)^{-\frac{1}{2}} \cdot (3) - \left(\frac{1}{2}\right)(2x^2+4)^{-\frac{3}{2}} \cdot 4x$

$= \frac{3}{2\sqrt{3x+2}} - \frac{2x}{(2x^2+4)^{\frac{3}{2}}}$

This is defined for all real numbers $x > -\frac{2}{3}$.