1. The problem is to find the derivative of $\sqrt{2x}$ with respect to $x$. 2. Recall the derivative rule for a function of the form $f(x) = \sqrt{g(x)} = (g(x))^{1/2}$: $$\frac{d}{dx} (g(x))^{1/2} = \frac{1}{2\sqrt{g(x)}} \cdot g'(x)$$ 3. Here, $g(x) = 2x$. We first find $g'(x)$: $$g'(x) = \frac{d}{dx} (2x) = 2$$ 4. Substitute $g(x)$ and $g'(x)$ into the derivative formula: $$\frac{d}{dx} \sqrt{2x} = \frac{1}{2\sqrt{2x}} \cdot 2$$ 5. Simplify the expression: $$\frac{1}{2\sqrt{2x}} \cdot 2 = \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$$ 6. Therefore, the derivative of $\sqrt{2x}$ with respect to $x$ is: $$\boxed{\frac{1}{\sqrt{2x}}}$$