\(\ln x\) is defined for \(x>0\), so for \(\ln\dfrac{2x}{\sqrt{2x}}\) to be defined, you have to require \(\dfrac{2x}{\sqrt{2x}}>0\).
Since \(x
eq0\) in this interval, you can simplify the argument slightly:
\(\dfrac{2x}{\sqrt{2x}}=\dfrac{2x}{\sqrt2\sqrt x}=\sqrt2\sqrt x=\sqrt{2x}\)
This means \(f(x)\) will be defined whenever \(\sqrt{2x}>0\), and this happens for all positive \(x\).
Computing the derivative is an exercise in applying the chain rule:
\(\dfrac{\mathrm d}{\mathrm dx}f(x)=\dfrac{\dfrac{\mathrm d}{\mathrm dx}\sqrt{2x}}{\sqrt{2x}}=\dfrac{\dfrac{\sqrt2}{2\sqrt x}}{\sqrt{2x}}=\dfrac1{2x}\)
