Answer
$$f'\left( x \right) = \frac{1}{{2x}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \sqrt {10x} \cr
& {\text{rewriting the radical by the property }}\sqrt u = {u^{1/2}} \cr
& f\left( x \right) = \ln {\left( {10x} \right)^{1/2}} \cr
& {\text{use the power property for logarithms}} \cr
& f\left( x \right) = \frac{1}{2}\ln \left( {10x} \right) \cr
& {\text{use the product property for logarithms}} \cr
& f\left( x \right) = \frac{1}{2}\left[ {\ln 10 + \ln x} \right] \cr
& f\left( x \right) = \frac{1}{2}\ln 10 + \frac{1}{2}\ln x \cr
& {\text{differentiate both sides}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln 10} \right] + \frac{d}{{dx}}\left[ {\frac{1}{2}\ln x} \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln 10} \right] + \frac{1}{2}\frac{d}{{dx}}\left[ {\ln x} \right] \cr
& f'\left( x \right) = 0 + \frac{1}{2}\left( {\frac{1}{x}} \right) \cr
& f'\left( x \right) = \frac{1}{{2x}} \cr} $$
