Answer
$$\frac{{dy}}{{dx}} = {\left( {\ln x} \right)^{x/2}}\left( {\frac{1}{{\ln x}} + \ln \left( {\ln x} \right)} \right)$$
Work Step by Step
$$\eqalign{
& y = 2{\left( {\ln x} \right)^{x/2}} \cr
& \frac{y}{2} = {\left( {\ln x} \right)^{x/2}} \cr
& {\text{Take the natural logarithm on both sides}} \cr
& \ln \frac{y}{2} = \ln {\left( {\ln x} \right)^{x/2}} \cr
& {\text{use the power property for logarithms}} \cr
& \ln \frac{y}{2} = \ln {\left( {\ln x} \right)^{x/2}} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{x}{2}\left( {\ln \left( {\ln x} \right)} \right)} \right) \cr
& {\text{use the product rule for derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{2}\frac{d}{{dx}}\left( {\ln \left( {\ln x} \right)} \right) + \ln \left( {\ln x} \right)\frac{d}{{dx}}\left( {\frac{x}{2}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{2}\left( {\frac{1}{{x\ln x}}} \right) + \ln \left( {\ln x} \right)\left( {\frac{1}{2}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2\ln x}} + \frac{{\ln \left( {\ln x} \right)}}{2} \cr
& {\text{solve the equation for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{1}{{2\ln x}} + \frac{{\ln \left( {\ln x} \right)}}{2}} \right) \cr
& {\text{replace }}y = 2{\left( {\ln x} \right)^{x/2}} \cr
& \frac{{dy}}{{dx}} = 2{\left( {\ln x} \right)^{x/2}}\left( {\frac{1}{{2\ln x}} + \frac{{\ln \left( {\ln x} \right)}}{2}} \right) \cr
& \frac{{dy}}{{dx}} = {\left( {\ln x} \right)^{x/2}}\left( {\frac{1}{{\ln x}} + \ln \left( {\ln x} \right)} \right) \cr} $$
