Answer
$$y'=\Big(\ln x\ln(\ln x)+1\Big)(\ln x)^{\frac{x}{2}-1}$$
Work Step by Step
$$y=2(\ln x)^{x/2}$$
Take the natural logarithm of both sides:
$$\ln y =\ln\Big(2(\ln x)^{x/2}\Big)=\ln2+\ln\Big((\ln x)^{x/2}\Big)$$ $$\ln y=\ln2+\frac{x}{2}\ln(\ln x)$$
Now using implicit differentiation and differentiate both sides with respect to $x$, we can find the derivative $y'$: $$\frac{y'}{y}=0+\Big(\frac{x}{2}\Big)'\ln(\ln x)+\Big(\frac{x}{2}\Big)\Big(\ln(\ln x)\Big)'$$ $$\frac{y'}{y}=\frac{\ln(\ln x)}{2}+\frac{x}{2}\times\frac{1}{\ln x}(\ln x)'$$ $$\frac{y'}{y}=\frac{\ln(\ln x)}{2}+\frac{x}{2}\times\frac{1}{\ln x}\times\frac{1}{x}$$ $$\frac{y'}{y}=\frac{\ln(\ln x)}{2}+\frac{1}{2\ln x}=\frac{\ln x\ln(\ln x)+1}{2\ln x}$$
Now replace $y=2(\ln x)^{x/2}$ and calculate $y'$: $$\frac{y'}{2(\ln x)^{x/2}}=\frac{\ln x\ln(\ln x)+1}{2\ln x}$$ $$y'=\frac{\Big(\ln x\ln(\ln x)+1\Big)\times2(\ln x)^{x/2}}{2\ln x}$$ $$y'=\Big(\ln x\ln(\ln x)+1\Big)(\ln x)^{\frac{x}{2}-1}$$
