Would it be $\displaystyle 0(2^xln2)$ or $\displaystyle 2(2^xln2)$
It is so confusing....
did you mean to type $\displaystyle 2^{2^x \ln 2}$ ?? if so, you need the chain rule. if you actually meant $\displaystyle 2(2^x \ln 2)$, then no chain rule necessary. just note that $\displaystyle 2 \ln 2$ is a constant, and so we just have to worry about the derivative of $\displaystyle 2^x$, hence
$\displaystyle \frac d{dx} 2(2^x \ln 2) = 2 (\ln 2^2) \cdot 2^x = 2^{x + 1} (\ln 2)^2$