Answer
$\frac{1+\ln 2}{u(1+\ln(2u))^2}$
Work Step by Step
$\frac{d}{du} \frac{\ln u}{1+\ln(2u)}$
$=\frac{(1+\ln 2u)\frac{d}{du}\ln u-\ln u\frac{d}{du}(1+\ln(2u))}{(1+\ln(2u))^2}$
$=\frac{(1+\ln(2u))\frac{1}{u}-\ln u*\frac{1}{2u}*\frac{d}{du}(2u)}{(1+\ln(2u))^2}$
$=\frac{\frac{1+\ln(2u)}{u}-\frac{\ln u}{u}}{(1+\ln(2u))^2}$
$=\frac{1+\ln(2u)-\ln u}{u(1+\ln(2u))^2}$
$=\frac{1+\ln 2+\ln u-\ln u}{u(1+\ln(2u))^2}$
$=\frac{1+\ln 2}{u(1+\ln(2u))^2}$
