Answer
$$ - \frac{5}{{x\ln {{\left( {2x} \right)}^6}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {\ln {{\left( {2x} \right)}^{ - 5}}} \right) \cr
& {\text{use the chain rule}} \cr
& = - 5\ln {\left( {2x} \right)^{ - 6}}\frac{d}{{dx}}\left( {\ln \left( {2x} \right)} \right) \cr
& {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = 2x \cr
& = - 5\ln {\left( {2x} \right)^{ - 6}}\left( {\frac{2}{{2x}}} \right) \cr
& {\text{simplify}} \cr
& = - \frac{5}{{x\ln {{\left( {2x} \right)}^6}}} \cr} $$
