Answer
$g'(t)=\frac{10}{\ln4}(\frac{1-\ln t}{t^2})$
Work Step by Step
$g(t)=\frac{10\frac{\ln t}{\ln4}}{t}$
$g(t)=\frac{10}{ln4}(\frac{lnt}{t})$
$g(t)=\frac{10}{ln4}(lnt)(t^{-1})$
let
$u=lnt$
$u'=\frac{1}{t}$
$v=t^{-1}$
$v'=-t^{-2}$
$g'(t)=\frac{10}{ln4}(\frac{1}{t^2}-\frac{lnt}{t^2})$
$=\frac{10}{\ln4}(\frac{1-\ln t}{t^2})$