Answer
$g’(t)=\frac{6t}{(t^2+7)(\ln 2)}$
Work Step by Step
$g(t)=\frac{3\ln(t^2+7)}{\ln 2}$
$u=3\ln(t^2+7)$
$u’=\frac{6t}{t^2+7}$
$v=\ln 2$
$v’=0$
$g’(t)=\frac{u’v-uv’}{v^2}$
$=\frac{\left(\frac{6t}{t^2+7}\right)(\ln 2)-(3)(\ln(t^2+7))(0)}{(\ln 2)^2}$
$g’(t)=\frac{6t}{(t^2+7)(\ln 2)}$