## Calculus 10th Edition

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