Answer
$f’(x)=\frac{2}{3(2x+1)\ln 2}$
Work Step by Step
$f(x)=\frac{(\frac{1}{3})\ln(2x+1)}{\ln 2}$
$u=\frac{1}{3}\ln(2x+1)$
$u’=\frac{2}{3(2x+1)}$
$v=\ln 2$
$v’=0$
$f’(x)=\frac{u’v-uv’}{v^2}$
$=\frac{\left(\frac{2}{3(2x+1)}\right)(\ln 2)-\left(\frac{1}{3}\right)(\ln (2x+1)(0)}{(\ln 2)^2}$
$f’(x)=\frac{2}{3(2x+1)\ln 2}$
