Answer
$\frac{\frac{1-v}{v}+\ln v}{(1-v)^2}$
Work Step by Step
$\frac{d}{dv}\frac{\ln v}{1-v}$
$=\frac{(1-v)\frac{d}{dx}\ln v-\ln v\frac{d}{dx}(1-v)}{(1-v)^2}$
$=\frac{(1-v)*\frac{1}{v}-\ln v*(-1)}{(1-v)^2}$
$=\frac{\frac{1-v}{v}+\ln v}{(1-v)^2}$
$=\frac{\frac{1-v}{v}+\ln v}{(1-v)^2}$