Answer
$$h'\left( x \right) = \frac{2}{{\left( {3x + 1} \right)\left( {x + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = \ln \left[ {\left( {3x + 1} \right)\left( { - x + 1} \right)} \right] \cr
& {\text{Using logarithmic properties }}\ln \left( {ab} \right) = \ln \left( a \right) + \ln \left( b \right) \cr
& h\left( x \right) = \ln \left( {3x + 1} \right) + \ln \left( { - x + 1} \right) \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \left[ {\ln \left( {3x + 1} \right)} \right]' + \left[ {\ln \left( { - x + 1} \right)} \right]' \cr
& h'\left( x \right) = \frac{3}{{3x + 1}} + \frac{{ - 1}}{{ - x + 1}} \cr
& {\text{Simplifying}} \cr
& h'\left( x \right) = \frac{3}{{3x + 1}} - \frac{1}{{x + 1}} \cr
& h'\left( x \right) = \frac{{3x + 3 - 3x - 1}}{{\left( {3x + 1} \right)\left( {x + 1} \right)}} \cr
& h'\left( x \right) = \frac{2}{{\left( {3x + 1} \right)\left( {x + 1} \right)}} \cr} $$
