Answer
$$\frac{{dy}}{{dx}} = \frac{{4{{\left( {\ln \left| {x + 1} \right|} \right)}^3}}}{{x + 1}}$$
Work Step by Step
$$\eqalign{
& y = {\left( {\ln \left| {x + 1} \right|} \right)^4} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {\ln \left| {x + 1} \right|} \right)}^4}} \right] \cr
& {\text{use the chain rule}} \cr
& \frac{{dy}}{{dx}} = 4{\left( {\ln \left| {x + 1} \right|} \right)^3}\frac{d}{{dx}}\left[ {\ln \left| {x + 1} \right|} \right] \cr
& {\text{use }}\frac{d}{{dx}}\ln g\left( x \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}} \cr
& \frac{{dy}}{{dx}} = 4{\left( {\ln \left| {x + 1} \right|} \right)^3}\left( {\frac{1}{{x + 1}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{4{{\left( {\ln \left| {x + 1} \right|} \right)}^3}}}{{x + 1}} \cr} $$
