Answer
$y' = - \frac{{10{x^4}}}{{ {3 - 2{x^5}} }}$
Work Step by Step
$$\eqalign{
& y = \ln \left| {3 - 2{x^5}} \right| \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {\ln \left| {3 - 2{x^5}} \right|} \right] \cr
& {\text{Use the formula }}\frac{d}{{dt}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dt}},{\text{ }}\left( {{\text{see page 218}}} \right),{\text{ }} \cr
& y' = \frac{1}{{\left| {3 - 2{x^5}} \right|}}\frac{d}{{dx}}\left[ {|3 - 2{x^5}|} \right] \cr
& y' = \frac{1}{{\left| {3 - 2{x^5}} \right|}}\cdot \frac{3-2x^5}{{\left| {3 - 2{x^5}} \right|}}\left( { - 10{x^4}} \right) \cr
& {\text{Simplify}} \cr
& y' = - \frac{{10{x^4}}}{{ {3 - 2{x^5}} }} \cr} $$
