Answer
$$y' = \frac{3}{{2x}} + \frac{{2{x^3}}}{{1 + {x^4}}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {{x^{3/2}}\sqrt {1 + {x^4}} } \right) \cr
& {\text{use product rule for logarithms}} \cr
& y = \ln \left( {{x^{3/2}}} \right) + \ln \left( {\sqrt {1 + {x^4}} } \right) \cr
& {\text{radical property }}\root n \of {{a^m}} = {a^{m/n}} \cr
& y = \ln \left( {{x^{3/2}}} \right) + \ln {\left( {1 + {x^4}} \right)^{1/2}} \cr
& {\text{power rule for logarithms }}\ln {a^m} = m\ln a \cr
& y = \frac{3}{2}\ln \left( x \right) + \frac{1}{2}\ln \left( {1 + {x^4}} \right) \cr
& {\text{find the derivative}} \cr
& y' = \frac{3}{2}\left( {\ln \left( x \right)} \right)' + \frac{1}{2}\left( {\ln \left( {1 + {x^4}} \right)} \right)' \cr
& {\text{use }}\left( {\ln u} \right)' = \frac{{u'}}{u}, \cr
& y' = \frac{3}{2}\left( {\frac{1}{x}} \right) + \frac{1}{2}\left( {\frac{{4{x^3}}}{{1 + {x^4}}}} \right) \cr
& y' = \frac{3}{{2x}} + \frac{{2{x^3}}}{{1 + {x^4}}} \cr} $$
