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