Answer
$$ = - 2\tan x + \frac{{2{x^3}}}{{1 + {x^4}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\ln \left( {{{\cos }^2}x} \right)\sqrt {1 + {x^4}} } \right] \cr
& {\text{radical property}} \cr
& = \frac{d}{{dx}}\left[ {\ln \left( {{{\cos }^2}x} \right){{\left( {1 + {x^4}} \right)}^{1/2}}} \right] \cr
& {\text{logarithm of a product}} \cr
& = \frac{d}{{dx}}\left[ {\ln \left( {{{\cos }^2}x} \right) + \log {{\left( {1 + {x^4}} \right)}^{1/2}}} \right] \cr
& {\text{power rule for logarithms}} \cr
& = \frac{d}{{dx}}\left[ {2\ln \left( {\cos x} \right) + \frac{1}{2}\log \left( {1 + {x^4}} \right)} \right] \cr
& {\text{differentiate}} \cr
& = 2\left( {\frac{{ - \sin x}}{{\cos x}}} \right) + \frac{1}{2}\left( {\frac{{4{x^3}}}{{1 + {x^4}}}} \right) \cr
& {\text{simplify}} \cr
& = - 2\tan x + \frac{{2{x^3}}}{{1 + {x^4}}} \cr} $$
