Answer
$$ - \tan x + \frac{{3x}}{{4 - 3{x^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\ln \frac{{\cos x}}{{\sqrt {4 - 3{x^2}} }}} \right] \cr
& {\text{radical property}} \cr
& \frac{d}{{dx}}\left[ {\ln \frac{{\cos x}}{{{{\left( {4 - 3{x^2}} \right)}^{1/2}}}}} \right] \cr
& {\text{logarithm of a quotient}} \cr
& = \frac{d}{{dx}}\left[ {\ln \left( {\cos x} \right) - \ln {{\left( {4 - 3{x^2}} \right)}^{1/2}}} \right] \cr
& {\text{power rule for logarithms}} \cr
& = \frac{d}{{dx}}\left[ {\ln \left( {\cos x} \right) - \frac{1}{2}\ln \left( {4 - 3{x^2}} \right)} \right] \cr
& {\text{differentiate}} \cr
& = \frac{{ - \sin x}}{{\cos x}} - \frac{1}{2}\left( {\frac{{ - 6x}}{{4 - 3{x^2}}}} \right) \cr
& {\text{simplify}} \cr
& = - \tan x + \frac{{3x}}{{4 - 3{x^2}}} \cr} $$
