Answer
$$ - \frac{{2\sin x}}{{\cos x}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {\ln \left( {{{\cos }^2}x} \right)} \right) \cr
& {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = {\cos ^2}x \cr
& = \frac{1}{{{{\cos }^2}x}}\frac{d}{{dx}}\left( {{{\cos }^2}x} \right) \cr
& {\text{by the chain rule}} \cr
& = \frac{1}{{{{\cos }^2}x}}\left( {2\cos x} \right)\frac{d}{{dx}}\left( {\cos x} \right) \cr
& = \frac{1}{{{{\cos }^2}x}}\left( {2\cos x} \right)\left( { - \sin x} \right) \cr
& {\text{simplify}} \cr
& = - \frac{{2\sin x}}{{\cos x}} \cr} $$
