Answer
$$y' = - \frac{{2\tan x}}{{\ln 10}}$$
Work Step by Step
$$\eqalign{
& y = \log \left( {1 - {{\sin }^2}x} \right) \cr
& {\text{differentiate}} \cr
& y' = \frac{{\left( {1 - {{\sin }^2}x} \right)'}}{{\left( {\ln 10} \right)\left( {1 - {{\sin }^2}x} \right)}} \cr
& {\text{chain rule }} \cr
& y' = \frac{{ - 2\sin x\left( {\cos x} \right)}}{{\left( {\ln 10} \right)\left( {1 - {{\sin }^2}x} \right)}} \cr
& {\text{simplify}} \cr
& y' = \frac{{ - 2\sin x\cos x}}{{\left( {\ln 10} \right)\left( {{{\cos }^2}x} \right)}} \cr
& y' = \frac{{ - 2\sin x}}{{\left( {\ln 10} \right)\left( {\cos x} \right)}} \cr
& y' = - \frac{{2\tan x}}{{\ln 10}} \cr} $$
