If \[y = \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ {\cos x}&{ - \sin x}&{\cos x}\\ x&1&1 \end{array}} \right|\]      then $$\frac{{dy}}{{dx}}$$ is

Solution :
\[\begin{array}{l} \frac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&{\cos x}\\ {\cos x}&{ - \sin x}&{\cos x}\\ x&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ { - \sin x}&{ - \cos x}&{ - \sin x}\\ x&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ {\cos x}&{ - \sin x}&{\cos x}\\ 1&0&0 \end{array}} \right|\\ = 0 - \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ {\sin x}&{\cos x}&{\sin x}\\ x&1&1 \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} {\cos x}&{\sin x}\\ { - \sin x}&{\cos x} \end{array}} \right| \end{array}\]
$$ = 0 + \left( {{{\cos }^2}x + {{\sin }^2}x} \right)$$