Question

If $y=\left|\begin{array}{ccc}\sin x& \cos x& \sin x\\ \cos x& -\sin x& \cos x\\ x& 1& 1\end{array}\right|,$ then the value of $\frac{dy}{dx}=$

Answer 1

We have,
$y=\left|\begin{array}{ccc}\sin x& \cos x& \sin x\\ \cos x& -\sin x& \cos x\\ x& 1& 1\end{array}\right|$
We know that,
if $\Delta =\left|\begin{array}{ccc}{f}_1\left(x\right)& f_2\left(x\right)& f_3\left(x\right)\\ g_1\left(x\right)& g_2\left(x\right)& g_3\left(x\right)\\ h_1\left(x\right)& h_2\left(x\right)& h_3\left(x\right)\end{array}\right|$
Then, ${\Delta }^{\prime }=\left|\begin{array}{ccc}{f}_1^{\prime }\left(x\right)& f_2^{\prime }\left(x\right)& f_3^{\prime }\left(x\right)\\ g_1\left(x\right)& g_2\left(x\right)& g_3\left(x\right)\\ h_1\left(x\right)& h_2\left(x\right)& h_3\left(x\right)\end{array}\right|+\left|\begin{array}{ccc}{f}_1\left(x\right)& f_2\left(x\right)& f_3\left(x\right)\\ g_1^{\prime }\left(x\right)& g_2^{\prime }\left(x\right)& g_3^{\prime }\left(x\right)\\ h_1\left(x\right)& h_2\left(x\right)& h_3\left(x\right)\end{array}\right|+\left|\begin{array}{ccc}{f}_1\left(x\right)& f_2\left(x\right)& f_3\left(x\right)\\ g_1\left(x\right)& g_2\left(x\right)& g_3\left(x\right)\\ h_1^{\prime }\left(x\right)& h_2^{\prime }\left(x\right)& h_3^{\prime }\left(x\right)\end{array}\right|$
$\therefore \frac{dy}{dx}=\left|\begin{array}{ccc}\cos x& -\sin x& \cos x\\ \cos x& -\sin x& \cos x\\ x& 1& 1\end{array}\right|+\left|\begin{array}{ccc}\sin x& \cos x& \sin x\\ -\sin x& -\cos x& -\sin x\\ x& 1& 1\end{array}\right|+\left|\begin{array}{ccc}\sin x& \cos x& \sin x\\ \cos x& -\sin x& \cos x\\ 1& 0& 0\end{array}\right|=0+(-1)\cdot 0+{\cos }^{2}x+{\sin }^{2}x=1$