Question

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

Answer 1

We have,
$y=\left(\begin{array}{ccc}\sin x& \cos x& \sin x\\ \cos x& -\sin x& \cos x\\ x& 1& 1\end{array}\right|\mathrm{Differentiating}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\begin{array}{ccc}\cos x& -\sin x& \cos x\\ \cos x& -\sin x& \cos x\\ x& 1& 1\end{array}\right|+\left(\begin{array}{ccc}\sin x& \cos x& \sin x\\ -\sin x& -\cos x& -\sin x\\ x& 1& 1\end{array}\right|+\left(\begin{array}{ccc}\sin x& \cos x& \sin x\\ \cos x& -\sin x& \cos x\\ 1& 0& 0\end{array}\right|\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=0-\left(\begin{array}{ccc}\sin x& \cos x& \sin x\\ \sin x& \cos x& \sin x\\ x& 1& 1\end{array}\right|+1\left({\cos }^{2}x+{\sin }^{2}x\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=0+1=1$