Question

If $y=\cos x,y_n=\frac{d^n\left(\cos x\right)}{d{x}^n}$, then $\left|\begin{array}{ccc}{y}_4& y_5& y_6\\ y_7& y_8& y_9\\ y_{10}& y_{11}& y_{12}\end{array}\right|=$_______

• A. $0$
• B. $-\cos x$
• C. $\cos x$
• D. $\sin x$

Answer 1

The correct option is A $0$

$y=\cos x$

$\Rightarrow y_n=\frac{d^n\left(\cos x\right)}{dx}$

$[\frac{d}{dx}\left(\cos x\right)=-\sin x]$

$[\frac{d}{dx}\left(\sin x\right)=\cos x]$

$\Rightarrow y_1=-\sin x$

$y_2=-\cos x$

$y_3=\sin x$

$y_4=\cos x$

$y_5=-\sin x$

So,

$\Rightarrow y_1=y_5=y_9...............=-\sin x$

$y_2=y_4=y_{10}...............=-\cos x$

$y_3=y_7=y_{11}...............=\sin x$

$y_4=y_8=y_{12}...............=\cos x$

So, $\left[\begin{array}{ccc}\cos x& -\sin x& -\cos x\\ \sin x& \cos x& -\sin x\\ -\cos x& \sin x& \cos x\end{array}\right]$

$=\cos ({\cos }^2+{\sin }^2)+\sin x(\sin x\cos x-\cos x\sin x)-\cos x({\sin }^{2}x+{\cos }^x)$

$=0$

Hence, the answer is $0$.