If f(x) = left| begin{array}{ccc} 2 cos x & 1 | vedclass

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(B) Given $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \ x - \frac{\pi}{2} & 2 \cos x & 1 \ 0 & 1 & 2 \cos x \end{array} ight|$.Using the pr

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(B) Given $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \ x - \frac{\pi}{2} & 2 \cos x & 1 \ 0 & 1 & 2 \cos x \end{array} ight|$.Using the property of differentiation of a determinant, $f^{\prime}(x)$ is the sum of three determinants where each row is differentiated one at a time:$f^{\prime}(x) = \left| \begin{array}{ccc} -2 \sin x & 0 & 0 \ x - \frac{\pi}{2} & 2 \cos x & 1 \ 0 & 1 & 2 \cos x \end{array} ight| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \ 1 & -2 \sin x & 0 \ 0 & 1 & 2 \cos x \end{array} ight| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \ x - \frac{\pi}{2} & 2 \cos x & 1 \ 0 & 0 & -2 \sin x \end{array} ight|$.Now, substitute $x = \pi$ (where $\sin \pi = 0$ and $\cos \pi = -1$):$f^{\prime}(\pi) = \left| \begin{array}{ccc} 0 & 0 & 0 \ \frac{\pi}{2} & -2 & 1 \ 0 & 1 & -2 \end{array} ight| + \left| \begin{array}{ccc} -2 & 1 & 0 \ 1 & 0 & 0 \ 0 & 1 & -2 \end{array} ight| + \left| \begin{array}{ccc} -2 & 1 & 0 \ \frac{\pi}{2} & -2 & 1 \ 0 & 0 & 0 \end{array} ight|$.The first determinant is $0$ because the first row is all zeros.The second determinant is $-2(0 - 0) - 1(-2 - 0) + 0 = 2$.The third determinant is $0$ because the third row is all zeros.Thus, $f^{\prime}(\pi) = 0 + 2 + 0 = 2$.

If $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$, then $f^{\prime}(\pi)$ is equal to

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