If  $f\left(x\right)=\left|\begin{array}{ccc}2\cos \mathrm{x}& 1& 0\\ 1& 2\cos \mathrm{x}& 1\\ 0& 1& 2\cos \mathrm{x}\end{array}\right|$ then $\left|{\mathrm{f}}^1\left(\frac{\mathrm{\pi }}{3}\right)\right|=$

The given function is $f\left(x\right)=\left|\begin{array}{ccc}2\cos x& 1& 0\\ 1& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$

Expand the determinant as 

$\begin{array}{rcl}f\left(x\right)& =& 2\cos x\left(4{\cos }^{2}x-1\right)-1\left(2\cos x\right)\\ & =& 8{\cos }^{3}x-2\cos x-2\cos x\\ & =& 8{\cos }^{3}x-4\cos x\end{array}$

Differenitate both  sides

$f\text{'}\left(x\right)=-24{\cos }^{2}x\sin x+4\sin x$

Substitute $x=\frac{\mathrm{\pi }}{3}$

$\begin{array}{rcl}f\text{'}\left(\frac{\mathrm{\pi }}{3}\right)& =& -24\left(\frac{1}{4}\right)\left(\frac{\sqrt{3}}{2}\right)+4\left(\frac{\sqrt{3}}{2}\right)\\ & =& -3\sqrt{3}+2\sqrt{3}\\ & =& \boxed{\mathbf{-}\sqrt{\mathbf{3}}}\end{array}$