If $f(x)=\sin \left\{\frac{\pi}{3}[x]-x^2\right\}$ for 2 < x < 3 and [x] denotes the greatest integer less than or equal to x, then $f'(\sqrt{\pi / 3})$ is equal to

$-\sqrt{\pi / 3}$

For 2 < x < 3, we have [x] = 2

∴ $f(x)=\sin \left(\frac{2 \pi}{3}-x^2\right)$

$\Rightarrow f'(x)=-2 x \cos \left(\frac{2 \pi}{3}-x^2\right)$

$\Rightarrow f'(\sqrt{\pi / 3}=-2 \sqrt{\pi / 3} \cos \pi / 3=-\sqrt{\pi / 3}$