If $f(x)=|x|^{|\tan x|}$, then $f'\left(\frac{-\pi}{6}\right)$ is equal to

$\left(\frac{\pi}{6}\right)^{1 / \sqrt{3}}\left\{\frac{-2 \sqrt{3}}{\pi}+\frac{4}{3} \log \frac{6}{\pi}\right\}$

In the neighbourhood of $x=-\frac{\pi}{6}$, we have

$|x| =-x$ and $|\tan x|=-\tan x$

∴  $f(x)=(-x)^{-\tan x}$

$\Rightarrow f(x)=e^{-\tan x . \log (-x)}$

$\Rightarrow f'(x)=(-x)^{-\tan x}\left\{-\sec ^2 x . \log (-x)-\frac{\tan x}{x}\right\}$

$\Rightarrow f'\left(\frac{-\pi}{6}\right)=\left(\frac{\pi}{6}\right)^{1 / \sqrt{3}}\left\{-\frac{4}{3} \log \frac{\pi}{6}-\frac{2 \sqrt{3}}{\pi}\right\}$

$\Rightarrow f'\left(\frac{-\pi}{6}\right)=\left(\frac{\pi}{6}\right)^{1 / \sqrt{3}}\left\{\frac{4}{3} \log \frac{6}{\pi}-\frac{2 \sqrt{3}}{\pi}\right\}$