If f(x)=|x|^|sin x|, then f'≤ft(-(π)/4) equals

In the neighbourhood of $-\pi / 4$, we have

$f(x) =(-x)^{-\sin x}=e^{-\sin x \log (-x)}$

$\Rightarrow f'(x) =e^{-\sin x \log (-x)}\left(-\cos x . \log (-x)-\frac{\sin x}{x}\right)$

$\Rightarrow f'(x)=(-x)^{-\sin x}\left(-\cos x . \log (-x)-\frac{\sin x}{x}\right)$

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

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