$(\sin x-\cos x)^{(\sin x-\cos x)}, \frac{\pi}{4}
Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$
Taking logarithm on both the sides, we obtain
$\log y=\log \left[(\sin x-\cos x)^{(\sin x-\cos x)}\right]$
$\Rightarrow \log y=(\sin x-\cos x) \cdot \log (\sin x-\cos x)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[(\sin x-\cos x) \log (\sin x-\cos x)]$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)+(\sin x-\cos x) \cdot \frac{d}{d x} \log (\sin x-\cos x)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot(\cos x+\sin x)+(\sin x-\cos x) \cdot \frac{1}{(\sin x-\cos x)} \cdot \frac{d}{d x}(\sin x-\cos x)$
$\Rightarrow \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}[(\cos x+\sin x) \cdot \log (\sin x-\cos x)+(\cos x+\sin x)]$
$\therefore \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)[1+\log (\sin x-\cos x)]$