Differentiate the function with respect to $x$: $(\sin x - \cos x)^{(\sin x - \cos x)}$,where $\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Let $y = (\sin x - \cos x)^{(\sin x - \cos x)}$.
Taking the natural logarithm on both sides,we get:
$\log y = \log [(\sin x - \cos x)^{(\sin x - \cos x)}]$
$\log y = (\sin x - \cos x) \cdot \log (\sin x - \cos x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x - \cos x) \cdot \log (\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{d}{dx} \log (\sin x - \cos x)$.
Since $\frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x$,we have:
$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) \cdot \log (\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{1}{\sin x - \cos x} \cdot (\cos x + \sin x)$.
Simplifying the expression:
$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) \cdot \log (\sin x - \cos x) + (\cos x + \sin x)$.
Factoring out $(\cos x + \sin x)$:
$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) [1 + \log (\sin x - \cos x)]$.
Multiplying by $y$:
$\frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [1 + \log (\sin x - \cos x)]$.