Find $f^{\prime}(x)$ if $f(x)=(\sin x)^{\sin x}$ for all $0 < x < \pi$.

Given the function $f(x) = (\sin x)^{\sin x}$.
Let $y = (\sin x)^{\sin x}$.
Taking the natural logarithm on both sides,we get:
$\ln y = \ln ((\sin x)^{\sin x}) = \sin x \cdot \ln(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x) \cdot \ln(\sin x) + \sin x \cdot \frac{d}{dx}(\ln(\sin x))$.
$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\sin x) + \sin x \cdot \frac{1}{\sin x} \cdot \cos x$.
$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\sin x) + \cos x$.
Factoring out $\cos x$:
$\frac{1}{y} \frac{dy}{dx} = \cos x (1 + \ln(\sin x))$.
Multiplying by $y$ to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = y \cdot \cos x (1 + \ln(\sin x))$.
Substituting $y = (\sin x)^{\sin x}$ back into the equation:
$f^{\prime}(x) = (\sin x)^{\sin x} \cos x (1 + \ln(\sin x))$.