Find f'(x) if f(x) = (sin x)^sin x for all 0 < x < π.

The correct answer is Option (3) → $(\sin x)^{\sin x} \cos x [1 + \ln(\sin x)]$ ##

The function $y = (\sin x)^{\sin x}$ is defined for all positive real numbers. Taking logarithms, we have

$\log y = \log (\sin x)^{\sin x} = \sin x \log (\sin x)$$

Then $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\sin x \log (\sin x))$

$= \cos x \log (\sin x) + \sin x \cdot \frac{1}{\sin x} \cdot \frac{d}{dx} (\sin x)$

$= \cos x \log (\sin x) + \cos x$

$= (1 + \log (\sin x)) \cos x $

Thus $\frac{dy}{dx} = y((1 + \log (\sin x)) \cos x) = (1 + \log (\sin x)) (\sin x)^{\sin x} \cos x$