Derivative of ln(sin(x))

Identify the outer function as $\ln(u)$ with $u = \sin(x)$ as the inner function. Since $\frac{d}{du}\ln(u) = \frac{1}{u}$, the chain rule gives $f'(x) = \frac{1}{\sin(x)} \cdot \frac{d}{dx}[\sin(x)]$.

Since $\frac{d}{dx}[\sin(x)] = \cos(x)$, substituting gives $f'(x) = \frac{\cos(x)}{\sin(x)}$. This can also be written as $f'(x) = \cot(x)$, valid on $(0,\pi)$ since $\sin(x) \neq 0$ there.