1. The problem is to find the derivative of the function $f(x) = \ln(\sin x)$.\n\n2. We use the chain rule and the derivative of the natural logarithm function. The derivative of $\ln u$ with respect to $x$ is $\frac{1}{u} \cdot \frac{du}{dx}$.\n\n3. Here, $u = \sin x$. The derivative of $\sin x$ is $\cos x$.\n\n4. Applying the chain rule:\n$$f'(x) = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x}.$$\n\n5. Simplifying, we get:\n$$f'(x) = \cot x.$$\n\n6. Therefore, the derivative of $f(x) = \ln(\sin x)$ is $f'(x) = \cot x$.