91Ó°ÊÓ

: First, we need to find the derivative of the natural logarithm function, which is given by the formula: $$\frac{d}{dx}(\ln(u(x)))=\frac{u'(x)}{u(x)}$$ In our case, $$u(x)=|\sin x|$$. So we first need to find the derivative of $$|\sin x|$$.

: The derivative of the absolute value function is determined by the sign of the input value. In this case, the input value is $$\sin x$$. For $$\sin x \ge 0$$, the derivative of $$|\sin x|$$ is equal to the derivative of $$\sin x$$: $$\frac{d}{dx} |\sin x| = \cos x$$ For $$\sin x < 0$$, the derivative of $$|\sin x|$$ is equal to the derivative of $$-\sin x$$: $$\frac{d}{dx} |\sin x| = -\cos x$$

: Now that we have the derivatives of the natural logarithm function and the absolute value of the sine function, we can apply the chain rule to find the derivative of the composite function: $$\frac{d}{dx}(\ln |\sin x|) = \frac{\frac{d}{dx}(|\sin x|)}{|\sin x|}$$ By using the derivatives of $$|\sin x|$$ we found earlier, we can write the final results for the two cases: For $$\sin x \ge 0$$: $$\frac{d}{dx}(\ln |\sin x|) = \frac{\cos x}{|\sin x|} = \frac{\cos x}{\sin x}$$ For $$\sin x < 0$$: $$\frac{d}{dx}(\ln |\sin x|) = \frac{-\cos x}{|\sin x|} = \frac{-\cos x}{-\sin x} = \frac{\cos x}{\sin x}$$ Since the derivatives match in both cases, the derivative of $$\ln |\sin x|$$ is: $$\frac{d}{dx}(\ln |\sin x|) = \frac{\cos x}{\sin x}$$