Find y' for y = ln sin x

Find $y'$ for $y = \ln\sin x$.

Explanation of Method

The logarithmic derivative of the sine.

Step 1: Apply the chain rule

With $u = \sin x$ and $u' = \cos x$, the rule $\bigl(\ln u\bigr)' = \dfrac{u'}{u}$ gives

\[ y' = \frac{\cos x}{\sin x} = \cot x. \]

Step 2: Verification

\[x = \frac{\pi}{2}:\quad y = \ln 1 = 0,\quad y' = \cot\frac{\pi}{2} = 0, \text{ the maximum of } y \;\;\checkmark\]

\[x = \frac{\pi}{4}:\quad y' = 1; \qquad \frac{y(\pi/4+10^{-5})-y(\pi/4-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

Result

Final answer:

\[\boxed{\,y' = \cot x, \qquad 2k\pi < x < (2k+1)\pi\,}\]

Remark

Remark. The domain is where $\sin x \gt 0$, i.e. the intervals $(2k\pi,\,(2k+1)\pi)$. On each of them $y$ rises from $-\infty$ to $0$ at the midpoint and falls back, exactly as the sign of $\cot x$ predicts.