To find the derivative of y = ln(sin x), I'll use the chain rule.
Given: y = ln(sin x)
Using the chain rule: dy/dx = d/dx[ln(sin x)]
The derivative of ln(u) is 1/u · du/dx, where u = sin x.
So: dy/dx = 1/(sin x) · d/dx(sin x)
Since d/dx(sin x) = cos x:
dy/dx = 1/(sin x) · cos x = cos x/sin x
Therefore: dy/dx = cot x
We can also write this as cos x/sin x, but cot x is the most simplified form.
Note: This derivative is defined wherever sin x ≠ 0, which means x ≠ nπ where n is any integer.
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