We know the derivative of lnx, which is ¹⁄ₓ. And also, the derivative cot is sec^{2}x.
(lnx)' = ¹⁄ₓ
(cotx)' = -csc^{2}x
We can find the derivative of ln(cotx) using chain rule.
If y = ln(cotx), find ᵈʸ⁄dₓ.
Let t = cotx.
Then, we have
y = lnt
By chain rule,
Substitute y = lnt and t = cotx.
Substitute t = cotx.
Therefore,
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