DERIVATIVE OF NATURAL LOG OF COTX

We know the derivative of lnx, which is ¹⁄ₓ. And also, the derivative cot is sec²x.

(lnx)' = ¹⁄ₓ

(cotx)' = -csc²x

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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