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,
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 11, 25 12:34 AM
May 10, 25 09:56 AM
May 10, 25 12:14 AM