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 22, 24 06:32 AM
May 17, 24 08:12 AM
May 14, 24 08:53 AM