We know the derivative of ln(x), which is ¹⁄ₓ.
[ln(x)]' = ¹⁄ₓ
We can find the derivative of ln(cscx - cotx) using chain rule.
Find ᵈʸ⁄dₓ, if
y = ln(cscx - cotx)
Let t = cscx - cotx.
y = ln(t)
Now,
y = ln(t) ----> y is a function of t
t = cscx - cotx ----> t is is a function of x
By chain rule, the derivative of y with respect to x :
Substitute y = ln(t) and t = cscx - cosx.
Substitute t = cscx - cosx.
Therefore,
[ln(cscx - cotx)]' = cscx
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 10, 25 09:56 AM
May 10, 25 12:14 AM
May 08, 25 12:28 PM