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)
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.
[ln(cscx - cotx)]' = cscx
Kindly mail your feedback to firstname.lastname@example.org
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 26, 23 12:27 PM
May 21, 23 07:40 PM
May 20, 23 10:53 PM