what is the derivative of (1 - ln(x))^2?

Answer 1

Just a simple Recall: given (a+bx)^n, the derivative is:
n(a+bx)^(n-1) X (a+bx)' = n(a+bx)^(n-1) X b
= nb(a+bx)^(n-1). so, we apply this principle is getting the derivative of (1-lnx)^2.
doing so, we have: 2(1-ln x) X (1-lnx)'
as we know, 1'=0 and -ln x' = -1/x. so we have:
-2(1-lnx)/x. i hope its clear enough..good day!:D

Answer 2

Hello

2(1-ln(x))*(-1/x) = (-2/x)*(1-ln(x))

Hope This Helps!!

Answer 3

f(x) = [1 - ln(x)]² = 1 - 2ln(x) + ln²x

Take the derivative of f(x).

f'(x) = 0 - 2/x + 2ln(x) / x = (-2/x)[1 - ln(x)]

Answer 4

-2(1-ln(x))(1/x)dx