Question

Differentiate the following function with respect to x
a) ln((2x-1/1+x)^1/2)

Answer 1

use your rules of logs to simplify this messy expression
I will assume you meant:
y = ln [ (2x-1)/(1+x) ]^(1/2)
= (1/2)( ln(2x-1) - ln(1+x) )

dy/dx = (1/2)( 2/(2x-1) - 1/(1+x) )
= 1/(2x-1) - 1/(2(1+x))

Answer 2

How do you change y= (1/2)( ln(2x-1) - ln(1+x) )

Answer 3

ln u^n = n * ln u
ln u*v = ln u + ln v
ln u/v = ln u - ln v

Better review the basic properties of logs.