What is the derivative of #ln((2x-1)/(x-1))#?

Question

7953 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-15T09:27:21

#dy/dx=1/((2x-1)(x-1))ln{(x-1)/(2x-1)}#

Let #y=ln{(2x-1)/(x-1)}#

Before Diif.ing #y#, we simplify #Y# using the Rules of Log. Funs.

#y={(2x-1)/(x-1)} rArr lny=ln(2x-1)-ln(x-1)#

Diff.ing both sides w.r.t. #x#,

#d/dxlny=d/dx{ln(2x-1)-ln(x-1)}=d/dxln(2x-1)-d/dxln(x-1)#

But, by Chain Rule, #d/dxlny=d/dylny*dy/dx=1/y*dy/dx#

Hence,

#1/ydy/dx=1/(2x-1)*d/dx(2x-1)-1/(x-1)*d/dx(x-1)#

#=2/(2x-1)-1/(x-1)={2(x-1)-(2x-1)}/{(2x-1)(x-1)}=-1/((2x-1)(x-1))#

#:. dy/dx=-y/((2x-1)(x-1))=-ln{(2x-1)/(x-1)}/((2x-1)(x-1)),# or,

#dy/dx=1/((2x-1)(x-1))ln{(x-1)/(2x-1)}#