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

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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,

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

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