How do you find the derivative of $y=1/2 ln[(1+x)/(1-x)]$ ?

Question

2799 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-09T12:48:57

$dy/dx = 1/(1-x^2)$

$y = 1/2ln((1+x)/(1-x))$
$:. y = 1/2{ln(1+x)-ln(1-x)}$ (law of logs)

Differentiating (using the chain rule) gives:

$dy/dx = 1/2{1/(1+x)-(-1)/(1-x)}$
$:. dy/dx = 1/2{ ( (1-x)+(1+x) )/((1+x)(1-x)) }$
$:. dy/dx = 1/2{ 2/(1-x^2) }$
$:. dy/dx = 1/(1-x^2)$

How do you find the derivative of y=1/2 ln[(1+x)/(1-x)]y=1/2 ln[(1+x)/(1-x)]?