How do you find the derivative of $y= sqrt((x-1)/(x+1))$ ?

Question

16586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-08-07T21:19:51

$y'=1/(sqrt(x-1)(x+1)^(3/2))$

Explanation

$y=sqrt((x-1)/(x+1)$

taking natural log of both sides,

$lny=1/2ln((x-1)/(x+1))$ ,

$lny=1/2(ln(x-1)-ln(x+1))$

differentiating both sides with respect to $x$ ,

$1/y*y'=1/2(1/(x-1)-1/(x+1))$

$y'=y/2((x+1-x+1)/(x^2-1))$

$y'=y(1/(x^2-1))$

plugging $y$

$y'=(sqrt((x-1)/(x+1)))(1/(x^2-1))$

$y'=(sqrt((x-1)/(x+1)))(1/((x-1)(x+1)))$

$y'=1/(sqrt(x-1)(x+1)^(3/2))$

How do you find the derivative of y= sqrt((x-1)/(x+1))y= sqrt((x-1)/(x+1)) ?