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

1 Answer
Aug 7, 2014

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))