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