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))#
