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

Answer 1

Christopher Allers

#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))#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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