How do you differentiate $y=sqrt ((x-1)/(x+1))$ ?

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Answer 1 · 2016-08-08T15:02:45

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

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

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

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

$d y=(cancel(2))/(cancel(2)*(x+1)^2*sqrt((x-1)/(x+1)))$

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

Answer 2 · 2016-08-08T18:45:32

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

Just another perspective...

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

simplifying the RHS
$y^2 = (x-1)/(x+1) = (x+ 1 -2)/(x+1) = 1 - 2/(x+1)$

implicit diff on LHS, power rule on RHS
$2 y \ y' = 2/(x+1)^2$

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

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

How do you differentiate y=sqrt ((x-1)/(x+1))y=sqrt ((x-1)/(x+1))?