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

Aug 7, 2014

$y ' = \frac{1}{\sqrt{x - 1} {\left(x + 1\right)}^{\frac{3}{2}}}$

Explanation

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

taking natural log of both sides,

$\ln y = \frac{1}{2} \ln \left(\frac{x - 1}{x + 1}\right)$,

$\ln y = \frac{1}{2} \left(\ln \left(x - 1\right) - \ln \left(x + 1\right)\right)$

differentiating both sides with respect to $x$,

$\frac{1}{y} \cdot y ' = \frac{1}{2} \left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)$

$y ' = \frac{y}{2} \left(\frac{x + 1 - x + 1}{{x}^{2} - 1}\right)$

$y ' = y \left(\frac{1}{{x}^{2} - 1}\right)$

plugging $y$

$y ' = \left(\sqrt{\frac{x - 1}{x + 1}}\right) \left(\frac{1}{{x}^{2} - 1}\right)$

$y ' = \left(\sqrt{\frac{x - 1}{x + 1}}\right) \left(\frac{1}{\left(x - 1\right) \left(x + 1\right)}\right)$

$y ' = \frac{1}{\sqrt{x - 1} {\left(x + 1\right)}^{\frac{3}{2}}}$