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

Aug 8, 2016

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

#### Explanation:

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

$d y = \frac{\frac{x + 1 - x + 1}{x + 1} ^ 2}{2 \sqrt{\frac{x - 1}{x + 1}}} \cdot d x$

$d y = \frac{\frac{2}{x + 1} ^ 2}{2 \sqrt{\frac{x - 1}{x + 1}}} \cdot d x$

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

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

Aug 8, 2016

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

#### Explanation:

Just another perspective...

$y = \sqrt{\frac{x - 1}{x + 1}}$

simplifying the RHS
${y}^{2} = \frac{x - 1}{x + 1} = \frac{x + 1 - 2}{x + 1} = 1 - \frac{2}{x + 1}$

implicit diff on LHS, power rule on RHS
$2 y \setminus y ' = \frac{2}{x + 1} ^ 2$

$y ' = \frac{1}{x + 1} ^ 2 \cdot \sqrt{\frac{x + 1}{x - 1}}$

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