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

Feb 25, 2016

$\frac{2}{x + 1} ^ 2$

#### Explanation:

Use the quotient rule, which states that

$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

Here, we see that

$f \left(x\right) = x - 1$
$g \left(x\right) = x + 1$

So both their derivatives equal

$f ' \left(x\right) = 1$
$g ' \left(x\right) = 1$

Thus, using the first equation,

$\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right) = \frac{\left(x + 1\right) \left(1\right) - \left(x - 1\right) \left(1\right)}{x + 1} ^ 2$

$\textcolor{w h i t e}{X X X X . l X X} = \frac{2}{x + 1} ^ 2$