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

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

#### Explanation:

Use $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} u - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{{v}^{2}}$

Let $u = x - 1$ and $v = x + 1$

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} u - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{{v}^{2}}$

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

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

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

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

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

God bless....I hope the explanation is useful.