# What is the derivative of 1/(x+1)?

$f ' \left(x\right) = - \frac{1}{x + 1} ^ 2$
You could write $f \left(x\right) = \frac{1}{x + 1} = {\left(x + 1\right)}^{- 1}$ and use the power rule and chain rule: $f ' \left(x\right) = \left(- 1\right) {\left(x + 1\right)}^{- 2} \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right) = - 1 {\left(x + 1\right)}^{- 2} \cdot 1 = - \frac{1}{x + 1} ^ 2$.
$f ' \left(x\right) = \frac{\left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} \left(1\right) - 1 \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right)}{{\left(x + 1\right)}^{2}} = \frac{0 - 1}{{\left(x + 1\right)}^{2}} = - \frac{1}{x + 1} ^ 2$