# How do you find the derivative of f(x)= 2/(x+1)^2?

Jul 7, 2016

$- \frac{4}{x + 1} ^ 3$

#### Explanation:

$u \left(x\right) = 2$
$v \left(x\right) = {\left(x + 1\right)}^{2}$

$f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$

((u(x))/(v(x)))^' = (u'(x).v(x) - u(x).v'(x))/(v(x)^2

$f ' \left(x\right) = \frac{0. {\left(x + 1\right)}^{2} - 2. \left(2 x + 2\right)}{x + 1} ^ 4$

$f ' \left(x\right) = - \frac{4 \left(x + 1\right)}{x + 1} ^ 4 = - \frac{4 \cancel{\left(x + 1\right)}}{\cancel{\left(x + 1\right)}} ^ 4 = - \frac{4}{x + 1} ^ 3$

Jul 8, 2016

$- \frac{4}{{\left(x + 1\right)}^{3}}$

#### Explanation:

An alternative view is the following

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

then we can use the chain rule and power rule

$f ' \left(x\right) = - 4 {\left(x + 1\right)}^{-} 3 = - \frac{4}{{\left(x + 1\right)}^{3}}$