# How do you differentiate (2x)/(x+1)^2?

Mar 30, 2018

$\left(\frac{f}{g}\right) ' = \frac{- \textcolor{red}{2} \left(x - 1\right)}{x + 1} ^ 3$

#### Explanation:

The quotient rule tells us that for a function in the form of $\frac{f}{g} , \left(\frac{f}{g}\right) ' = \left(g f ' - f g '\right) {g}^{2}$

Here, $g \left(x\right) = {\left(x + 1\right)}^{2}$

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

$f \left(x\right) = 2 x$

$f ' \left(x\right) = 2$

So,

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

Simplify.

$\left(\frac{f}{g}\right) ' = \frac{2 \left({x}^{2} + 2 x + \textcolor{red}{1}\right) - \left(4 {x}^{2} + 4 x\right)}{x + 1} ^ 4$

$\left(\frac{f}{g}\right) ' = \frac{2 {x}^{2} + 4 x + \textcolor{red}{2} - 4 {x}^{2} - 4 x}{x + 1} ^ 4$

$\left(\frac{f}{g}\right) ' = \frac{\textcolor{red}{\left(2 - 2 {x}^{2}\right)}}{x + 1} ^ 4$

$\left(\frac{f}{g}\right) ' = \frac{- \textcolor{red}{2} \left({x}^{2} - 1\right)}{x + 1} ^ 4$

$\left(\frac{f}{g}\right) ' = \frac{- \textcolor{red}{2} \cancel{\left(x + 1\right)} \left(x - 1\right)}{x + 1} ^ \left(\left(\cancel{4}\right) 3\right)$

$\left(\frac{f}{g}\right) ' = \frac{- \textcolor{red}{2} \left(x - 1\right)}{x + 1} ^ 3$

Mar 30, 2018

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

#### Explanation:

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

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

$\mathrm{du} = 2 , \mathrm{dv} = 2 \left(x + 1\right) \cdot 1 = 2 \left(x + 1\right)$

Using quotient rule, $f ' \left(x\right) = \frac{v \mathrm{du} - u \mathrm{dv}}{v} ^ 2$

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

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

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

f'(x) = (2 cancel(color(brown) (1 + x)) * (1-x) ) / (x+1)^cancel(color(brown)4^color(green)(3)#

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