# How to differentiating using quotient rule?

## How to differentiate $\frac{x}{x + 1}$

Mar 24, 2016

$\setminus \frac{1}{{\left(x + 1\right)}^{2}}$ for all values of $x$ except for $x = - 1$

#### Explanation:

If a function, $f \left(x\right)$, can be written as

$f \left(x\right) = \setminus \frac{g \left(x\right)}{h \left(x\right)}$

where $h \left(x\right) \ne 0$, then the Quotient rule states that the derivative of $g \frac{x}{h \left(x\right)}$ is

$f ' \left(x\right) = \setminus \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{{\left[h \left(x\right)\right]}^{2}}$ ............(1)
In the problem $f \left(x\right) = \setminus \frac{g \left(x\right)}{h \left(x\right)} = \frac{x}{x + 1}$

Inserting appropriate functions in (1)
$f ' \left(x\right) = \setminus \frac{\left(x + 1\right) \frac{d}{\mathrm{dx}} \left(x\right) - x \frac{d}{\mathrm{dx}} \left(x + 1\right)}{{\left(x + 1\right)}^{2}}$

$f ' \left(x\right) = \setminus \frac{\left(x + 1\right) - x}{{\left(x + 1\right)}^{2}}$
$f ' \left(x\right) = \setminus \frac{1}{{\left(x + 1\right)}^{2}}$ for all values of $x$ except for $x = - 1$