# How do you find the first and second derivative of x^2-2x-3?

May 12, 2016

If $f \left(x\right) = {x}^{2} - 2 x - 3$
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 2 x - 2$ and
$\textcolor{w h i t e}{\text{XXX}} f ' ' \left(x\right) = 2$

#### Explanation:

The first derivative:
$f ' \left(x\right) = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\mathrm{dc} o l \mathmr{and} \left(red\right) \left({x}^{2}\right)}{\mathrm{dx}} - \frac{\mathrm{dc} o l \mathmr{and} \left(b l u e\right) \left(2 x\right)}{\mathrm{dx}} - \frac{\mathrm{dc} o l \mathmr{and} \left(g r e e n\right) \left(3\right)}{\mathrm{dx}}$

$\textcolor{w h i t e}{\text{XXXXXXXXX}} = \textcolor{red}{2 {x}^{1}} - \textcolor{b l u e}{2 {x}^{0}} - \textcolor{g r e e n}{0}$

$\textcolor{w h i t e}{\text{XXXXXXXXX}} = \textcolor{red}{2 x} - \textcolor{b l u e}{2}$

The second derivative
The second derivative is just the derivative of the first derivative.
$f ' ' \left(x\right) = \frac{\mathrm{df} ' \left(x\right)}{\mathrm{dx}} = \frac{\mathrm{dc} o l \mathmr{and} \left(red\right) \left(2 {x}^{1}\right)}{\mathrm{dx}} - \frac{\mathrm{dc} o l \mathmr{and} \left(b l u e\right) \left(2 {x}^{0}\right)}{\mathrm{dx}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = \textcolor{red}{2 {x}^{0}} - \textcolor{b l u e}{0}$

$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = \textcolor{red}{2}$