# How to find the derivative of x^4?

Oct 15, 2015

$\frac{d \textcolor{red}{{x}^{4}}}{\mathrm{dx}} = \textcolor{b l u e}{4 {x}^{3}}$

#### Explanation:

In general
$\textcolor{w h i t e}{\text{XXX}} \frac{d \textcolor{red}{a {x}^{b}}}{\mathrm{dx}} = \textcolor{b l u e}{b \cdot a {x}^{b - 1}}$

This is a standard formula, but you could derive it from basic definitions of derivative:
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

In this case:
$\textcolor{w h i t e}{\text{XXX}} \frac{{\mathrm{dx}}^{4}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \frac{{\left(x + h\right)}^{4} - {x}^{4}}{h}$

$\textcolor{w h i t e}{\text{XXXXX}} = {\lim}_{h \rightarrow 0} \frac{\cancel{{x}^{4}} + 4 {x}^{3} h + 6 {x}^{2} {h}^{2} + 4 x {h}^{3} + {h}^{4} \cancel{- {x}^{4}}}{h}$

$\textcolor{w h i t e}{\text{XXXXX}} = {\lim}_{h \rightarrow 0} 4 {x}^{3} + 6 {x}^{2} h + 4 x {h}^{2} + {h}^{3}$

$\textcolor{w h i t e}{\text{XXXXX}} = 4 {x}^{3}$