# How do I find the derivative of f(x) = 6x^2 - 1 using the definition of a derivative?

## I understand that I am to plug-in components to make it look like f'(x) = 6(x+h)^2 - 1 - (6x^2 - 1)/h. But from this point onward I begin to get stuck. I am unsure as to what I am supposed to do, and the few places explaining what the next step is do not explain HOW to do that step.

May 28, 2018

$f ' \left(x\right) = 12 x$

#### Explanation:

By definition of a derivative

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

We want differentiate color(blue)(f(x)=6x^2-1, thus

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{6 {\left(x + h\right)}^{2} - 1 - \left(6 {x}^{2} - 1\right)}{h}$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\lim}_{h \to 0} \frac{6 \left({x}^{2} + {h}^{2} + 2 x h\right) - 1 - 6 {x}^{2} + 1}{h}$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\lim}_{h \to 0} \frac{6 {x}^{2} + 6 {h}^{2} + 12 x h - 1 - 6 {x}^{2} + 1}{h}$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\lim}_{h \to 0} \frac{6 {h}^{2} + 12 x h}{h}$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\lim}_{h \to 0} \frac{6 {h}^{2}}{h} + {\lim}_{h \to 0} \frac{12 x h}{h}$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\lim}_{h \to 0} 6 h + {\lim}_{h \to 0} 12 x$

$\textcolor{w h i t e}{f ' \left(x\right)} = 12 x$