# How do I use the limit definition of derivative to find f'(x) for f(x)=x^3-2 ?

Jun 5, 2018

$f ' \left(x\right) = 3 {x}^{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} - 2$

The limit definition states that:

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

In our case, this is:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\left({\left(x + h\right)}^{3} - 2\right) - \left({x}^{3} - 2\right)}{h}$

Straight away, the $- 2$s cancel with each other, we also expand the term in the brackets to get:

$= {\lim}_{h \to 0} \frac{{x}^{3} + 3 {x}^{2} h + 3 x {h}^{2} - {x}^{3}}{h}$

The ${x}^{3}$ cancel each other leaving us with:

$= {\lim}_{h \to 0} \frac{3 {x}^{2} h + 3 x {h}^{2}}{h}$

Cancel the $h$s on the top with the bottom to get:

$= {\lim}_{h \to 0} \left(3 {x}^{2} + 3 x h\right)$

Evaluating this limit, there is no $h$ on the first term so it will stay put. As $h$ on the 2nd term tends to 0, the second term will vanish completely to leave us with:

$f ' \left(x\right) = 3 {x}^{2}$