# Question #1d4dc

Jan 27, 2017

There is more than one method called the three-step method. (The steps are described differently.) They have the following in common.

#### Explanation:

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

We'll need to find $f \left(x + h\right)$ and $f \left(x + h\right) - f \left(x\right)$ and $\frac{f \left(x + h\right) - f \left(x\right)}{h}$

Some call this steps 1 and 2, others call all of this step 1.

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

$= 2 {x}^{3} + 6 {x}^{2} h + 6 x {h}^{2} + 2 {h}^{3}$

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

$= 2 {x}^{3} + 6 {x}^{2} h + 6 x {h}^{2} + 2 {h}^{3} - 2 {x}^{3}$

$= 6 {x}^{2} h + 6 x {h}^{2} + 2 {h}^{3}$

$\frac{f \left(x + h\right) - f \left(x\right)}{h} = \frac{6 {x}^{2} h + 6 x {h}^{2} + 2 {h}^{3}}{h}$

$= \frac{h \left(6 {x}^{2} + 6 x h + 2 {h}^{2}\right)}{h}$

$= 6 {x}^{2} + 6 x h + 2 {h}^{2}$

Now we need to find ${\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

${\lim}_{h \rightarrow 0} \left(6 {x}^{2} + 6 x h + 2 {h}^{2}\right) = 6 {x}^{2} + 0 + 0 = 6 {x}^{2}$

So $f ' \left(x\right) = 6 {x}^{2}$