Understanding of Derivatives via Limits The book I'm reading introduces derivatives via limits.  It gives the following example:
$f(x) = 12x-3x^3$
$f'(x)=\lim{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$


*

*$=\lim_{\Delta x\rightarrow 0}\frac{12(x+\Delta x)-(x+\Delta x)^3-(12x-x^3)}{\Delta x}$

*$=\lim_{\Delta x\rightarrow 0}\frac{12x+ 12\Delta x -x^3 -3x^2\Delta x -3x(\Delta x)^2-(\Delta x)^3-12x+x^3}{\Delta x}$

*$=\lim_{\Delta x\rightarrow 0}(12 - 3x^2 - 3x\Delta x - (\Delta x)^2)$

*$=12-3x^2$


I'm having trouble with how they got from step 2 to step 3.  Where did $\Delta x$ on the bottom go?  
 A: You have:
$$=\lim_{\Delta x\rightarrow 0}\frac{12x+ 12\Delta x -x^3 -3x^2\Delta x -3x(\Delta x)^2-(\Delta x)^3-12x+x^3}{\Delta x}$$
Note that $\color{#bb0000}{12x} + 12\Delta x - \color{#00bb00}{x^3} - 3x^2\Delta x - 3x (\Delta x)^2 - (\Delta x)^3 - \color{#bb0000}{12x} + \color{#00bb00}{x^3} $ $= 12 \Delta x - 3x^2 \Delta  x - 3x (\Delta x)^2 - (\Delta x)^3$
So they just cancel out a $\Delta x$ everywhere.
A: First, I think you made a little typo.
I think it should be $f(x)=12x−x^3$
That being said, the problem is quite simple:
$$
\begin{align}
f'(x) &= \lim_{\Delta x\rightarrow 0}\frac{12x+ 12\Delta x -x^3 -3x^2\Delta x -3x(\Delta x)^2-(\Delta x)^3-12x+x^3}{\Delta x}\\
&= \lim_{\Delta x\rightarrow 0}\frac{ 12\Delta x -3x^2\Delta x -3x(\Delta x)^2-(\Delta x)^3}{\Delta x}\\
&= \lim_{\Delta x\rightarrow 0}\frac{ (12 -3x^2 -3x\Delta x-(\Delta x)^2)\cdot \Delta x}{\Delta x}\\
&= \lim_{\Delta x\rightarrow 0} (12 -3x^2 -3x\Delta x-(\Delta x)^2)
\end{align}
$$
The rest I assume is clear to you.
