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The book I'm reading introduces derivatives via limits. It gives the following example:

$f(x) = 12x-3x^3$

$f'(x)=\lim{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

  1. $=\lim_{\Delta x\rightarrow 0}\frac{12(x+\Delta x)-(x+\Delta x)^3-(12x-x^3)}{\Delta x}$
  2. $=\lim_{\Delta x\rightarrow 0}\frac{12x+ 12\Delta x -x^3 -3x^2\Delta x -3x(\Delta x)^2-(\Delta x)^3-12x+x^3}{\Delta x}$
  3. $=\lim_{\Delta x\rightarrow 0}(12 - 3x^2 - 3x\Delta x - (\Delta x)^2)$
  4. $=12-3x^2$

I'm having trouble with how they got from step 2 to step 3. Where did $\Delta x$ on the bottom go?

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2 Answers 2

up vote 0 down vote accepted

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.

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OK, that was pretty dumb of me not to follow that... thanks. –  Robert S. Barnes Jul 15 '12 at 9:40

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.

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