# The derivative is a real limit?

In my classes, the derivative is usually defined as "the limit of the fractional incremental ratio". But I found out another way to define the derivative from an old book from Gardner & Thompson's "Calculus Made Easy". For example, if we have $f(x) = x^2$ and we want to calculate the derivative:

$$f(x) = x^2$$

So our y is

$$y = x^2$$

Right now we are only considering the increment of y and x so we can calculate that this way:

$$y+dy = (x+dx)^2$$ $$y+dy = x^2+dx^2+2xdx$$

I can remove $$dx^2$$ because it's a very small quantity related to our magnitude.

The result is

$$y+dy = x^2+2xdx$$

I subtract the original quantity $$y+dy-y=x^2+2xdx-x^2$$ $$dy=2xdx$$ $$dy/dx=2x$$

The derivative is equal to $$2x$$ and I calculate that without using any limits. So, my question is: the derivative is a real limit? What about the orders of magnitude? A limit is the representation of a value related to our magnitude?

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I don't really consider that as a "definition" for a derivative. How do you quantify when some term is small and you can leave it away? – Listing Nov 29 '11 at 10:08
This is basically how calculus was done in the early days (before Weierstrass, essentially). It's fast, easy, extremely non-rigorous and happens to give the right answer for simple problems (such as polynomials). We have come a long way since then. – kahen Nov 29 '11 at 10:20
This formal derivation hides the limit in the phrase "I can remove $dx$ because it's a very small quantity related to our magnitude". – Dirk Nov 29 '11 at 10:23
@kahen: but we've also circled back to that method, albeit with much more knowledge about how to make that type of arguments precise and rigorous. – Willie Wong Nov 29 '11 at 10:31
@Micro you might find the answers to this question quite relevant. – Bill Cook Nov 29 '11 at 15:00

That's exactly what limits are about: what happens when h (or dx, or whatever you call it) is very small.

More precisely, recall that $\lim f(h)=a$ iff $f(h)=a+o(1)$ (little-o notation); so $\lim\frac{f(x+h)-f(x)}h=a$ iff $\frac{f(x+h)-f(x)}h=a+o(1)$ or equivalently iff $f(x+h)=f(x)+ah+o(h)$. The last statement is just more formal way to say that $f(x+h)$ is equal to $f(x)+ah$ "up to next orders of magnitude".

This point of view of yours is, actually, not only more natural but also easier to generalize (in the direction of, say, multivariable calculus).

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An example, where thinking in terms of linear approximation is especially convenient: math.stackexchange.com/a/3105 – Grigory M Nov 29 '11 at 10:33
ok, but considering a more generic approach, what about trascendental function? what about when an order of magnitude is difficult or impossible to define? I made an example about a really simple function, also i know for sure that this a crescent function and it's always positive, how can i know more about this kind of approach with derivatives and limits, something that can be applied to each and all the several families of functions? – Micro Nov 29 '11 at 10:35
@Micro Let me stress, that this "little-o-style definition" is precisely equivalent to the "limit-style definition" -- so if a function has derivative it can be approximated by a linear function. But, of course, finding such an approximation is not always that easy (compared to what you describe in your post). For example $e^h$ is $1+h+o(h)$ but it's quite non-trivial theorem (which one can't prove just by writing down all terms and throwing away higher-order ones). – Grigory M Nov 29 '11 at 10:40
ok, assuming that it's a long story for a topic on the internet, at least i like to have the right big picture in mind, could you say what are the theorems that i can try to revisit to better understand this with this particular approach about the calculus ? – Micro Nov 29 '11 at 10:44
@Micro Oh... what would be really useful is some reference. Unfortunately I know nothing about English-language calculus text(book)s (and Knuth's comment suggests that maybe there is some problem here), but maybe someone else could help. – Grigory M Dec 1 '11 at 7:47

You actually are doing exactly the limiting process, only it's a bit hidden. If you keep that $dx^2$ term for a bit longer, we get $$dy = 2x dx + dx^2$$ and so $$\frac{dy}{dx} = 2x + dx.$$ Taking "$dx^2$ to be small" amounts to setting $dx=0$ on the right hand side of that last equation. Now let's compare this to the usual limiting method: $$\lim_{h\to 0} \frac{(x+h)^2-x^2}{h}$$ $$= \lim_{h\to 0} \frac{x^2+2hx+h^2-x^2}{h}$$ $$= \lim_{h\to 0} \frac{2hx+h^2}{h}$$ $$= \lim_{h\to 0} (2x+h).$$ In that last expression we can plug in $h=0$ to get that the derivative of $x^2$ is $2x$.

We get to the exact same expression (except instead of a $dx$ we have an $h$) in which we set our little increment to $0$.

In general, many limits can be evaluated by simplifiying algebraically and then just plugging in the point you're looking at. When you are taking the limit as something goes to zero, lots of times the "plugging in" part boils down to "this gets really small so let's ignore it."

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