I’ve gotten back into my quest to learn calculus, but I’ve run into a little hiccup in the form of what appears to be a fundamental error in the book I’m using (Silvanus P. Thompson’s *Calculus Made Easy*, Second Edition (the PDF version at that link).

On page 18, we see this paragraph:

What does (dx)^2 mean? Remember that dx meant a bit – a little bit – of x. Then (dx)^2 will mean a little bit of a little bit of x; that is, as explained above (p. 4), it is a small quantity of the second order of smallness.

It may therefore be discarded as quite inconsiderable in comparison with the other terms.

Emphasis mine. This is the part that really made me go “Huh?”, but I figured that the author knew more than I do, and decided to leave it.

Then later, on pages 19 & 20, we are given an example of differentiating y=x^3, and just to make sure I was understanding things correctly I tried to work it out. I decided I would have x=3 and dx=1. Here is the example from the text:

This ends with *dy/dx = 3x^2*, but that part won’t copy for some reason.

But, there’s a problem. If, as I did, you set *x=3* and *dx*=1*, *then you end up with *y=27* and *y+dy=64*, making *dy=37*. But, if you take *(3x^2)(dx)* when *x=3* and *dx=1*, you end up with *dy=27*.

So, I tried working it out without discarding *(dx)^2* and *(dx)^3*, and got:

*y + dy = x ^{3} + 3x^{2}(dx) + 3x(dx)^{2} + (dx)^{3}*

*y = x ^{3}*, so they cancel and give:

*dy = 3x ^{2}(dx) + 3x(dx)^{2} + (dx)^{3}*

*dy = 3(3 ^{2})(1) + 3(3)(1^{2}) + (1)^{3}*

*dy = 27 + 9 + 1 = 37*

The right answer.

A 10 out of 37 error is *not* negligible as the book asserts. So, can anyone who knows calculus tell me who is wrong here? Me, or the book? And why?

Help!

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## wfgodbold

/ April 25, 2012dx isn’t really a specific number. And as you’re using it, it’s more like Δx (from how I remember calculus explaining things). Your book explains differently from how I learned it.

If you substitute Δx in for dx in your equations, then with a large Δx (large change in x), you’ll get large variations.

The derivative (dx) is what you get when you take the limit as Δx approaches 0.

So for Δx=1, you get a large Δy, since the difference is large.

So (y+Δy) = (x+Δx)^3 = x^3 +3x^2 *Δx + 3x *Δx^2 + Δx^3.

Since y=x^3, you can cancel those terms out, leaving you with Δy=3x^2 *Δx +3x *Δx^2 + Δx^3.

Now you take Δx=1; this is large. If x=3, and Δx=0.1, you get Δy = 2.791.

Basically, the smaller Δx gets, the faster the last two terms become negligible.

The first derivative is the limit as Δx approaches 0 of ((y+Δx) – y)/(x+Δx – x). since you can cancel out the y and x terms, this becomes the limit as Δx approaches 0 of Δy/Δx

Now, if you divide the Δy=3x^2 *Δx +3x *Δx^2 + Δx^3 by Δx, you get:

Δy/Δx = 3x^2 + 3x*Δx +Δx^2.

Since the derivative is the limit as Δx approaches 0, the last two terms drop out (because they have Δx terms and the 3x^2 term does not) and you get dy/dx = 3x^2. You can test this out by calculating (3+Δx)^3-(3^3))/(Δx); the closer Δx gets to zero, the closer you get to the actual value of dy/dx at x=3, which is 27.

I had an eventual point, but it kind of got away from me. I don’t know if any of that helped.

Good luck!

## DirtCrashr

/ April 25, 2012Good luck with that, I failed pre-Calculus, twice! After they finished experimenting on us lab-rats with “New Math,” I could barely pass Algebra.

## aretae

/ May 6, 2012are you still looking for help?

## Jake

/ May 7, 2012Right now I’m trying to go on further into the book, and see if this bit falls into place later on.

Thanks for the offer! I certainly plan to post any other questions I have here as they come up (and I’m sure they will!). The most irritating part right now is that I keep having to go back and refresh my memory of algebra and trig rules and principles that I haven’t needed in 10-15 years.

## aretae

/ May 7, 2012The key here is that dx is talking limit cases. Did the first few chapters of the calculus book talk limits? Usually it does…and that’s a huge part of the thought process here. dx is a limit case…when you think dx, you can’t think 3 or 1, you have to think 0.00001. If you think dx=0.00001, it all works how you want. Your dx is too large to see the limit.

GL. Overall, I’m happy to help…math degree…I’ve taught calculus…

## Jake

/ May 7, 2012Nope. It starts with differentials. I just skimmed through, and don’t see any mention at all of limits until chapter 14. The idea of the “limiting case” gets a whole two sentences on page 15, and I think that’s it.

## aretae

/ May 7, 2012Jake,

There’s 3 big ideas in Calculus.

1. Limits.

2. Derivatives

3. Integrals.

Limits discusses the idea of what happens when you look at tiny changes…changes that are arbitrarily close to zero.

The question of what happens arbitrarily close to zero on the function (x^3+2x)/x is different than the question…what happens *at* zero.

The first question is the one we’re asking about derivatives…and the one about dx. dx means a _tiny_ change in x. if you are working with whole numbers…and you think about dx=0.001 then you’re pretty close.