You Already Know Calculus: Derivatives

Calculus is made to be a whole big hoopla in high school and first year university. It’s supposedly the hardest math class in high school, notoriously complicated and unintuitive.

I blame this on bad education, not just because I’ve observed so much bad math education at the high school level, but because I can’t see any other way anyone could conclude that calculus is difficult. Because I believe that everyone already knows calculus. They just never connect what they already know to the symbols they’re manipulating in math class.

(I’m writing this series of posts — yes, this is only the first of a number of posts! — in a didactic form, but I think they may be of interest to mature mathematicians. While they are (will be?) informal and unrigorous, they provide intuitive reasons for why everything in basic calculus is true. The results in it are certainly valuable to me: they’re the result of me spending some time trying to answer ‘why’ everything is true at a compelling level. If nothing else, they may be useful in teaching calculus.)

So what is calculus? It’s the mathematical study of rates of change, nothing more and nothing less. We call the rate of change of a function its derivative.

The fact of the matter is that you deal with derivatives all the time. They’re part of the way you think. For example, the speed of your car is the derivative of the distance you’ve traveled and acceleration is the derivative of the speed.

A large part of the calculus curriculum is concerned with the “rules of differentiation” which are just a couple of identities regarding derivatives that are useful in calculating them for a function. With the possible exception of one, you already know all of them, though some may take a tiny bit of prodding to draw out. We shall begin with a few rules that you not only know, but intuitively use on a regular basis.

Sum rule is an identity for when two functions are added together. For example, consider a person walking on a boat: their distance from the shore is the distance of the center of the boat from the shore plus the amount they’ve walked away from the shore relative to the mast..

Well, their speed is the speed of the boat plus their speed relative to the boat! Obvious enough, and that’s sum rule. To write it out, we will use the prime notation, with $f'$ representing the derivative of $f$ . Then the sum rule can be stated as $(a+b)' = a' + b'$ . Just don’t let the symbolic version make you forget the real thing!

But suppose they’d been walking in the opposite direction. Then their position would have been the distance from the shore to the boat’s mast minus the distance they’ve walked from the mast towards the shore. And obviously their speed would be the speed of the boat minus their relative speed! That’s difference rule: $(a-b)' = a' - b'$ .

As another example, consider the amount of water in a bath tub: It’s the amount that has gone in minus the amount that has gone out. And clearly the derivative of the amount of water in the tub, that is the rate at which the amount of water in the tub is changing, is the rate at which it is entering minus the rate at which it is leaving!

Now lets consider a very different scenario. You’re laying back on your couch and watching a movie. A runner dashes through the screen!

Now things are slightly tricky because movie time can flow differently than normal time (fastforward, rewind, etc). So at any point in real time t, the movie player is at a certain point in movie time, $m(t)$ . Normally video time and real time flow at the same rate — one movie second passes per real second — so the derivative of movie time, $m'(t)$ , is 1. But when we fastforward, it might be $4$ and when we rewind it might be $-4$ .

Now consider our runner friend. Let’s call him Bob. Bob is running at $5 m/s$ in the movie, and normally that is what we observe him doing as well. But when we hit fastforward he moves (to us) at $20 m/s$ , even though in movie time he is only going at $5m/s$ . That’s perfectly natural, the speed Bob seems to be going at is the speed he moves at in the movie times the rate movie time is passing relative to real time.

Let’s try and formalize this a little bit, let Bob’s position be describe by a function $p$ . $p(m(t))$ is his position at a point in real time. $p'(m(t))$ is Bob’s in-movie speed he at a point in real time. $m'(t)$ is the rate movie time passes with respect to real time. And, as we agreed, his in-reality speed is those two multiplied: $p'(m(t))*m'(t)$ .

More generally, the rate at which $f(g(x))$ changes (ie. $f(g(x))'$ ) is the rate at which $f$ changes at $g(x)$ (ie. $f'(g(x))$ ) times the rate $g(x)$ changes (ie. $g'(x)$ ).

This is what we call chain rule: $f(g(x))' = f'(g(x))g'(x)$

(If you don’t immediately get the chain rule, don’t worry! While you already know the underlying idea, it is more complicated to formalize than the previous ones. Go back over the last few paragraphs and pay special attention to what each part of the formula means. )

Now, when you were watching Bob, you were fairly into the movie and so you were thinking about his movement in terms of in-movie meters. But now imagine you are somewhat more detached. He’s moving across the screen at 2 (real) cm/s. Now you go watch the same scene at your friend Sally’s house because her television screen is twice as big (on both axis, not area! So it’s a massive difference!). And — jolly gosh! — Bob now moves 4 (real) cm/s. Imagine that!

This seemingly obvious fact is given the (extremely creative) name ‘constant multiple rule’ and goes $(af(x))' = af'(x)$ . If you multiply a function by a constant amount, you multiply its rate by the same amount. If the world becomes twice as big, everything goes twice as fast!

(Together with sum rule, the constant product rule means that differentiation is linear. That is to say $(af(x) + g(x))' = af'(x) + g(x)$ .)

Another question: at what rate does something that doesn’t change change? How fast does a person who is standing still move? If you answered, nada, it wasn’t changing, zero, they didn’t move, zero meters per second or any similar answers, you were correct. The derivative of a constant is zero. That is to say, if $k$ is a constant, $k'=0$ . This fact is called constant rule, and it is definitely in the running for the most obvious rule of differentiation.

Now, lets consider something else entirely. You may not know this, since the way we teach trigonometry approaches being brain-dead, but cosine and sine are, respectively, the horizontal and vertical components of the point at angle theta on the unit circle. (As far as I’m concerned, this is how we should define cosine and sine. I can’t really go into detail with this, though, since it is beyond the purposes of this post.)

Our units are radians, the distance we’ve swept out on the circle. So if we consider our point on the circle as a function of the angle, we’re moving at unit speed.

So now, imagine yourself running along the perimeter of a circle with a radius of one meter. You’re going at 1 m/s. At every point, you’re moving tangent to the circle, at a speed of one meter a second.

That arrow is the derivative! Imagine moving it so that it’s base was at the center of the circle.

So you see, the derivative of moving in a circle is the position a quarter of a turn in front of you (with appropriate changes in units). If we split up the axes we get:

$\cos'(\theta) = \cos(\theta + \frac{\tau}{4}) = \cos(\theta + \frac{\pi}{2}) = -\sin(\theta)$

$\sin'(\theta) = \sin(\theta + \frac{\tau}{4}) = \sin(\theta + \frac{\pi}{2}) = \cos(\theta)$

If your having trouble following, focus on the underlying idea: If you walk around a circle with a radius of one meter, at a speed of one meter a second, you are walking tangent to the circle (and, get this, your speed is one meter a second!). If you imagine this as an arrow pointing out from the center of the circle, it lands on the circle a quarter turn ahead of you.

And thats it for the post! We just went through about a third of a high school calculus course, and you knew it all!

In the next post, I will introduce an alternative way to look at derivatives, differential forms, in light of which the remaining rules of differentiation will become obvious.

Continued in: You Already Know Calculus: Differential (One) Forms

Some Notes

Please leave a comment below if there was something you couldn’t follow or a question you’d like to ask. I’m far from a perfect author, and my explanations often leave much to be desired.

This first post didn’t have the most interesting content, but things should get better. My current plans for the series is:

Derivatives (Sum/Difference rule, chain rule, trigonometric derivatives)
Differential Forms (Product Rule, Power Rule, exponentials)
Integration (from differential forms perspective)
Partial Derivatives, Jacobian, and Determinants
n-forms
Stokes Theorem

Several of these are mostly written and just need some editing. The first three were originally going to be a single post, but I was persuaded that shorter is better.

I also want to talk about gradients, but I’m going to do that in a different context: the geometry of implicit functions. (I have plans for a CAD program based on this that I’m working on implementing.)

Regarding whether people already know calculus… I find it difficult to imagine it not being the case. People deal with rates all the time.

But I have to admit that I really don’t have much insight into how most people think. Most people aren’t really interested in talking about this kind of stuff with me (even when they want me to help them with their homework, they usually want to finish the problem set, not understand the concept) which leaves me with only my friends to probe, and they aren’t exactly representative of society at large.

For now, I’m going to stick with my claim. I certainly think it is true for the large majority of the population. And if nothing else, it’s an awesome title!

Tags: calculus, math

This entry was posted on July 31, 2011 at 05:48 and is filed under YAKC. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

17 Responses to “You Already Know Calculus: Derivatives”

Anonymous Says:
July 31, 2011 at 06:48 | Reply
Is f'(g(x)) the same as f(g(x))’ ?
- George G. Says:
  July 31, 2011 at 13:40 | Reply
  No. Suppose f(x):=x^2, g(x):=3x
  
  then:
  
  f'(g(x))=2(3x)=6x and f(g(x))’=3*2*(3x)=18x.
- christopherolah Says:
  July 31, 2011 at 14:00 | Reply
  Nope, not quite. f’(g(x)) is the value of (the rate f is changing) at g(x), where as f(g(x))’ is the rate f(g(x)) is changing at x.
- Anonymous Says:
  July 31, 2011 at 19:55 | Reply
  Thanks for the clarification. It was not clear if the ‘ placement before or after the function arguments was just a stylistic choice.
  - christopherolah Says:
    July 31, 2011 at 20:37 | Reply
    And thank *you* for pointing out to me that this wasn’t obvious to a reader unfamiliar with the notation. I’ve added the paragraph:
    
    >More generally, the rate at which $f(g(x))$ changes (ie. $f(g(x))'$ ) is the rate at which $f$ changes at $g(x)$ (ie. $f'(g(x))$ ) times the rate $g(x)$ changes (ie. $g'(x)$ ).
    
    Which will hopefully clarify things.
    
    If you could give some more in depth comments on how you found this essay, it would be greatly appreciated.
tolomea Says:
July 31, 2011 at 07:49 | Reply
+1 for the use of tau

Also you have two obvious errors in the notes, the first is “my explanations often leave ??? to be desired”, somthing has to be left there, generally “much”.

The other is “exponentials_” that “_” should be a “)”
- christopherolah Says:
  July 31, 2011 at 13:36 | Reply
  Thanks for catching that. This is what I get for not proof reading after writing that last section… 🙂
- Anonymous Says:
  August 11, 2011 at 23:20 | Reply
  Seems to have successfully left you desiring it — whatever it is
George G. Says:
July 31, 2011 at 10:52 | Reply
Nice entry! +epsilon for using tau.

I am especially intrigued now that you’ve promised to write about differential forms and Stokes theorem, some of my favorite subjects. (On a somewhat unrelated note, you might find this book useful: http://arxiv.org/abs/math/0306194 it seems to be similar in tone and approach to your post.)

Of course, I thought your was post clear and perfectly intuitive. But the real question is: what would a [poor, miserable] person that doesn’t know calculus make of it?
- George G. Says:
  July 31, 2011 at 10:55 | Reply
  Oh, I forgot to ask: got any nifty, intuitive explanations for
  (fg)’=f’g+g’f and (f/g)’=f’g-g’f/g^2? I can think one, but it’s not truly all that clear.
  - christopherolah Says:
    July 31, 2011 at 14:02 | Reply
    I have a very nice one for product rule, in the context of differential forms. And quotient rule follows simply from product rule.
    
    You’ll see in the next post 🙂
- christopherolah Says:
  July 31, 2011 at 13:57 | Reply
  > +epsilon for using tau.
  
  Yay Tau!
  
  >On a somewhat unrelated note, you might find this book useful…
  
  Oooh, just a quick glance and I like this textbook. In the course I took at UofT we used Munkres’ Analysis On Manifolds, which I’ve come to rather dislike. I already knew a substantial part of the content and that book confused me — it took hours to figure out the connection between alternating tensors and differential forms, which were treated as a simple matter of definition. (On the other hand, I liked Munkres’ Topology, though that may say more about the previous textbooks I used…)
  
  > what would a [poor, miserable] person that doesn’t know calculus make of it?
  
  This is indeed the question. Once I have a few more posts up, I’m going to post this here and there. With luck, I’ll be able to attract a few such people.
Sacha Chua Says:
August 1, 2011 at 21:43 | Reply
Thanks for the teaching tips! I’ll see if I can sneak calculus into the everyday math we’ve been teaching J-, who’s 13. =)
- christopherolah Says:
  August 2, 2011 at 00:02 | Reply
  Awesome! Please tell me how it goes. 🙂
Ivan Avery Frey Says:
August 8, 2011 at 03:21 | Reply
Is there a syntactical mistake above when you wrote “And, as we agreed, his in-reality speed is those two multiplied: p'(m(t)) * m'(t)”?

Didn’t you mean to say his apparent speed?
- christopherolah Says:
  August 8, 2011 at 12:22 | Reply
  I’m trying to contrast in-movie and in-reality speed. I guess I could also have described it as apparent speed, but then what’s his real speed — the one in the movie?
  
  In any case, I don’t see anything grammatically wrong with the way I handled it. Could you be more explicit with where you see the syntax mistake?
  - Ivan Avery Frey Says:
    November 25, 2011 at 20:06 | Reply
    Hi Chris,
    I’m sorry if I wasn’t explicit enough. Upon reflection what I found confusing was the use of the terms of in-movie speed and in-reality speed.
    
    In movie-speed obviously refers to the Bob’s speed when the movie is played back at normal speed and in-reality speed refers to Bob’s speed when the movie is played back at four times normal speed.
    
    I would have preferred to use Bob’s apparent speed when the movie is played back at normal speed and Bob’s apparent speed when the movie is played back at four times normal speed. Maybe the terms B1x and B4x could be used.