Tag Archives: implicit differentiation

Tales from Zombieland, Calculus Edition, Part 2

The comments on part I have been fascinating. I want to reiterate that my math zombie’s teacher is not encouraging this behavior; I have no idea if she lectures or teaches using a more “progressive” style, but she certainly doesn’t believe that “procedural fluency leads to conceptual understanding”. A commenter also argues that “We Are All Math Zombies”. No. “Zombie” doesn’t mean “ran into the math ability wall”, nor does it mean someone who struggles with a topic and decides to forge through an obstacle, putting a black box around the difficulty to be returned to later, with more experience. I refer readers to the Brett Gilland definition of “math zombies” who “who can reproduce all the steps of a problem while failing to evidence any understanding of why or how their procedures work”.

Back to it–we are now into the “rules” questions, 3 through 8. She did question 3 easily. Please remember that my knowledge of calculus is being pushed to the limit in this entire sequence. I found this nifty derivative calculator so non-calculus folks can see how much rote algebra my zombie was doing, mostly correctly, again with no understanding.

Problem up: question 4: g(x) = (x2 + 1)(x2 – 2x)

She began by just taking the derivative of both terms and multiplying them.

“Um, no.”

“You don’t just multiply them?”

“Didn’t you do a bunch of rules? Product, Power, Chain, Quotie….”

She looked vague, but I was pretty firm on this point. “Look, you have to stop being so helpless. This math hasn’t been imposed on you by some fascist regime. Turn back a page or two in the book again.”

And then, a page or two back, when she spotted the product rule, “Oh, yeah.”

And she instantly started into the procedure.

“Stop. STOP!!! What the heck are you doing?” She looked at me in confusion.

“You’ve done this before. You have no memory of doing this before. Now you’re all oh, yeah, mindlessly working a routine you didn’t even recognize 30 seconds ago. Your next two years are going to be a case of lather rinse and repeat if you don’t start forging some memories, some connections.”

“I’ll just forget it again.”

“Then stop making yourself crazy and go take actual pre-calc.”

“I don’t even think that exists in my school.”

“Then listen up. What you know how to do is find derivatives of individual terms added together. First step is to realize that multiplying, dividing, or exponentially changing functions is more complicated. So there are separate rules that build on the easier, basic task of finding derivatives of individual terms.”

I wish I could say I broke into her drive for “just do something”, but at least she slowed down a bit. “But I wrote it down.”

“You did that the first time. So let’s try something different. Repeat this. The Product Rule: multiply the derivative of the first term by the second. Add it to the derivative of the second term times the first.”

“Yeah, I wrote it down.”

“No, you wrote down an abstraction. Say it.”

“What, like in words?” I looked at her sternly.

“Okay, I take the derivative of the first term. Then I…multiply it…”

“Stop. You’re into memorization, so memorize. But words, not symbols. The Product Rule: multiply the derivative of the first term by the second term. Add it to the derivative of the second term times the first.”

She repeated it patiently; I made her do it two more times.

“Okay, now you can work the problem.”

(I have no evidence for the notion that auditory/oral repetition helps, but intuitively, it seemed to me that the many rules are easier to remember by focusing on what the actions are, rather than what they look like. I lunched a few days later with my friend the real mathematician and department head, who told me that he requires his students to write–yea, write, Barry and Katherine!–a description of the product, quotient, and chain rules in addition to the algorithms. “Whenever I had to recall them in college, I remembered them verbally first.”)

Did you know there were online derivative calculators? So for those who want some kind of idea what she did, I’ll link these in.

“I always wondered if you can just distribute the product and use the power rule,” I mused, scratching through the steps. “Looks like you can. (x2 + 1)(x2 – 2x) expands to x4-2x3+x2-2x which…has a derivative of 4x3-6x2+2x-2.”

“That’s what I got. But why would you multiply it out when you can use the Product Rule?”

“Oh, I dunno. Maybe some people forget the Product rule temporarily. But if they actually understood the math, they could just think hey, no problem. I’ll just expand the terms until I can look up the rule. Or until it occurs to me to look up the rule, since you were stuck on that step until I showed up.”

She allowed as that was true. “But you can’t do that with the quotient rule.”

“I’m not good enough at this to know for sure. But most of the time you’d have a remainder, which would be expressed as a quotient, so it’s kind of reiterative. Question 5 is a fraction that is, I think, always going to be less than 1, so I’ll take a crack at doing the division on question 6 while you work out the quotient rule on both problems.”

“But how can I find a derivative of a cube root?”

“Gosh, wouldn’t it be great if there were a way to express a root as an exponent?”

“Oh, that’s right.” And she set to work on some rather complicated algebra and then stopped. “How do you know that this will always be less than 1?”

“Well, look at it. I’m dividing the cube root of a number and dividing it by its square. So think about taking the cube root of, say, 8? which is 2. Then dividing it by 8 squared + 1, which is 65. Even if x is less than 1, I’m adding 1 to the square of the fraction, so that sum will always be greater than the cube root of a positive fraction less than 1. I think, anyway.” Her eyes had long since glazed over, but I confess–I graphed it just to brag.


“I finished question 5, but it doesn’t match the book.”

I looked. “No, you didn’t drop the power on the cube root. It’s going to be negative two-thirds, which will move it to the denominator.”

She redid the problem while I did long division on problem 6, getting -1 with a remainder of -2x+2. Since the derivative of the constant was zero, I then had to take the derivative of the remainder (divided by x2-1).

“It just occurred to me I could use the Chain Rule here, too. Huh. I wonder if that means all quotient derivatives could be worked with the chain rule.”

Our answers to number 6 matched up, and my student was mildly interested. “So I can find derivatives with more than one method?”

“As is usually the case with demon math. But file this away with ‘repeat the processes verbally’ as a means of survival strategy.”

She worked her way through the next group, enduring my comments patiently but with little interest. I kept plugging away, trying to get her to think about the math–not because I wanted her to share my values, but I thought the conversations might create some memory niches.

So when she worked the derivative for problem 10: “hey, that’s interesting. That graph will always be negative, which means the slope at any point on the original graph will be negative.”

“What? How can you tell?”

“No, you can figure this out. Look at it closer.”

“It’s negative 8 divided by…oh, I see. Squares are always positive. So it’s a negative divided by a positive.”

“So that means that no matter what point we put in…” I prompted.

“Wait. Every slope is negative? No matter what?”

“I wonder if it’s always true for reciprocal functions. Huh.”

“Is that a reciprocal or a hyperbola.”

“Huh. I….think… they’re the same thing? Or a reciprocal is a type of hyperbola? Not sure. Good question. A hyperbola is a conic, I know, and I’m more familiar with transformations than conics.” (Answer is yes, a reciprocal function is a rectangular hyperbola.)

Then, when we got to problems 11 and 12: “Look, you need to remember that a square root function will in all cases turn into some sort of reciprocal function. You keep on messing up the algebra and aren’t catching it because you aren’t thinking big picture.”

“I don’t see why it’s a negative exponent.”

“What do you always do with exponents in derivatives?”

“You subtract….oh! I’m always subtracting 1 from a fraction.”

“Bingo. And negative exponents are..”

“they’re reciprocals, you’re dividing. Okay.”

“But look at the bright side. You actually understood this question.”

“I do! You really have helped.” I beamed. And she was able to work problem 13, finding a derivative given a graph, without help when an hour earlier she couldn’t. Progress, at least in the short term.

Problem 14 was interesting. “Determine the points at which the graph of f(x) = 1/3x3 – x has a horizontal tangent line.”

“Should I use implicit differentiation?”

“What? No. Well. I don’t really grok implicit differentiation, but that’s not what this one is asking. What does a horizontal line have to do with slopes?”

“Horizontal lines have a slope of zero. So the rate of change is zero? It’s asking where the rate of change is zero? The derivative is….x2 – 1.”

“Which factors to (x-1)(x+1). Hmmm.”

“So it is implicit differentiation?”

“No. Look, I don’t know what implicit differentiation is specifically, but it always involves y. This is….I’m just confused, because the point at which this parabola has a slope of 0 is the vertex, which is x=0.”

“Yeah, the slope of the parabola isn’t what I’m looking for, right? That means the slope of the other graph is 0 and I should plug in 1 and -1.”

I looked at her, impressed. “My work here is done.”

“What, I’m wrong?” She quickly worked the problem. “It’s positive and negative 2/3. That’s what the book says, too.”

“You’re not wrong at all. I was the one who was confused and you spotted the problem. Very good!”

“But why couldn’t I have used implicit differentiation?”

“Look, you need to talk to your teacher about that because it’s at the edge of my knowledge. I know that working the math of implicit differentiation is easier than understanding it. But at 90,000 feet, what you need to remember is that you use implicit differentiation when you can’t isolate y, so your equation has two variables. Circles and ellipses, for example. Or some of those other weird circular graphs. Look at problems 16-19, for example. Anyway, the derivative on this one was simple. The crux of the question was the link between the zeros of the parabola and the rate of change on the original graph.”

And with that, our ninety minutes were up. I tried to talk the mom out of paying me, since I’d learned a lot and wasn’t an expert, but she insisted.

Some observations:

She was capable of some pretty brutal algebra without any real understanding of what she was doing, time and again. That’s the zombie part–that and the fact that she really didn’t much care about anything other than plowing through. She wasn’t ever really interested but hey, all this stuff the tutor was saying seemed to help, so play along.

I learned a great deal, in ways that will further inform my pre-calculus class curriculum. Can’t wait to try it out. I also wrote out a lot of equations and may have made typos, so bear with me. And yeah, that’s how I remember implicit differentiation–it’s the one with “y”. I get the basics–normally it’s just x changing, this is saying they both change with respect to each other, or something. Implicit differentiation is the point at which I start to realize that the algebra of the differentiation language (dy/dx) has its own logic and wow, a chasm of interesting things of which I know nothing about opens and threatens to swallow me up so I look away.

I’ve really increased my understanding in advanced (high school) math over the past few years, and going back into calculus armed with that additional knowledge has led me to think—really, for the first time—about the lunacy involved in high school calculus instruction. I am starting to understand how math professors could be dismayed at the total ignorance demonstrated by students who scored 5 on the BC Calc test.

Finally, consider that this student is taking pre-calculus. Her transcript reflects pre-calculus. Yet the content is clearly calculus. Meanwhile, I teach a lot of second year algebra with an analytic geometry spin in my pre-calc class. Most schools fall somewhere in between. This is why I laugh when people propose doing away with tests and using grades and transcripts. I still believe in good tests, despite my increased awareness of cheating and gaming.

This enormous range of difficulty and subject matter reflects the bind faced by high schools kneecapped by our education policy. We must offer all students “college level” material, and our graduation and class enrollments are scrutinized closely by the feds and civil rights attorneys ever in search of a class action suit. So we have to move kids along, since we can’t fail them and can’t offer them easier courses. So we have to try and teach good, solid math that isn’t too much of a lie. That’s what I do, anyway.

Maybe things will change with the new law. I’m not counting on it.