As with most people that have studied mathematics beyond high school, I first encountered implicit differentiation in my initial semester of calculus. It seemed like magic then. In the intervening years, I have seen the idea and the underlying theory pop up again a few times, and it still seems like some kind of magic. I have worked through the appropriate proofs, and am continually astonished that it works. I probably shouldn’t admit this in public, but even today I feel like I have no intuition regarding the proof. The whole argument is a bit of a black box.

On the bright side, I didn’t have to prove the theorem to my calc I students (and, frankly, the proof probably shouldn’t be done in public—it should only be performed behind closed doors by small groups of hooded monks as a kind of hazing ritual for young math majors). For them, it is sufficient to see the power of the result.

### What I Taught

To get some idea of the incredibly fast pacing of the class, here is a very brief synopsis of what went on: I gave a quiz on trig functions (20 minutes), discussed homework problems which addressed the chain rule (25 minutes), introduced implicit differentiation (30 minutes), and used the theorem to derive the derivatives of a couple of inverse trig functions (20 minutes). In the normal course of a semester-long class, this probably would have been split over two or three days, with both the quiz and the homework questions taking place in a recitation section run by a grad student. Whee!

The main thrust of the day was introducing implicit differentiation. The basic idea is that an equation like \(x^2+y^2=1\) defines a curve in the plane (in this case, a circle of radius 1 centered at the origin). It is reasonable to ask what the slope of a line tangent to that curve at a particular point will be. Using the techniques developed prior to this moment, this involves solving the equation for \(y\) (which gives two results in this case—we have to choose the correct result before continuing), finding the derivative function, and evaluating that function for the value of \(x\) that is of interest.

This can be quite tedious, and can also be essentially impossible (consider a fifth degree polynomial in \(x\) and \(y\), for instance, which cannot be solved by radicals). Thus, instead of solving for \(y\) as an explicit function of \(x\), we note that the original equation implicitly defines a function in terms of \(x\). That is, the equation tells us that \(y=f(x)\) for some function \(f\), even if we don’t know exactly what the function is. (The existence and differentiability of \(f\) is the part that is pure magic, by the way.) To find the slope, we note that if \(x^2 + y^2 = 1\), then the derivatives of the left- and right-hand sides must also be equal, i.e.

\[

\frac{d}{dx}(x^2+y^2) = \frac{d}{dx}1.

\]

Since \(y\) is a function of \(x\), we apply chain rule to obtain

\[

2x + 2yy’ = 0,

\]

then solve for the derivative, getting

\[

y’ = \frac{-x}{y}.

\]

Now finding the slope is a “simple” matter of substitution.

In addition to this example, I also worked through the folium of Descarte and an example involving an astroid (not an ast*er*oid, but an astroid; namely \(x^{2/3}+y^{2/3} = 4\)). I finished up by showing that

\[

\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}

\qquad\text{and}\qquad

\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}.

\]

I stated the derivatives for the remaining inverse trig functions without further derivation.

### What Worked

When discussing the folium of Descarte, I worked through much of the problem explicitly using Maple. The goal was to impress upon the students how difficult a problem it is without implicit differentiation. When I issued the command for Maple to solve the equation for \(y\), the old computer that I was working on chugged for a bit, then displayed 10 lines of output, all gnarly notation with lots of radicals. It looked quite impressive, and made my point quite well, I think.

I am also pleased with the time management. While there was a lot to do, I never felt rushed, had time to address questions and concerns, and still managed to wrap up everything with two or three minutes to spare at the end of class to go over administrative stuff.

### What Didn’t Work

I am concerned that the lecture was a little scattered. I generally try to model clear, linear derivations and arguments, as those are the kinds of things that I want to see on quizzes and exams. However, there are several points in the lecture when I seem to have done too much algebra all at once, or skipped some minor simplification detail to the confusion of the students. This often involved backtracking to fill in details in the margins of the board, which was neither clear nor linear.

The biggest problem was taking the derivative of functions of \(y\), where \(y\) is understood to be a function of \(x\). In the first problem, I began by very explicitly writing \(y = f(x)\), where \(f\) is some unknown function. Then \(y^2 = [f(x)]^2\), and, using the chain rule, we have

\[

\frac{d}{dx} y^2

= \frac{d}{dx} [f(x)]^2

= 2f(x)\cdot\frac{d}{dx}f(x)

= 2f(x)f'(x).

\]

Recalling that \(f(x) = y\), this implies that \(\frac{d}{dx} y^2 = 2yy’\). I had planned to go through all of these details only on one problem, then elide them a bit (with more elision as the class went on) on future problems. Turns out, this was a bad plan. Every time that we took the derivative of some function of \(y\), we had to work through all of these details. Every. Single. Time.

Had I planned for it, I would have left space on the board. I didn’t, and it became marginalia, probably causing as much confusion as it cured. On the bright side, there will be homework questions to answer during the next class, so hopefully some of that confusion can be cleared up.