Category Archives: Derivative Theory

Foreshadowing the Chain Rule

I assigned another very easy but good problem this week. It was simple enough, but it gave a hint of things to come.

Use the Product Rule to find the derivative of {{\left( f\left( x \right) \right)}^{2}}.

Since we have not yet discussed the Chain Rule, the Product Rule was the only way to go.

\frac{d}{dx}{{\left( f \right)}^{2}}=\frac{d}{dx}\left( f\cdot f \right)=f\cdot {f}'+{f}'\cdot f=2f\cdot f'

 And likewise for higher powers:

\frac{d}{dx}{{f}^{3}}=\frac{d}{dx}\left( f\cdot f\cdot f \right)=f\cdot f\cdot {f}'+f\cdot {f}'\cdot f+{f}'\cdot f\cdot f=3{{f}^{2}}{f}'

If you just look at the answer, it is not clear where the {f}'  comes from. But the result foreshadows the Chain Rule.

Then we used the new formula to differentiate a few expressions such as {{\left( 4x+7 \right)}^{2}} and {{\sin }^{2}}\left( x \right) and a few others.

Regarding the Chain Rule: I have always been a proponent of the Rule of Four, but I have never seen a good graphical explanation of the Chain Rule. (If someone has one, PLEASE send it to me – I’ll share it.)

Here is a rough verbal explanation that might help a little.

Consider the graph of y=\sin \left( x \right). On the interval [0,2\pi ] it goes through all its value in order once – from 0 to 1 to 0 to -1 and back to zero. Now consider the graph of y=\sin \left( 3x \right). On the interval \left[ 0,\tfrac{2\pi }{3} \right] it goes through all the same values in one-third of the time. Therefore, it must go through them three times as fast. So the rate of change of y=\sin \left( 3x \right) between 0 and \tfrac{2\pi }{3} must be three times the rate of change of y=\sin \left( x \right). So the rate of change of  must be 3\cos \left( 3x \right). Of course this rate of change is the slope and the derivative.

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At Just the Right Time

This is about a little problem that appeared at just the right time. My class had just learned about derivatives (limit definition) and the fact that the derivative is the slope of the tangent line. But none of that was really firm yet. I had assigned this problem for homework:1

Find (3) and f ‘ (3), assuming that the tangent line to y = f (x) at a = 3 has equation y = 5x + 2

To solve the problem you need to realize that the tangent line and the function intersect at the point where x = 3. So (3) was the same as the point on the line where x = 3. Therefore, (3) = 5(3) + 2 = 17.

Then you have to realize that the derivative is the slope of the tangent line and we know the tangent line’s equation and we can read the slope. So f ‘ (3) = 5

In my previous retired years I wrote a number of questions for several editions of a popular AP Calculus exam review book.2 I found it easy to write difficult questions. But what I was after was good easy questions; they are more difficult to write. One type of good easy question is one that links two concepts in a way that is not immediately obvious such as the question above. I am always amazed at the good easy questions on the AP calculus exams. Of course they do not look easy, but that’s what makes them good.

Now a month from now this question will not be a difficult at all – in fact it did not stump all of my students this week. Nevertheless, appearing at just the right time, I think it did help those it did stump, and that’s why I like it.


1From Calculus for AP(Early Transcendentals) by Jon Rogawski and Ray Cannon. © 2012, W. H. Freeman and Company, New York  Website p. 126 #20

2 These review books are published by D&S Marketing Systems, Inc. Website


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Filed under AP Calculus Exams, Derivative Theory, Derivatives

Lin McMullin’s Theorem

Mathematics more often tends to delight when it exhibits an unanticipated result rather than conforming to … expectations. In addition, the pleasure derived from mathematics is related in many cases to the surprise felt upon the perception of totally unexpected relationships and unities.

- Mario Livio,  The Golden Ratio

I have a theorem named after me. I did not name it, but I did prove it – well more like I tripped over it. It is a calculus related idea. Here is how it came about. I say came about because as you will see I did not set out to prove this it. I just sort of fell in my lap as I was working on something else.

I was trying to do an animation of an idea that I had heard about: If you have a fourth degree, or quartic, polynomial with a “W” shape it has two points of inflection. If you draw a line through the points of inflection three regions enclosed by the line and the polynomial’s graph are formed. The areas of these regions are in the ratio of 1:2:1. In order to make the animation work I needed the general coordinates of the 4 points where the line intersects the quartic.

The straightforward way to proceed would be to write a general fourth degree polynomial,

f\left( x \right)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{2}}+{{c}_{1}}x+{{c}_{0}}

differentiate it twice to find the second derivative. Then find the zeros of the second derivative (by the quadratic formula), write the equation of the line through them, and then find where else the line intersects the quartic. Without even starting I realized that even with a CAS the algebra and equation solving was going to be really fun (Not!). So I decided on an alternative approach.

I decided to let the zeros of the second derivative be x = a and x = b, then at least they would be easy to work with. Then the second derivative is \displaystyle {{f}'}'\left( x \right)=12{{c}_{4}}\left( x-a \right)\left( x-b \right) where the {{c}_{4}}  is the leading coefficient of the quartic and the 12 comes from differentiating twice.

I integrated to get the first derivative and added {{c}_{1}}, the coefficient of the linear term, as the constant of integration. I integrated again and added {{c}_{0}}, the constant term. as the constant of integration. This resulted in the original quartic function:

\displaystyle f\left( x \right)={{c}_{4}}{{x}^{4}}-2\left( a+b \right){{c}_{4}}{{x}^{3}}+6ab{{c}_{4}}{{x}^{2}}+{{c}_{1}}x+{{c}_{0}}

Then I wrote the equation of y(xthe line through the points of inflection. It is too long to copy, but you may see it in the screen capture at the end of the post. (That is what is nice about a CAS: you really do not have to worry about how complicated things are.)

Then I solved the equation f\left( x \right)=y\left( x \right). Two of the solutions are x = a and x = b as I expected. The other two I did not expect. They are:

\displaystyle {{x}_{1}}=\frac{1+\sqrt{5}}{2}a+\frac{1-\sqrt{5}}{2}b and \displaystyle {{x}_{2}}=\frac{1+\sqrt{5}}{2}b+\frac{1-\sqrt{5}}{2}a

And that’s when I stopped astonished! Those numbers are the Golden Ratio \Phi =\frac{1+\sqrt{5}}{2},  and its reciprocal \phi ={{\Phi }^{-1}}=\frac{1-\sqrt{5}}{2}. So the roots are \Phi a+\phi b and \Phi b+\phi a. How did they get there?

To this day I have no idea why the Golden Ratio should be so involved with quartic polynomials, but there they are in every quartic!

There were no assumptions made about a and b – they could be Complex numbers. In that case there are no points of inflection, but the “line” and the quartic still will have the same value at the 4 points.

This was all about 10 years ago and until just this year I never checked the ratio of the areas. (They check.)

Here is a CAS printout of the entire computation.



Speaking of the Golden Ratio, the Calculus Humor website has a nice feature on the Golden Ratio in logos. To view it click here.

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Experimenting with CAS – Chain Rule

Discovering takes good data and a computer algebra system (CAS) can provide good data. You can do “experiments” by producing the results with a CAS and looking for patterns. As an example let’s look at how you and your students might discover the chain rule for derivatives.

One of the ways you could introduce the chain rule is to ask you class to differentiate something like (3x + 7)2. Not knowing about the chain rule, just about the only way to proceed is to expand the expression to 9x2 + 42x + 49  and differentiate that:  18x + 42 and then factor 6(3x + 7). Then you show how this relates to the power rule and where the “extra” factor of 3 comes from differentiating the (3x + 7).  You really cannot a much more complicated example, say a third or fourth power, because the algebra gets complicated very fast.

Or does it?

Suggest your students use a CAS to do the example above this time using the third power. The output might look like this:


But even better: what we want is just the answer. Who cares about all the algebra in between? Try a few powers until the pattern become obvious.


Now we have some good data to work with. Can you guess the pattern?

Nor sure where the “extra” factor of 3 comes from? Try changing the 3 in the original and keep the exponent the same.


Now can you guess the chain rule? See if what you thought is right by changing only the inside exponent.


Then you can try some others:


You can count on the CAS giving you the correct data (answers). Do enough experiments until the chain rule pattern becomes clear.

But I think the big thing is not the chain rule, but that the students are learning how to experiment in mathematics situations. In these we started by changing only the outside power. Then we kept the power the same power and changed the coefficient of the linear factor. Then we changed the power inside power, each time seeing if our tentative rule for differentiating composite function was correct and adjusting it if it was not. Finally we tried a variety of different expressions. You change things. Not big things but little things. You don’t jump from one trial to something very different, only something a little different.

You can do the same thing for the product rule, the quotient rule, maybe some integration rules and so on. You have accomplished your goal when the students can produce the data they need without your suggestions.

But be aware: sometimes this can lead to unexpected results. Does the pattern hold here?


Or here?


Hint: \frac{7}{3.2}=2.1875  and 3{{\left( 3.2 \right)}^{3}}=98.304

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