Trig Pictures, Part III

We almost have the whole picture in view.

Here’s where we left off:


In this diagram, the secant line segment is adjacent to the central angle θ.  And the tangent line segment is opposite that same angle.

But what if we want to draw our secants and tangents relative to the central angle 90-θ, the complement?  Let’s add those segments to the diagram and call them the cosecant and the cotangent.



Find similar triangles and then use ratios to establish each of the following:


(If you need help getting started, I did the first two of these for you in the last post.)


Find right triangles and then apply the Pythagorean theorem to establish each of the following:


Those two exercises reveal a lot of information all derived from that one diagram.  Once again…it’s better with pictures.

Picturing the Trigonometric Ratios: Step-by-step


Let’s start with the unit circle and a central angle θ that meets an arc on the circle:


The blue segment is called the “sine” of the angle.  It gets its name from a word that means “bowstring”.  Here is a diagram of a bow string truss.


So if you think of the intercepted arc as the bow and focus only on the upper half…


… you can see how they first decided to call that segment the “bowstring.”

But you do not have to hold your bow vertically.  If you use the complementary angle, 90-θ, you get the horizontal segment I have drawn in red (and re-drawn on the x-axis as well…you’ll see why soon.)


We can call that red segment the “bowstring of the complement”, or the “complement’s sine.”  Or we could just call it the cosine!

Now we are going to add two segments to the diagram: one that just touches the circle and one that crosses the circle.  We will use the same names we used for these back in geometry class.  As you can see from the labels, the orange one is called the tangent and the green one is the secant.


Now, as we examine this diagram, we can find a number of interesting pairs of similar triangles.  To discuss them, it would be helpful if we label some points of interest:


Now we have arrived at the key step.

Take a moment to convince yourself that ΔEBC is similar to ΔBDC. 

Then, since similar triangles have proportional sides, it must also be true that:


These are NOT arbitrarily chosen definitions.  They follow by necessity from the geometric definitions of these terms.

I will pause here to allow everyone to catch their breath.  In the next post, we will finish developing the diagram and add in the Pythagorean identities.

Also Better With Pictures — The Trigonometry Ratios

One of the digressions we follow almost every time I teach AP Physics is intended to give my students a better visual understanding of the 6 basic trig ratios.  But it also resolves several mysteries left over from their first introduction to trig.

When you first learned the trig ratios back in geometry class, you were probably taught the mnemonic SOHCAHTOA.  You can also find this many places on the web.  For example, here is a summary from


Already, we have a bit of a mystery.  Students don’t know where the word “sine” comes from, but that’s fine.  Lots of new topics introduce new vocabulary without going into the etymology of every new term.  But what about “cosine”?  What is co- about it?  A minor mystery to be sure, but even so…

Then we tell them about “tangent”.  We say it is “opposite over adjacent”, which is the TOA in SOHCAHTOA.  And a tiny bit of algebra shows that the tangent can also be written as sine over cosine.  But we still have a mystery:

What is “tangent” about the tangent? 

This is not just an etymology question.  The word “tangent” is already part of our math vocabulary.  We saw in geometry that at tangent line is a line that touches but does not cross a circle.  There are all kinds of rules about the angles and the lengths associated with tangent lines (and secant lines too, for that matter).  But what does that use of the word have to do with the way we are using it now, here in trigonometry?

From there, we move on to the reciprocal trig ratios:


And now you have more mysteries:

What does this secant have to do with the secant we learned about in geometry?

In geometry class, a secant is a line that crosses a circle.  Is there some reason we are re-purposing this vocabulary?

And as long as we are asking questions:

Why did you match “secant” with “one over COSINE”? And “cosecant”  with “one over SINE”?  Isn’t that making things unnecessarily complicated?

And once again:

What is “co-” about the cosecant?  And the cotangent?

All of these mysteries can be resolved with a single diagram.  I will develop the diagram step-by-step in my next post.  But for now, here it is in the finished form.  Ponder it well…


As you will see in the next posts, there are lots of similar triangles waiting to be discovered in this diagram.  Also, the Pythagorean theorem makes several useful appearances.

But for now, let’s resolve one of the mysteries:

Look at the segments labeled sine, secant and tangent.  Note their positions relative to the central angle, θ.

Now look at the segments labeled cosine, cosecant and cotangent.  Note their positions relative to the central angle 90-θ.  Each of these segments relate to 90-θ the way their corresponding segments relate to θ. And remember that 90-θ is called the “complement” of the angle.  So that’s where these segments get their names:

“the complement’s sine” —> cosine

“the complement’s secant” —> cosecant

“the complement’s tangent” —> cotangent

One mystery solved: “co-” is for “complement” !

More mysteries to be unraveled soon.


Those Flipping Inverse Graphs

"Sometimes the sophisticated thing is easier to understand 
than the kludgey thing. The flipped plots in the books were 
driving me crazy. They don't even pass the dimensional analysis 
 -- John Denker

I have been known to rant occasionally about some of the curriculum choices we have made for high school mathematics.  I feel the math we teach should be useful or interesting, and ideally both.  But it should also be true!  I think of that as the teacher’s equivalent of medical rule: first, do no harm.

What brings this to mind is an email discussion I have been having with Dr. John Denker, a physicist who I have “met” through the email list.  He has, as you see above, some very reasonable objections to the way that my book repeatedly points out the symmetry between the graph of a function and its inverse.  I would like to discuss his objections here in this post and also to show his suggested alternative. Then, in the spirit of this blog, our email discussion led me to remember a nice method for using graphs to evaluate iterated functions.  I will write about that in another post soon.

[By the way, to any current or future physics teachers out there: I strongly recommend joining the email list Phys-L.  It may be the single most helpful resource I know of, populated by many patient and enthusiastic experts, including Dr. Denker.  My current students can attest to the many occasions when my teaching has been assisted and influenced by ideas I encountered on Phys-L.]

OK, on to those flipping graphs…

CLAIM:  If you graph a function and its inverse on the same axes, the two graphs will have an interesting and esthetically pleasing symmetry with respect to the line y = x.

I am by no means the only one who makes this claim!  Search the web for “inverse function graph symmetry” and you will find that claim stated quite frequently.

For example, consider the squaring function, with its domain restricted to the non-negative real numbers.  It has an inverse function, the square root function, and sure enough, if you graph both functions on the same axes, you see the symmetry that I am talking about:


The graph above was made at, a very user-friendly graphing site.  And the graph certainly provides evidence in favor of the claim.  And I personally find this to be, as I said, esthetically pleasing.  Also, this idea is not without application.  When you learn a formula for finding derivatives of inverse functions, this symmetry will present a graphical way to understand that formula.  Still, if I had to guess, I suspect that most of the people who write about this claim do it because they like how the graphs look.  They look nice.


“If you graph a function and its inverse…”  But why would you ever do that?  As JD pointed out to me:

I don't think anybody ever needs to flip the axis to form 
the inverse function. Just read the axes in the other order.

That’s a good point.  For example, let’s look again at the squaring function by itself this time.


The graph is a convenient way to represent the relationship between the length of the side of a square and its corresponding area.  But if you already knew the area and wanted the length, you would not need to draw a new graph!  You would just need to know how to read the graph you already have!  You’d start on the vertical, area axis, go over to the curve and then read down to the horizontal axis where you would read the length.  Same graph, different purpose.



OK, I don’t need to graph the inverse function.  But what if I just want to because I like how it looks.  Is it wrong?  Does it violate the teacher’s version of the Hippocratic Oath?  Alas, not always, but often, and in a way that I should have considered, seeing as a major theme in my book is how math is actually used in science.  My current students may find it amusing that what I have overlooked are the units!  I may feel like graphing the two functions ON THE SAME AXES, but in most cases, it would be wrong.  Which brings us to…


The flipped graphs do not have the right units on each axis!

You don’t need to use sophisticated physics or chemistry to show this.  Our first graph above with the squaring function and the square root function illustrates the problem sufficiently.  Look again at the vertical axis.  Pick a number.  OK, 2.  But 2 WHAT? Is it inches or square inches? Meters or square meters?  D’oh.

The bottom line is that if your functions represent physical quantities, then those quantities have units.  And unless the units happen to be the same, then you cannot just flip the graph.  You have to flip the axes as well. You just can’t throw those graphs on the same axes.  Look at them side-by-side:


If you want to show the symmetry of the original claim, you could take these two graphs, put them on overhead transparencies and lay one on top of the other:


It’s not as pretty, but at least its not wrong.

When you are focusing on the math, it is easy to forget that these quantities have physical meaning and units that go with them.  For example, there are many books and websites where you can see a function and its derivative graphed on the same axes.  (It’s even done that way on some of the websites that I used in earlier posts about derivatives.)  But that makes no sense!  For example, how can you graph position and velocity on the same axes?  They don’t have the same units.

I did a quick web search for “function derivative graph applet”.  Here is one site (of many) I found that puts the function and derivative on the same axes:


Here’s one that correctly puts them on different axes:


These are both very clever websites, good for developing a feel for what a function’s derivative looks like.  But the second one is better — it remembers that these are quantities with units.  (Also, the shading on the derivative graph is a neat bit of fore-shadowing.  That’s a topic for another upcoming post as well.)

Getting back to the Flipping Graphs…

If you don’t like the looks of the two overlaid transparencies, there are some other ways to visualize these functions and their inverses.  John Denker presents a few of them on his website here:

Here is one diagram (figure 10) from his site that I especially liked:


The “reflector panel” reminded me of another clever method of interpreting graphs.  More on that in a future post…


Getting Back to Work

And an early review of my book!

I admit that my blogging efforts have been side-tracked for a while.  All of the posts since last summer have really been directed at my AP Physics students to give them a quick intro to the calculus we will be using this year.  That course is well under way.  And they are a fine bunch of students, doing quite well.  Also, many of them are now at the point in their calculus classes where they no longer need my help.

I will eventually add one more post about using anti-derivatives to find areas, but for now, I am eager to get back to the topics that originally motivated this blog.  I want to write about ideas that didn’t quite fit into my book but that would still fit the overall theme of a first course in algebra.  And I want to discuss questions that come up as people read my book.  The book is now available,as you may have guessed from the subtle ads on the right.   And I am hoping readers will ask questions, make comments, offer suggestions. You can email me or use the “Questions” tab above.  I’ve already received some very interesting feedback which I will discuss in the next post.  But for now, I just want to share the first review.  Boy, it’s a good one!

Zach note

Thank you Zach!


And a first look at differential equations

At some point in the past, I believe you learned a little about radioactive decay and half-life.

Here are some things that you may remember:

1. The size of a radioactive sample is can be expressed as a function of time:


2. The graph of this function looks like this:


Note that at time = 0, the value is No and then the value decays asymptotically toward zero.

3. Graphs like this show up in many other contexts.  For example, this looks a lot like the graph of current vs. time for a charging capacitor.  It also looks like the graph of acceleration vs. time for an object in free-fall with air resistance.  And there are a number of graphs associated with motors and generators that also have this shape.

4. When a quantity decays exponentially, in equal time intervals it will decrease by the same factor.  Most notably, the time it takes to decrease by 50% is called the “half-life”.  There is a useful formula relating the half-life to the decay constant:


Do NOT just memorize this – derive it for yourself.  If you need help getting started, you can begin by letting N(t)= ½ No.

However, all of this assumes that we already know that the function is N(t)=Noe-kt . But how do we know that?  When you see exponential decay, you should suspect that somewhere behind the scenes lurks something called a differential equation.  And that brings us (finally) to the topic of this post.


The rate at which a given radioactive sample decays is proportional to the size of the sample.”

I hope that sentence makes sense to you, intuitively.  It is saying that when there are more nuclei available to decay, the decay rate is faster.  As the sample shrinks in size, there are fewer available nuclei and so the rate slows down.

But that sentence can be re-written using mathematical symbols:


The resulting equation relates a function, N(t), to one or more of its derivatives.  In this case, the highest order derivative is the first derivative so this is called a “first order differential equation”.  We are going to encounter a number of these in AP Physics and also a handful of second order differential equations as well.  You will learn how to solve some of these in math class this year.  If you continue in math and science, you may spend a number of semesters learning more about this topic.  For now, I am going to show you a method that will be sufficient for our specific needs.


When you solve an algebraic equation, you find a number that you can use in place of the variable, thus obtaining a true statement.  For example, x= 3 is a solution to the equation 2x + 4 = 10 because when you replace x with 3 in that equation, you get a true result.  And even if you didn’t learn the step-by-step method of solving that equation, you could still verify that x=3 works.  It wouldn’t matter if the solution came to you in a dream!  Once you verified it, you’d know you were right.

Now we have a different kind of equation:


We are not looking for a number.  We are looking for a function, one that will make a true statement when we use it to replace N(t) in that differential equation.  Here is how we are going to do this:

(Don’t worry if this doesn’t “click” at first.  We will walk through this several times.)

1. Based on our intuition, draw a graph of what the function N(t) will look like.

2. Use our extensive knowledge of pre-calculus to guess a general form of a function that has that a graph shaped like the one we just drew.

3. Take the derivative of our guess (and second derivative if needed).

4. Substitute back into the equation we are trying to solve to see if we get a true statement.

If we have guessed correctly (as we often will), we will actually pick up some bonus information.  Follow along with me and see what I mean.  We’ll start from the beginning.

We are seeking a function N(t) that will satisfy this differential equation:


1. Our intuition about radioactive decay suggests that when we find the solution, its graph, as we already noted, will look like this:


2. There are a number of functions that have that shape.  For example, it could be that:


It’s not a terrible guess. It has the right value at t=0 and it approaches zero asymptotically.  I don’t remember seeing that function used to model decay before and it would be all wrong for t approaching -1, but let’s try it anyway.

3. Taking the derivative, we get…


4. Now, substituting the function and its derivative back into the original differential equation, we get:


Hmm…can this be true?  For a given value of k, it is true at some particular time. But we want a solution that is ALWAYS true.  There is nothing we can do to fix this one.  It turns out that our initial guess, though not terrible, was wrong.  OK, new guess.  Let’s try an exponential function.  Our memories from pre-calc  give us high hopes for this one:


In this case, we would have the derivative:


(because the derivative of “e to the something” is “e to the something” and then there is that chain rule multiplier.)

But what happens when we substitute these expressions back into the original differential equation?


Can this be true? Why yes, but only if c = k.  In other words, we have just learned that we had correctly guessed the general form of the solution but it can’t be just any old exponentially decaying function.  The decay constant in the function has to match the constant of proportionality in the original differential equation.  So with that adjustment, we have our solution:


This kind of thing happens a lot when we use this technique.  At the end of the process, when you ask, “can this equation be true?” you get as the answer: “yes, but only if…” followed by some new information that tells you the required value of some constant.

Closing remarks

Please do not worry if you have found this post to be challenging.  In the next practice set, I will give you a couple of additional examples and walk you through this process step-by-step.  And we will also do a couple together in class.  But review this a few times and you may realize that it is just another way to make use of derivatives and we still have not used any rules beyond the ones I showed you in the earlier posts.  But that does not mean that this was easy!

Making the Best of Things

A Very Useful Thing About Slope Formulas

As you have worked through the previous posts and exercises, I hope you have noticed something special about places on a graph where the derivative is zero.  These places have horizontal tangent lines.  And in their own neighborhoods, they are often a local maximum or local minimum.  This piece of information is very helpful when you are trying to solve an “optimization” problem.

For example…

Suppose you are firing a projectile on a horizontal surface at a fixed launch speed, v.  And let’s also suppose that air friction is negligible.  What angle will maximize the range (horizontal distance) of the projectile?

You may remember the answer from your first year of physics.  It is intuitive and there was a way to get the answer without calculus.  But we can still use this example to illustrate the calculus method.


(Graphic from where you can also see a review of the projectile physics)

STEP 1: Express the quantity to be optimized as a function of a single variable.

In this case, a review of that website or last year’s notes will remind you that the distance varies with the launch angle by the rule:


In brief, we resolved the launch velocity into components, used the vertical component to determine time in air and then multiplied that time by the horizontal component of the velocity.  But now that we have this expression, for our current purposes it would be helpful to make use of the double angle identity for the sine function.  We can write:


Don’t worry if you would not have remembered how to do all that without notes.  The point is to learn what we can do with this expression now that we have it.


STEP 2: Find the derivative

This is one we know how to do:  we know the derivative of sine functions and we know the special case of the chain rule.


STEP 3: What is the domain of interest to us?

When a math book asks for the domain of a function, they usually mean the largest, most inclusive domain possible.  For a sine function, that would be all real numbers.  But that is not what we are after here.  We need to think about the portion of the domain that is relevant.  To fire this cannonball as far as we can, clearly we will choose an angle between 0 and π/2 radians.

STEP 4: Where in that domain is the derivative equal to zero?

In this case…


(OK, call it 45 degrees if you prefer.)

 STEP 5: (Your math teacher may not like this) Convince ourselves that we are done.

In math class, once you have found where the derivative is zero (or undefined), you are not nearly done.  You have to investigate each of these “critical points” to see what kind of extreme it is and to justify your conclusion with such things as sign charts and 2nd derivative tests.  But in physics class, you are going to get away a little easier because of the following claim:

If a “smooth” function has only one critical point on some interval and the value of the function is higher at that critical point than at either of the endpoints, then that value is the “absolute maximum” value.

So in the example at hand, we know the cannonball’s range is zero when the angle is zero, and its zero when the angle is 90 degrees (as the ball goes straight up and down).  But it is NOT zero when the launch angle is 45 degrees.  But the tangent line is horizontal at 45 degrees.  So that’s the “best” angle.

You still have to learn to do things the math-class way.  There are lots of functions with multiple critical points.  And there are functions that have horizontal tangent lines at places that turn out not to maxima or minima.  So you do have to know how to investigate further.

But the problems you will see in AP Physics don’t turn out that way.  Most often, there will be only one critical point and it will be the one we were looking for.  Here is another example I have always liked.


Suppose you have a large cylindrical tank full of water, open at the top.  You are going to punch a hole in the tank somewhere along its side and you want the resulting jet of water to land as far from the tank as possible.  Where should you make the hole?  And where will that initial jet of water land?

Let’s say you make the hole y units from the ground and the initial jet lands x units from the tank.

STEP 1: Determine x as a function of y

First we need the “speed of efflux” which you can think of as the horizontal launch velocity of the water coming out of the hole.  There is a neat result from fluid dynamics that shows that the water leaves the tank at the same speed as if it had free-fallen from the top of the tank.  It’s called Torricielli’s Law.


You can see more about this calculation here:’s_law  (where I also got the graphic above)

Now, we just find the time in air and multiply by that launch speed.

For time in air, we can use the fact that


and solve for time to get:


So for the distance as a function of the height of the hole we get:



STEP 2: Find the derivative

Well, it seems we need the more advanced version of the chain rule. But there is a trick to make things easier.  The distance x will be at its maximum when the term inside the square root is at its maximum.  So let’s ignore the square root and just maximize the innards.


STEP 3: Domain of interest?  We can’t make a hole below the ground or above the height of the cylinder so we only have to consider y values between 0 and h.

STEP 4: Set the derivative equal to zero


We need a hole in the middle!

 STEP 5:  Convince ourselves that we are done

There was only one critical point.  And the endpoints both lead to a distance of zero.  So the hole at y=h/2 is the one that gives us the farthest landing point.

And what is that landing point? Substitute y=h/2 into our original function, which was:


I will leave the substitution and clean-up to you.  I will say that it cleans up quite nicely and has a memorable result.  The first of my students to post the correct result in a comment wins some silly thing or other.

Quick Recap and Closing Remarks

In AP Physics, when you want to find the maximum or minimum possible value of some quantity, most of the time it will just be a matter of taking a derivative, then setting it equal to zero and then waving your hands around to convince us that you are done.

The hardest part of the process is not going to be the calculus.  It’s going to be setting up the variables and finding the expression for the quantity you hope to optimize.  That’s why you had all those homework problems back in pre-calc that said “Express this thing in terms of that thing”.

I guess I should mention that if you are having trouble either finding the derivative or solving after you set it equal to zero, you may find your TI82 to be of some use.  But honestly, sometimes it feels like the calculator only gives intelligible answers to people who already know what the solution looks like.



(and some loose ends)

Before we get to the practice questions…

1.  I want to introduce anther notation for derivatives.

Suppose we are working with a function y=f(x)

Instead of writing the derivative as y’ we often write:


The expression is telling you to find the derivative of the function f with respect to the variable x.  For now, this seems like a more cumbersome notation but there will be times when you appreciate its clarity.  It leaves no doubt about what variable you are working with.  This will be critical when you are working with functions of more than one variable.


2. And yet another notation for derivatives, this time a lazier one…

In many instances, the functions we are considering will be functions of time, t.  When we want to indicate that we are taking the derivative with respect to time, we just write a dot over the variable.  So for example, let x(t) be the position of an object as a function of time.  In that case, the velocity would be:


Typically, if we are being this terse, we may as well be briefer and write:


Here we are figuring that by using the dot notation, we are implicitly saying that these are both functions of time.

And then we can say things like:


This is just a concise short-hand for “acceleration is the derivative of velocity with respect to time and the second derivative of position with respect to time.”


OK, this brings us to the practice set.  I have posted it here as a pdf.  Note to my Holmdel students:  I have also emailed it to your school account.  So if you did not get it, you should email me to ask to be added to the mailing list.  If you have any questions, post them here as comments (Holmdel student or not).

Practice Set #1 AP Physics Math Preview


Simple Version of the Chain Rule

Suppose we want to find a derivative for this function:  h(x) = sin(2x).

The sine function is on our list – we know that the derivative of sin(x) is cos(x).  But we are not dealing with just the sine function.  There’s another function inside the function!  We don’t just have “sine of x” but rather we have “sine of something”, where the “something” is itself a function.

This means that to find this kind of derivative, it is essential that you have a clear understanding of the composition of two functions.  You have to look at  h(x) = sin(2x) and see it as the composition of functions and you have to recognize the building blocks:

g(x) = 2x and f(x) = sin(x)

We are not going to just add them or multiply them.  Instead, we use the output of one of them as the input to the other. So we build:

  h(x) = f(g(x)) = sin(2x)

The g function is the inner function.  It starts off by doubling its input.  Then that output is handed off to the “outer function”,  f  which finds the sine.

And now, it we want to find the derivative, we need to learn…


 Once again, I have to pause and remind myself of the purpose of these posts. If I show you  “the chain rule”, we can then use it to start finding derivatives of long, challenging “chains” of compositions of multiple functions, such as:


But then I remember two things:

1. This is beyond what we will actually need in the opening months of AP Physics.

2. Your math teacher deserves to have some of the fun as well.

So I am only going to introduce a lesser version of the chain rule, one that is just enough to handle a special case, one that we will actually need in class very early in the year.


For example, look at each of these:


In each case, the “outer” function is one from our list of well-known functions.  And the “inner” function is a constant multiplier.

There is an easy rule that handles all of these:


In other words, go ahead and take the derivative using the familiar rule, but then multiply your answer by the constant you see “inside” the original expression.

With this rule we can find all three of the derivatives above:



The derivative of “e to the something” is “e to that something”, and the 3 is a multiplier.



The derivative of “the cosine of something” is “the minus sine of that something” and the 5 is a multiplier.



The derivative of “the sine of something” is “the cosine of something” and the 2 is a multiplier.


Why do we multiply by that same number you see in the inner function?  It turns out that there is an intuitive way to understand this rule.  Once again, we have to look at graphs and their transformations.

In the last post, we saw that compared to the graph of y = f(x), the graph of y = f(x) was stretched vertically when a>1, and that the vertical stretch affects the tangent line as well, also increasing its slope by that same factor, a.

But now we consider the graph of y = f(a·x), we see that a>1 leads to a horizontal compression, and that compression affects the tangent line as well.  Slope is rise over run, so reducing the run increases the slope, again by that same factor, a.  That’s why our derivatives all have that multiplier: compressing the graphs horizontally makes the slopes steeper.

For example, here are graphs of f(x) = ex and f(x) = e3x.



You can see that the graph is horizontally compressed.

And when you compare the slopes of the tangent lines, you see that the second one is 3 times as steep as the first. [But notice: the first tangent line is at x=.6 and the second one is at x=.2.  Why did I do that?  Post a comment with your answer…]  Until your math teacher unveils the chain rule in its full glory, please let the diagram above serve as a proof of our simpler version.


At this point, we have assembled some basic derivative formulas and some rules about combining them.  This would be a good place to pause and see if you are keeping up.  So in the next post, you will find a set of practice questions.  Once you have mastered them, you will be ready to find out what we do with these slope-finding formulas once we have derived them.

Derivatives, Part II: Combining the Functions

In the last post, we compiled a small collection of derivative formulas, formulas which will enable you to find slopes of tangent lines.  (Why you would want to do such a thing is a discussion that is coming soon.)

You will see all of those functions in various settings as you study AP Physics.  But they don’t always act alone.  You have to be able to work with combinations of those functions in some ways that are intuitive but others that are, well, less intuitive.

1. Adding and Subtracting — Just as you would guess

Suppose f(x) =x3 +x2

and you would like to find the derivative.  You might be thinking:

“Wait — I have a rule for each of those terms, separately.  Can I just add them? ”

In other words, is f'(x) = 3x2 + 2x ?   Yes it is.  We could say this more officially:

If h(x) = f(x) + g(x) then h'(x) = f'(x) + g'(x).

But all that means is that if you have functions added together, you can use their separate derivative formulas and then add them.  And it works the same way for subtraction.

2. Multiplying by a Constant — Also as you would guess

Given f(x) = 3sin(x)…yes, it’s true: f'(x) = 3cos(x).  Multiplying the original function by a constant just multiplies its derivative by the same constant.

There is an intuitive way to see why this must be true:

Consider a function, g(x) = a·f(x), where a>0.   How does the graph of g compare to the graph of f? [SAT fun fact!]  It is vertically stretched or vertically compressed, depending on whether a is greater or less than one.  Now, how would that affect the tangent line?  A vertical stretch by a factor, a, will increase the slope of that tangent line by that same factor, a.

Here is an illustration of what I mean:


On the left, you see the graph of f(x) = √x and its tangent line at x = 1.  The slope of that tangent line is .5.

And on the right, you see the graph of f(x) = 3√x and its tangent line, also at x = 1.  The graph has been stretched vertically by a factor of 3.  And sure enough, the tangent line is 3 times as steep, or 1.5.

We’ll call that a “proof” and state the rule officially:  if g(x) = a·f(x) then g'(x) = a·f'(x).


3. Multiplying and Dividing Functions: NOT what you would expect

Well, it’s a bad new/good news kind of thing…

Bad news: you can’t just multiply the derivatives.  In other words, if h(x) = f(x)·g(x), it would be convenient if  h'(x) = f'(x)·g'(x).  But, alas.  Though we don’t know what the rule is, we know enough to see what it isn’t.

All we have to do is let f(x) = x3 and let g(x) = x2.  Then h(x) = f(x)·g(x) = x.

We also already know the derivatives: f'(x) = 3x2 and g'(x) = 2x.  When you multiply them, you get. 3x2·2x = 6x3  But we already know that h'(x) = 5x4.  So the product of the derivatives did not match the derivative of the products.  That’s the bad news. And a similar argument can be constructed to show that quotients don’t work that way either.

So what’s the good news?

1.  There are rules that handle these situations.  They are called the product rule and the quotient rule.  I am sure that your math teacher will be happy to teach them to you.

2.  This is where I pause and remind myself what these posts are for.  I am trying to get you ready for your year of AP Physics.  We may need the product rule and the quotient rule before the year is over, but we won’t need them right away.  By the time we do need them, you will have seen them in math class.

On the other hand, there is one more way to combine functions that we will need almost immediately.  So in the next post, we are going to look at something called “the chain rule”.