Lesson 1: Limits
One standard concept in pre-calculus that we never got around to teaching you last year involves the idea of
limits. Believing, as we do, in the ancient adage, “Better late than never”, we will start our year of calculus with
a lesson on limits.
The good news is this—a terribly complicated notion called “formal definition of limits” is included in most
calculus books, and it’s how most college calculus classes start—but we’re blowing off the really tough “formal
definition of limits” part of the show. I figure about one third of the kids who drop college calculus do so
because they can’t wrap their heads around the formal definition of limits—which is a drag, because you don’t
need the formal definition of limits to understand calculus (or to understand limits, for that matter).
With that in mind, here are the big ideas of limits in mathematics.
Big Idea One: There are actually two kinds of limits out there. We’ll call them “false limits” and “true limits.”
You won’t see these phrases anywhere else, but they make sense to us, and we know they’ll make sense to you,
too. True Limits are really nothing more than numbers in most cases: Specifically, the number a function’s Y
value gets close to as its X value approaches a given value. False limits look just like true limits, but they do
not generally represent a number. Instead, False Limits represent the derivative of a function (most of the time).
Since false limits and true limits have completely different meanings and since you use completely different
methods to evaluate them, it’s critically important to understand at a glance which type of limit you’re looking
at when confronted with a limit problem. Perhaps the illustration below will be enough for you to figure out
how to tell them apart.
Here are some examples of True Limits:
Here are some examples of False Limits:
So, is it obvious to you what constitutes a false limit and what represents a true limit? Here’s what we’re
hoping you noticed after staring at the limits from the previous page:
False limits always have the limiting variable approaching zero…AND
False limits are always fractions with the limiting variable all alone in the denominator…AND
False limits always contain an even number of terms in the numerator…AND
False limits’ numerator terms always occur in opposite pairs with “x+h” replaced with “x”. If an (x+h)2 term
exists, it must be matched later by a -(x) term. If a -2(q+m) term exists, it must be matched later by a 2q term.
If a –7 term exists, it must be matched later by a +7 term, and so forth.)
We will begin learning what True Limits are really all about in a moment, but before we leave this first big idea,
take a moment and classify the following limits as either False or True based on the informal definition/rules
listed above. It is very important that you learn to do this—otherwise you will have no way of knowing what
approach to take on limit problems in calculus later on!
We hope you classified the limits this way: 1=T, 2=T, 3=F, 4=F, 5=F, 6=T, 7=T, 8=T, 9=F, 10=F.
In our next lesson we will go into great detail about False Limits. We will come to understand that they
represent nothing more than the first derivative of a particular function every time, and we will explain where
these false limits come from (of all places, the slope formula from back in Algebra 1), but for now we will leave
false limits behind and embark upon a deep understanding of True Limits.
Big Idea 2: From here on out this lesson deals strictly with true limits, so we’ll drop the “true” and just call
them limits for the duration. The most important thing to understand about limits is this—a limit is just a
number. The second big thing to understand about limits is this: The limit of some function as x approaches
blob is simply the y value that function’s graph is approaching as x approaches blob (where “blob” is the math
term for “a given numeric value.”)
With this in mind, it’s perhaps easy to understand that the statement
in the box at right is really just a way of saying, “As x gets closer and
closer to 2, what y value does the function y = 6x + 5 get close to?
Because the function y = 6x + 5 is a plain old linear function with
no jumps or holes or anything, the y value at x = 2 is what the
function approaches as x approaches 2. So, this limit simply
equals 6*2 + 5 or 17.
Most functions (like polynomial functions, sine and cosine functions, absolute value functions and so forth)
have limits that can be evaluated by simply plugging in the number x is approaching and chugging out the y
value. Other functions, however, have gaps or holes or jumps or vertical asymptotes—it’s these functions’
limits that give us the most trouble, and that’s where we’ll go next. But first a quick summary of this big idea:
If y = f(x) is a simple polynomial function, trig function or other function type with no holes, vertical
asymptotes or jumps, the limit of f(x) as x approaches any number a is simply f(a)—the y value of f(x) when
x=a.
(Functions of this variety, by the way, are called Continuous Functions. The easiest way to explain continuous
functions is this: Can you trace the entire graph without ever picking up your pencil? If so, it’s a continuous
function).
Big Idea 3: When a function has holes, gaps, jumps, vertical asymptotes or other odd characteristics that make
the function non-continuous (or discontinuous as the math books say…when you can’t trace the graph entirely
without picking up your pencil), our limit work gets more complicated. Such is the case with rational functions,
piecewise functions and other function types we will work with this year. There are some pretty clear-cut rules
about limits of discontinuous functions, but before explaining those rules we need some new terminology.
Left Hand Limits and Right Hand Limits are terms we use to denote limits coming from one side only—not
approaching an x value from both sides at once. The sketches below explain these concepts.
In this sketch, the left hand limit as x approaches 2 means “as x
approaches 2 from the left.” This limit equals 4, because that’s the y
value f(x) approaches as x approaches 2. It’s important to note that
the limit equals 4 even though the function’s actual value at x=2 is
-3.
-3.
In this sketch, the right hand limit as x approaches 3 means “as x
approaches 3 from the right.” This limit equals 1, because that’s the
value f(x) approaches as x approaches 3. It’s important to note that
the limit equals 1 even though the function’s actual value at x=3 is 5.
-3.
As we’re sure you’ve already figured out, there is a shorthand terminology for left and right hand limits:
To see if you’ve mastered this skill, please look back at the last two sketches and determine the RH limits as x
approaches 2 in the first sketch and the LH limit as x approaches 3 in the second sketch.
Big Idea 4: Your answers above were –3 and 5, we hope? Good…let’s continue. With these left and right
hand limits under our belt, we can now define the existence of a limit at any given point for any given function.
Here it is:
The function f(x) will have a limit at any x value for which the left hand and right hand limits are equal. The
limit at that point will be the value of the left and right hand limits. Sometimes this definition is written out in
the following limit notation:
Likewise, we also are now ready for the definition of when a function does NOT have a limit. If the left hand
and right hand limits of a function do not agree at a particular x value, the function does not have a limit at that
point. In limit notation form:
Check out the sketch below and explain using the language of limits why the function does not have a limit
at x=5.
Big Idea 5: We hope your response to the previous question was a little something like this: “F(x) has no limit
at x=5 because the left hand limit there is 2 and the right hand limit there is –1 and these are not the same
value”.
Now we come to the biggie: Continuity. With the rules above about the existence of limits under our belts, we
can nail down the more important concept of where a function is or is not continuous. Here’s the rule: If the
limit of f(x) as x approaches a exists, and if this limit equal f(a), then F(x) is continuous at x=a.
In the language of limits, here’s how we’d write that:
Later in this class we will learn that a function cannot have a derivative at a point where it is not continuous—
that’s why we’re laboring to create a definition of continuity. But for now, use the continuity definition above
to explain why the following function is not continuous at x=3.
Big Idea 6: We hope you said, “At x=3 the left and right hand limits agree, so the limit exists and it is 4. But
the function’s value at that point is 2, and since 2 and 4 are not the same number, there is a discontinuity at
x=3.”
Our last topic in this lesson concerns limits involving infinity. One way limits can involve infinity is if x is
approaching infinity (or negative infinity). In these cases we are simply talking about the end behavior of the
graph. As x gets really big, what does the graph do? Does the graph get higher and higher as is the case with
f(x) = x2. If so, we say the limit is infinity (and since infinity is not a number, we say the limit does not exist.)
If, on the other hand, as x approaches infinity the graph gets lower and lower (as is the case with f(x)=-2x+1),
we say the limit is negative infinity (or the limit does not exist).
A third possibility is this—as x approaches infinity, the graph has a horizontal asymptote—as is the case with
f(x) = 1/x. Since this graph hugs the x axis as x gets bigger and bigger, we say the limit as x approaches infinity
of 1/x is zero.
There’s another way infinity can get involved in limits: as x approaches a certain number, the graph goes crazy
and gets really big or really small. This is what happens when we have a vertical asymptote—and just about the
only time we see this happen is when the denominator of a fraction gets closer to zero. Consider the following
function with its graph shown:
As x approaches 2 the denominator approaches zero and we have a vertical asymptote situation on our hands.
In this case we say the limit as x approaches 2 is infinity (or fails to exist).
The pre calculus summer packet information on limits and end behavior (found in sections 13 and 14 of those
packets) contains really valuable information you ought to review if any of this seems a little unclear to you.
Please classify these statements as true or false before trying the problems we assigned for this lesson!
1=t, 2=f, 3=f, 4=t, 5=t, 6=t, 7=f, 8=f, 9=f, 10=t, 11=f, 12=t.
Lesson 2: Derivatives as Limits
We’re starting out the year by telling you all you really need to know about calculus. Are you ready?
Calculus is a branch of mathematics that deals with how things change over time.
There. That’s it. We’re done!
Well, not so fast, grasshopper… There’s much to learn.
The good news is that you’ve already dabbled in change over time in algebra when you studied slope. You’ll
see this concept again in calculus. In fact, you’ll see it right away in this lesson. As the year progresses, we’ll
continue to develop that concept. In the end, however, it really all boils down to that fundamental idea.
Remember that.
When we talk about how things change over time, we can be concerned with how something is changing at
some specific point in time, or we can be concerned with how much change has accumulated over time.
There are two main branches of calculus that deal with these ideas. They are:
 derivatives, which tell us the rates of change of functions, and
 integrals, which tell us the accumulated amount of change in a function over time.
What do these two things mean? What are these two things useful for in the real world? Well, that, my friends,
is what you’ll be learning over the next several months.
Derivatives:
Let me start out by making this connection explicit. DERIVATIVES ARE RATES OF CHANGE. SLOPES
ARE RATES OF CHANGE. DERIVATIVES ARE SLOPES. There, I said it. But, you have no reason to
understand why this is true. Yet.
To understand this fundamentally important concept, we need to take a trip back to our wonderful days in
algebra where we studied slopes.
Example: Find the slope of the line that passes through the points (-4, 5) and (1, 7).
You all should be able to figure this out easily enough. Show your work and answer here.
Now, onto more interesting problems.
Example: Find the rate of change of the function y = x2 between x = 1 and x = 5.
Again, not a tough problem, but you’re now working with a curve. This doesn’t change the steps you use, but it
does change what your answer represents. We’ll talk about that in a second. For now, show your work and
answer for this problem below.
This type of problem is actually asking you to find, what a calculus student would call, the average rate of
change of the function between x = 1 and x = 5. Certainly you can appreciate the fact that the function isn’t
changing at the same rate over the entire time period, as it is a curve not a line. Remember only lines have
constant rates of change. The average rate of change you found above is actually the slope of the segment
connecting the two points. Can you picture that?
Here’s another problem for you. Before working it out, however, hypothesize whether your answer will be
larger or smaller than the answer you got for the previous problem, and explain why you think that way.
Example: Find the rate of change of the function y = x2 between x = 2 and x = 4. Show your work below.
We could go on with more problems like these, and in fact you will. Please fill in the following table. You
should notice that the function we’re dealing with is still y = x2, and that you’ve already done the work for the
first two rows.
P1
(1, 1)
(2, 4)
(2.5, 6.25)
(2.8, 7.84)
(2.9, 8.41)
(2.99, (8.9401)
P2
(5, 25)
(4, 16)
(3.5, 12.25)
(3.2, 10.24)
(3.1, 9.61)
(3.01, 9.0601)
y
x
y/x (Ave. Slope)
OK. Answer these three questions about your table.
1. What single point are we “closing in on?”
2. What value is x approaching?
3. What is happening to the average slope as we “close in” on the point you mentioned in #1?
Hopefully you said we were closing in on the point (3, 9). Also, you should have noticed that the average slope
is approaching 6. Ah, yes. It is approaching 6. Where have we heard that terminology before? You bet.
LIMITS!!! (Good thing we did all that work with limits already, right?!)
The Limit Definition of a Derivative:
Before we get into this too deeply, a quick check of your knowledge is in order. Above, you saw that as we
made the distance between two points smaller and smaller, the average slope between those two points
approached some limit. This is the fundamental idea behind derivatives. That is, you allow the distance
between two points to become infinitesimally small, and calculate the slope of the segment (which is really,
really short) that connects these two points.
Well, allowing the distance between the two point to become infinitesimally small really means that you are
letting the change in x (x) become closer to what value?
The answer you gave was hopefully zero. That’s important idea #1 in the limit definition of a derivative. You
will always find the limit of some function as h approaches 0. (Don’t worry about the fact that we changed the
letter to an h, yet. You’ll understand why shortly).
The second important idea in the limit definition of a derivative, we’ve already stated, and that is that
DERIVATIVES ARE SLOPES. Now you certainly remember how to find the slope between two points, you
did that when you filled in the chart on the previous page. It is simply the change in the y-values divided by the
change in the x-values, or rise over run—whatever you want to call it.
Also remember that as we moved the two points closer together, as you did in the chart you filled in above, that
the change in the x-values approached 0. In other words, the denominator in the fraction that represents slope
gets closer to 0.
Let’s think about the numerator now for a minute. The numerator in the slope fraction represents the change in
the y-values. Now, if we know the x-values that we are using, how would we find the corresponding y-values?
Hopefully you immediately said, “Use the function!” In other words, we’re going to use f(x).
The first y-value is just that, f(x). The second y-value comes from an x-value that’s just a tiny bit different from
the first x-value (remember that the difference in the x-values is approaching zero). We’re going to let h
represent that tiny bit of difference between the two x-values.
If we do that we get the following limit.
lim
h 0
f ( x  h)  f ( x )
h
Notice that the numerator is finding the difference in the two y-values. The denominator is still finding the
difference in the two x-values—a difference we said all along approaches 0.
OK. Did you follow all of that? Need a clear, concise, step-by-step recap? If so, try this on for size.
1.
We started by saying derivatives mean change in y over time (x).
change over time
2.
This can be written in delta (  ) notation.
y
x
3.
Of course, this is simply a slope since  y = y2  y1 and  x = x2  x1 .
y 2  y1
x2  x1
4.
This idea applies to curves as well as lines if you use limits.
lim
y
x
lim
y 2  y1
x
x  0
Now, all we need is a little algebra…
5.
Since  y = y2  y1 , we get:
6.
Since y 1 = f(x1) and y2 = f(x2), we get:
x 0
lim
x 0
f ( x2 )  f ( x1 )
x
7. Since  x = x2 – x1, then x2 =  x + x1, so we get:
lim
x 0
f ( x1  x)  f ( x1 )
x
8. Since  x can be renamed “h”, we get:
lim
f ( x1  h)  f ( x1 )
h
9. Since there’s only one x-value left, we can use x, instead of x1 and get:
lim
f ( x  h)  f ( x )
h
h0
h 0
And, finally, since this all started by declaring that derivatives are change over time and in every step we simply
rewrote “change over time” in a new form, that final limit is another way of saying “the derivative of f(x)”.
If y = f(x), then the derivative of f(x) or f’(x) = lim
h 0
f ( x  h)  f ( x )
h
Finding Derivatives Using the Limit Definition:
We won’t pretend that you don’t already know how to find some derivatives. From pre-calculus, you know:
1. That the derivative of y = x 2 is y’ = 2x (power rule);
2. That if f(x) = 5x 4 2x 3 then f’(x) = 20x 3 6x 2 (power rule, again);
3. That velocity is the first derivative of position and acceleration is the first derivative of velocity (or
the 2nd derivative of position).
4. That curves have slopes at points. For example, if y = x 3 then at x = 5 the curve has a slope of y’ =
3x 2 = (3)(52) = (3)(75) = 225.
5. That the sign of the derivative tells you something about the shape (behavior) of the function.
6. You can use techniques like the product, quotient, and chain rules to find derivatives of slightly more
complex functions.
But, because the limit definition of a derivative is so important (there will be several questions on the AP test
specifically about this concept), we’re going to assume you don’t know how to do those things and have you
use the limit definition of a derivative to find one.
In the first part of this lesson, there was an excellent discussion of “true” limits and “false” limits. By now,
you’ve hopefully equated “false” limits with the limit definition of a derivative. In fact, we’ll drop the
terminology “true” and “false” limits going forward. The following point is still important, however:
1. That you can recognize when a limit represents a derivative (when it’s a “false” limit).
And, by time we’re done with this lesson, we’ll add a second important point:
2. That you can recognize within this limit the function you are finding a derivative for.
Hopefully, you’ve got point 1 down by now (if not go back and review the first part of this lesson on “false”
limits). We’re going to concentrate on point 2. From the limit definition of a derivative, you see that the
function, f(x), is the last term in the numerator. ALWAYS LOOK HERE TO DETERMINE THE FUNCTION
YOU ARE DERIVING. Be careful with functions that have more than one term. Parentheses are important!
In the previous lesson, you practiced evaluating many different types of limits. When evaluating the limits that
represent derivatives, you should have developed a strategy that went something like this:
1. First simplify the numerator using algebra.
2. Factor out an h.
3. Simplify again.
Example: Find the derivative of y = x2 using the limit definition of a derivative.
Start with the limit definition of a derivative:
Since f(x) = x2, the limit becomes:
f ( x  h)  f ( x)
h 0
h
2
( x  h)  ( x ) 2
lim
h 0
h
lim


( x 2  2 xh  h 2 )  x 2
h 0
h
Simplifing the numerator gives us:
lim
Factoring out an h, and “canceling” gives us:
lim 2 x  h
And, as h approaches 0, we get:
2x + 0 = 2x
2 xh  h 2
h 0
h
 lim

h 0
In the space below, we’re going to work through a few more examples. Be sure to write in each step in a neat
and organized manner.
Example: If f(x) = x3, find f ’(x) using the limit definition of a derivative.
Example: If f(x) = x2 + 5x, find f ’(x) using the limit definition of a derivative.
That certainly was fun now wasn’t it!
Remember point #2 above; that you need to be able to recognize within the limit definition of a derivative what
function you are finding a derivative for? Hopefully you can see now that the power rule is one technique that
makes finding derivatives much easier.
Example: Give the function that you are finding a derivative for with the limit below.
( x  h) 2  5( x  h)  x 2  5 x
h 0
h
lim
Remember we said parentheses would be important!
Where’s the x?!
We should make one last point. If there is a number where you normally see an x, we’re just asking you to
evaluate the derivative of the function when x = that number!
Example: Find lim
h0
(3  h) 2  (3) 2
h
Because of its basic structure, this looks like the limit definition of the derivative. It is!! This is nothing more
f ( x  h)  f ( x )
than lim
with f (x) = x2 and x = 3.
h0
h
I’ve left room below for you to find this limit. Have fun! 
Some final thoughts:
Notation:
The first derivative of a function can be written many ways. If y = f(x) = x 2 , the derivative is 2x and we write:
y’ = 2x or f’(x) = 2x or
dy
d
= 2x or
f(x) = 2x.
dx
dx
This lesson’s important idea based on the fact that derivatives are slopes gave us:
The Limit Definition of a Derivative
The derivative of f(x) is: lim
h 0
f ( x  h)  f ( x )
.
h
Lesson 3: The Power Rule
Mathematicians are fundamentally lazy. They continually look for (and find) elegant shortcuts to tedious
algorithms. People might have used the limit definition of derivatives to find derivatives for awhile, but
eventually someone noticed a pattern.
If y = x 2 , then y’ = 2x
If y = x 3 , then y’ = 3x 2
If y = 6x 2 + 2x, then y’ = 12x + 2
If y = ax n bx m , then y’ = anx ( n 1) bmx ( m1)
This pattern is called the power rule. You’ll be asked to find derivatives using this quick, slick method on
upcoming assignments. For now, check out these trickier examples:
1.
2.
3.
Find y’ if y = 3x
Find y’ if y = 7
Find y’ if y = x



y’ = 3
y’ = 0
y= x
1
y=x2
1
1
y’ = x 2
2
A quick note on simplifying in calculus. Generally, you simplify things at your own risk. This is especially
true if you’ve already found a derivative for example. However, there are times when you must simplify. Most
notable is on the multiple-choice section of the AP test.
If you look at example 3 above, you’ll notice that the final derivative could be rewritten so there is no negative
exponent. It also could be written with a radical sign rather than an exponent. Our point is that you may need
to “simplify” (change the form of) your answer so it matches one of the choices they give you. Since this relies
on your algebra skills and not calculus, we’ll leave it to you to make sure you’re able to do this. BE SURE TO
NOT MAKE MISTAKES WHEN YOU SIMPLIFY. You could have found the correct derivative, but if you
simplify incorrectly, you’ll lose valuable points.
Before we wrap up this lesson, let’s revisit a problem from the last lesson.
Example: Find lim
h0
(3  h) 2  (3) 2
h
Because of its basic structure, this looks like the limit definition of the derivative. It is! Remember that the
function you are finding a derivative for appears after the subtraction sign. In this case, there is no variable,
there’s just a number. That’s ok. The number simply replaces the variable in the function (the input, x, is given
to you). For this problem the function is f(x) = x2 and x = 3.
In other words we’re finding the derivative of y = x2 when x = 3. Here’s how to work it, using the power rule:
y = x2
y’ = 2x
y’(3) = 2(3) = 6
That’s right; this big ugly thing: lim
h0
(3  h) 2  (3) 2
is simply 6.
h
An Historical Perspective
There’s not an assignment that goes along with this material. In fact, you don’t really have to “know” this stuff.
However, we know that some of you will find the information interesting; if you don’t, please humor us.
History
Winston Churchill said, “Study history. Study history.” Just because this is a math class, don’t think you’re off
the hook. Listen up because calculus plays a huge role in human history and we want you to know something
about it.
Around 2500 years ago in Greece, modern scientific thought was born. Philosophers gathered the lore of
cultures older to them (then) than they are to us (now) and produced the flowering of inquiry and reason which
fueled Athens, then Rome, the Catholicism, then European capitalist expansion. The Greek notion of a static,
unchanging universe held sway for ages and was among the hallmarks of their legacy.
Calculus changed all that.
The independent discovery (invention?) by Newton and Leibniz of calculus in the 17th century worked like a
karate chop on the tired Aristotelian world-view. The mathematics of motion—of change over time—paved the
way for a new cosmology, the industrial revolution, breakthroughs in art, the whole shebang. With calculus in
his back pocket, Newton was able to model gravity, explore the properties of light and propose a “clock work
universe” whose workings, to the last detail, could be known to the mind of humanity.
His ideas necessitated the concepts of infinite space (a really big universe) and deep time (a really old universe)
which eventually led to the number-one big-time most-important invention (discovery) of all time: Darwin’s
theory of evolution by natural selection.
Sure, Newtonian physics has its limits of applicability—as Einstein pointed out with his Theory of Relativity
which proves that new rules apply near the speed of light. But Einstein didn’t prove Newton wrong—he just
proved there were “places” in the universe Newton never imagined existed.
Similarly, just when Einstein was being proven 100% right regarding relativity, the quantum physics revolution
(behind Bohr, Planck, and others) came along and proved relativity breaks down at the smallest scales. In none
of these cases did a new theory come along and disprove an existing theory. Relativity and quantum mechanics
merely extended thought into new realms the old Newtonian physics couldn’t touch.
And calculus was there every step of the way. Open a modern text on quantum chromodynamics—integrals
abound. In biology, genetics, business, engineering, psychology, you name it—calculus is used to model
quantities changing over time.
Because of its startling originality, its unmatched applicability to so many fields, its inherent simplicity and its
remarkable staying power over the past 300 years, calculus deserves to be viewed among the most important
achievements of humanity.
But perhaps we babble too much. Churchill probably studied history so hard only because he stunk at math and
perpetually put off starting his calculus homework. You will, of course, take a more enlightened track.
Lesson 4: More on Derivatives as Slopes
Back in lesson 1, we saw repeatedly that derivatives give us a way to find the slope of a curve at any given
point. We call this the instantaneous rate of change (vs. the average rate of change).
Let’s start off by reviewing a concept you should already have learned.
Slopes of Curves at Particular Points:
Example A: Find the slope of the curve given by f(x) = x2 + 5x + 6 at the point (-2, 0).
Even though the parabola y = x2 + 5x + 6 is definitely curved, it does have a distinct slope at every point. To
find the slope of this curve at the point (-2, 0) we do the following:
1. Find the derivative (use the power rule in this case)
2. Evaluate the derivative at x = -2.
Please show the work necessary to find the slope below.
Note that if they don’t give you the point at which to find the slope, they’ll have to give you at a minimum the
x-value. Of course, if they give you the x-value, finding the y-value is easy, right?! (How would you do it?)
Let’s move onto more intriguing problems.
Tangent Lines:
One way to visualize this “slope of a curve at a point” idea is to draw a tangent line to the curve. When we do
this we need to remember two important characteristics of a tangent line.
A tangent line:
1. Shares exactly one point with the curve (in an interval arbitrarily close to the point)
2. Has the same slope as the curve at that point
You’ll use these two ideas in the next two examples.
Example A: Find the equation of a tangent line to the curve y = x2 + 5x + 6, at the point (-2, 0).
Relying on point #1 above, we know that in the equation of a line (y = mx + b), that x = -2, and y = 0 since the
line must also go through the same point given for the curve.
Relying on point #2 above, we can calculate the value of m in the equation for the tangent line. This would
leave b, the y-intercept of the tangent line, as the only unknown. We can then solve for that variable to get our
final equation.
So, we know that the slope of the tangent line matches the value of the derivative at x = -2. That means we
need to find the derivative of the function. Then we can substitute x = -2 into the derivative to find the slope at
that point.
y’ = 2x + 5 (use the power rule)
y’ = 2(-2) + 5
y’ = -4 + 5 = 1
Since the slope of the tangent line is 1, this is also the value of m. Also remember that x = -2 and y = 0. This is
the point of tangency. So we get:
y  mx  b 
y  1x  b 

0  1 (-2)  b  y = 1x + 2
0  (-2)  b 


b2
Example B: Find the equation of the tangent line for y = x3 at x = 0.5.
I’ve included a graph of the curve and the tangent line to the right.
Notice the tangent line crosses the graph twice. That’s NOT breaking
any rules as long as the tangent line only shares one point with the
graph in an interval arbitrarily close to (0.5, 0.125).
Show your work below (using example A as a guide, if necessary).
In general, to find the equation of a line tangent to a curve f(x) at point (c, d):
1. Find f ’(x)
2. Evaluate f ’(a) (this means that you are finding the slope of the function at x = a)
3. Use this value as m in y = mx + b
4. Use (c, d) as x and y in y = mx + b
Normal Lines:
In a way, you can think of tangent lines as being “parallel” to a curve at a point
(parallel lines have the same slope, after all). If a line is “perpendicular” to a
curve at a point, on the other hand, we call it a normal line. Its slope will be
the opposite reciprocal of the curve’s slope at that point.
Example A: Find the equation of the line normal to y = x2 at (3, 9).
1. The derivative (slope) of y = x2 is y’ = 2x
2. At (3, 9) the slope is 2(3) or 6.
3. The opposite reciprocal of 6 is –1/6. This is the slope of the normal line.
4. The equation for the normal line is:
y = mx + b
y = –1/6 x + b
9 = –1/6 (3) + b
9 = -1/2 + b
9.5 = b
So, the equation is y = –1/6 x + 9.5.
Some TI-83 (TI-84) Stuff:
Your calculators are powerful tools to help you in finding derivatives. This section will introduce you to
several of these features. Please keep in mind that your calculator is not a substitute for you knowing how to
find derivatives by hand.
a. nDeriv tells you the derivative of any function for any x value.
Example A: Find the value of the first derivative of y = 3x2 at x = 5.
1.
2.
3.
4.
Hit MATH
Choose Option #8: nDeriv(
Enter the following: “3x2, x, 5)”.
Hit Enter
b. Tangent draws the tangent line to a functions graph at a specified point
1.
2.
3.
4.
c.
Hit 2nd PRGM (for the “Draw” menu)
Choose Option #5: Tangent(
Enter the function, a comma, and the x-coordinate of the point of tangency.
Hit Enter
dy
tells you the value of the first derivative of a function from its graph (the slope at a specified point).
dx
1. Graph the function
2. Hit 2nd Trace (for the “Calculate” menu).
dy
3. Choose Option #6:
dx
4. Type in the value for x that corresponds to the point where you want to find the slope of the graph
5. Hit Enter. The derivative (slope) appears in the lower left-hand corner.
OK. Time for one final thought.
All Curves are “Locally Straight”:
Well, actually that’s a bit of a lie. But, most curves are. Here’s what this means. If you graph a function, then
graph it’s tangent line at a point, they might look very different. In other words, it might be a curvy function,
but it will always be a straight tangent line.
But, if you magnify your view, the curve starts to “straighten out” and it is soon indistinguishable from the
tangent line!
Try this:
1. Graph y = cos x in the standard viewing window (set in radians, of course!)
2. Draw in the tangent line at x = .3174603175 (Use the steps in item “b” on the bottom of the previous
page).
3. Curvy cosine; straight tangent line, right?
4. Now zoom in around the point of tangency (option 2 in the “Zoom” menu).
5. Redraw the tangent line by going back to the home screen and repeating step 2 above.
6. Repeat step 5 three times. After the third time you can barely tell the difference between y = cos x and
its tangent line.
This concept is in many ways related to the idea that derivatives—instantaneous rates of change—are
equivalent to the limit of the average rate of change as the change in x becomes infinitesimally small. Think
about this. If you are finding the average rate of change between two points, you are essentially finding the
slope of a segment that connects the two points. If the points are far apart and the function really curvy, there
will be quite a bit of space between the two functions. (Can you picture this?)
But as the two points get closer together, the amount of space between the segment connecting these points and
the curve will get smaller. Draw yourself a picture if you don’t believe me.
It is important that you are able to understand this point, and really all points about derivatives, and the fact
that they represent slopes, from multiple perspectives—algebraically, and graphically.
Later we will use this “local straightness” of curves to shortcut some of our work. For now, just remember that
(almost) all curves are straight if you zoom in closely enough.
Use the space below to draw yourself some pictures/graphs that represent the concepts we’ve developed
algebraically in the past few examples.
Lesson 5: Derivatives—To Exist or Not to Exist
Don’t worry. Even though this sounds like it could get philosophical, we’ll resist the urge.
But, we do need to talk about situation where derivatives simply do not exist. “What’s that?” Well, remember
that derivatives represent the slope of a function at a specific point (the instantaneous rate of change). You’ll
also remember that to find the coordinates of any point on a function, you simply substitute a value for x into
the function and solve for y.
Well, there are many functions that are undefined for certain values of x. For example, f(x) =
x
.
x 1
When Derivatives Exist (or Don’t Exist):
The idea above gets at the idea behind the continuity of a function. This is one of the critical requirements for a
function to have a derivative at a certain point—the function must be continuous there.
Remember when we talked about limits and we tried to make the point that the limit of f(x) as x  a exists
only if the left-hand limit exists and the right-hand limit exists AND both limits are equal. Written out in math
form:
lim f(x) exists iff:
xa
1. lim f (x) = L
x a
2. lim f (x) = L
x a
where L is some real number.
Furthermore, for a function to be continuous at the point x = a, the limit as x  a must exist and it must equal
the value of the function at x = a. In math form:
f (x) is continuous at x = a iff
1. lim f (x) = L and
xa
2. f(a) = L
where L is some real number.
Knowing these two things allows us to make a huge leap forward. Here it is: f(x) is differentiable (has a
derivative) at x = a only if f (x) is continuous at x = a and the left-hand derivative (slope) equals the right-hand
derivative (slope) at x = a. In math form:
f ’(a) exists iff:
1. f (x) is continuous at x = a
2. the slope of f(x) to the left of a = the slope of f(x) to the right of a in an arbitrarily small interval
around a.
Yes, this is all very technical. But, it’s a necessity for you to know. Perhaps if we look at several concrete
examples and see how these requirements play out, you’ll appreciate them more.
Let’s start by talking about polynomials. The good news is that all polynomial functions have derivatives at all
points. So, rather than babble on about those boring polynomials which are differentiable everywhere, we’ll
show you some graphs of functions that fail to have derivatives at certain points and attempt to explain why
they fail to have derivatives at these points.
Example A: y = 1/x has no derivative at x = 0, because it is not continuous there.
Any time a graph has a vertical asymptote at x = a, it is not differentiable at x = a.
Look at the derivative of y = 1/x (or y = x-1). Using the power rule, y’ = -1x-2.
1
At x = 0, y’ = -1 (0)-2 = 2 which is undefined (division by zero).
0
x2  5x  6
Example B: y = 2
has no derivative at x = -1 and at x = -2, because
x  3x  2
( x  3)( x  2)
this function factors into y =
, and at x = -1, there is a vertical
( x  1)( x  2)
asymptote (discontinuity) and at x = -2, a hole (removable discontinuity) exists.
Later, when we learn to take derivatives of rational expressions, we’ll see that y’
is undefined at x = -1 and at x = -2 for this function.
Example C: y = |x| has no derivative at x = 0, because the left-hand slope does not match
the right-hand slope.
Remember that absolute value functions can also be written as piece-wise functions. For
this function,
x; if x  0
y= 
- x; if x  0
1; if x  0
and y’ = 
- 1; if x  0
In other words, the right-hand slope is a constant 1, while the left-hand slope is a constant –1. In fact, at x = 0,
this function is NOT locally straight. The “sharp corner” remains under any magnification (try it!).
To review, derivatives always exist at all points (assuming no absolute values are involved) for:
1. Polynomial functions
2. Exponential functions
3. Logarithmic functions (if x > 0) -- we’ll talk more about these later in the year
Derivatives do NOT exist, in other words, y = f (x) will not be differentiable at x = a if:
1. A “corner” occurs at x = a where the right-hand derivative is not equal to the left-hand derivative
2. A vertical asymptote occurs at x = a
3. A “hole” (removable discontinuity) occurs at x = a
4. A “jump” occurs at x = a
Please sketch a graph of each one of these types of functions below. You need to be able to recognize these
functions as non-differentiable and explain where and why they don’t have a derivative.
More specifically, the three types of functions we’ll deal with that have no derivatives at certain points are:
1. Absolute value functions
2. Rational functions
3. Piecewise functions
The first two have “sharp corners”, asymptotes or “holes”, and are relatively easy to identify. On the other
hand, piecewise functions can have several interesting characteristics that make them non-differentiable at
specific point(s). Let’s study piecewise functions more fully now.
x 2 ; if x  0
Example A: Does y = 
have a derivative at all points?
3x; if x  0
Well, if x < 0, it does (y = x2 is a polynomial) and if x > 0 it does (y = 3x is a polynomial), but x = 0 may
present a problem. We will need to check:
1. if continuity exists at x = 0
2. if the right- and left-hand limits match at x = 0?
This function will be continuous at x = 0 since 02 = 3(0). Continuity exists (there is no “hole” nor asymptote).
Now, let’s check the limits. To find the right-hand limit: y = 3x, so y’ = 3 (at all points).
To find the left-hand limit: y = x2, so y’ = 2x. At x = 0, y’ = 0.
Since the limits don’t match, the function is not differentiable at x = 0.
The next example presents a problem that has appeared on numerous AP tests in the past.
x 2 ; if x  2
Example B: What values of m and b must the function y = 
have to ensure that the function is
mx  b; if x  2
differentiable at all points?
To find out, we need the right- and left-hand slopes to match at x = 2 and we need continuity at x = 2.
To make the slopes match, derive both parts:
y = x2, so y’ = 2x
y = mx + b, so y’ = m
At x = 2, the two slopes from above must be equal. In other words, 2x must = m.
At x = 2, 2x = 2(2) = 4. So, m = 4.
x 2 ; if x  2
So far the function must look like this: y = 
.
4x  b; if x  2
To ensure continuity (no holes or jumps), we need the y values to be equal at x = 2. If y = x 2, then at x = 2, y =
4. So in y = 4x + b, we need y = 4, when x = 2.
4 = 4(2) + b
4=8+b
-4 = b
x 2 ; if x  2
Viola!! Our function, y = 
will be continuous and differentiable if m = 4 and b = -4.
mx  b; if x  2
Lesson 6: The Product Rule and the Quotient Rule
Here’s BIG IDEA #1, right off the bat. It’s called the Product Rule. You must memorize this fact.
If y = f(x)  g(x), then y’ = f’(x)  g(x) + g’(x)  f(x).
Example: If y = 3x2  4x5, find y’.
Let f(x) = 3x 2 . So f’(x) = 6x
Let g(x) = 4x 5 . So g’(x) = 20x 4
y’ = 6x  4x 5 + 20x 4  3x 2
 
 
f’(x) g(x) g’(x) f(x)
This simplifies to 24x 6 60 x 6  84 x 6 .
Why not just simplify first, you ask? Well, you could. If you did you’d get:
y = 3x 2  4x 5 or y = 12x 7 . Using the power rule, y’ = 84x 6 .
You get the same answer more quickly if you simplify first, but I’m trying to illustrate a new rule here, and you
have to admit that the Product Rule does get you the right answer! We will come across problems later where
the Product Rule is the only way to find f’(x). So if these examples seem “contrived” because an easier way
exists, just roll with it for now!
Example: Find y’ if y = (x 2 3)( x 3  2 x  1)


f(x)
g(x)
y’ = (3x 2 + 2)(x 2 + 3) + (2x)(x 3 + 2x + 1)




g’(x)  f(x) + f’(x)  g(x)
You could expand this all out and simplify, but
you don’t need to do so for me. If I ask for y’,
just give it to me in the first complete form rather
than hassling with a bunch of algebraic drudgery.
Of course, if you’re trying to check your answer against an answer key, and that answer is in simplified form,
then knock yourself out. You might also have to simplify to make your answer match one that’s listed in a
multiple-choice problem like those found on an AP Test.
A quick note on notation:
In calculus, the variables x and y typically represent the independent and dependent variables in a function. The
variables u and v are used to represent functions of x.
The product rule as it’s written above, takes up a lot of space. But if we let u = f(x) and v = g(x), then we get:
If y = uv, then y’ = u’v + v’u
This is much more pleasing to the eyes!
Next up is the Quotient Rule. You must memorize this fact also!
If y =
u
vu'uv'
where u and v are functions of x and v  0, then y’ =
.
v
v2
3x 2  1
Example: Find y’ if y =
.
4x3
First write down u and v.
u = 3x 2 + 1 v = 4x 3
Then find their derivatives.
u’ = 6x
v’ = 12x 2 .
Then “fill in” the Quotient Rule.
4 x 3  6 x  (3x 2  1)  12 x 2
y’=
(4 x 3 ) 2
Again, this could be simplified, but let’s not.
x5
Example: Find y’ if y = 2
.
x 1
u = x5
v = x 2 1
u’ = 5x 4
v’ = 2x
vu'uv'
( x 2  1)  5 x 4  x 5  2 x
y’ =
so
y’
=
. Again, don’t bother simplifying.
v2
( x 2  1) 2
There’s one more quick fact in this lesson you’ll want to know. It won’t come in too handy yet, but it will later!
It is the Constant Times a Function Rule, for lack of a better name.
If y = cf(x), then y’ = cf’(x).
If y = cf(x)g(x), then y’ = c(f’(x)g(x) + g’(x)f(x)).
In other words, just multiply the derivative by the constant. Notice that in the second statement above, the
product rule is also used.
Examples:
If y = 3(x 2 + 5x), then y’ = 3(2x + 5) or 6x + 15.
If y = 7(3x 2  4x 5 ), then y’ = 7(6x  4x 5 + 3x 2  20x 4 )
Again, it would be easier to just distribute the 3 into the parentheses in the first example, but later when we deal
with trig derivatives and such, this little item will be more useful.
Lesson 7: Derivatives of Trigonometric Functions
Derivatives of the 6 Trig Functions:
We start with an amazing fact—one that, yes, you must memorize.
If y = sin(x) then y’ = cos(x)
Yup! You got it. The derivative of sin(x) is cos(x). Put another way, the function y = sin(x) has a slope at
every point and that slope is the cosine of x at that point!
Phenomenal, isn’t it? Boy, I sure think so. I mean periodicity exists in nature, right? The sunlight emerges and
recedes, the seasons cycle through, strings vibrate—wave patterns exist all over the place, and the perfect
mathematical expression of this wavy periodicity is the function y = sin(x) [or y = cos(x), depending on which
starting point you choose]. And y = sin(x) has a slope—a derivative—which turns out to be none other than
y = cos(x) itself! WOW!!! Someone must have marveled at this fact when it was first discovered.
Anyway, I digress. Back to business!
It should come as no surprise that y = cos(x) has a derivative too. It has a slope at every point after all. Even
more remarkable is the fact that the slope of y = cos(x) can be found by evaluating –sin(x).
If y = cos(x) then y’ = -sin(x)
OK. So it’s not perfect symmetry (because of the negative sign) but it’s still pretty cool!
Armed with these two facts, we can use the power and quotient rules to find derivatives of the other 4 trig
functions. Let me show you one example. I’ll leave the others for you to do as an exercise.
Example: Find y’ if y = sec x
We know that sec x =
This makes y’ =
1
. Using the quotient rule with u = 1 and v = cos (x), we get u’ = 0 and v’ = -sin (x).
cos x
vu'uv' (cos x)(0)  (1)(  sin x)
sin x
sin x
1
=
=
=

= tan x  sec x
2
2
2
v
cos x
cos x cos x
(cos x)
This leads us to another important fact.
If y = sec(x), then y’ = sec(x)  tan(x)
We could easily repeat this process to get the following:
If y = csc (x), then y’ = -csc (x) cot (x)
Either trust me, or try it yourself (it’s good practice!).
So, should you memorize these two facts? That’s up to you. You certainly can if you want. But if you don’t,
or you start to panic on a test because you’re not sure if you are remembering it correctly, don’t forget that you
can always figure out the derivatives of these trig functions using the quotient rule.
OK. I’ll show you one more application of the quotient rule to help us find another derivative of one of the trig
functions. Here goes…
y = tan x =
sin x
cos x
Using the quotient rule with u = sin (x) and v = cos (x) we get:
1
cos 2 x  sin 2 x
cos x  cos x  (sin x)  ( sin x)
=
=
= sec2 x
2
2
cos 2 x
(cos x)
cos x
y’ =
So, if y = tan (x), then y’ = sec2 x
By the same method, if y = cot (x), then y’ = -csc2 x
For your convenience here are the derivatives of the 6 trig functions again. Memorize the first two! If you
want, memorize the other 4 also. Remember that you can find them in a pinch by using the quotient rule.
Function
y = sin x
y = cos x
y = sec x
y = csc x
y = tan x
y = cot x
Derivative
y’ = cos x
y’ = -sin x
y’ = sec x tan x
y’ = -csc x cot x
y’ = sec2 x
y’ = -csc2 x
Using the Derivatives of the 6 Trig Functions:
With these 6 trig derivatives in hand we can now derive some new types of functions.
Example A:
y = (sin x)(3x2)
Using the product rule with u = sin x, and v = 3x2 we get:
y’ = u’v + v’u = (cos x)(3x2) + 6x sin x
Example B:
y=
sec x
x2
Using the quotient rule with u = sec x, and v = x2 we get:
x 2 sec x tan x  (sec x)( 2 x)
x 2 sec x tan x  2 x sec x
y’ =
=
( x 2 )2
x4
Why the derivatives of the trig functions are what they are
A few of you may be dying to know why the derivative of y = sin x is the beautiful and elegant y’ = cos x. If
you are, see one of us. We can go through a (rather complex) proof with you. Rest assured, you won’t have to
know this for the AP test in May. It used to be a mandatory part of the curriculum, but the AP Calc gods no
longer require it.
These things you must know, however. First are the derivatives of the six trig functions. Second is how to use
the product and quotient rules to find derivatives of functions involving sin(x) and cos(x). Finally, is the idea
that:
lim
x0
sin x
=1
x
For some reason, this seems to crop up from time to time in our later work. If you want to see the proof of the
limit above, your textbook has it on page 63, as Theorem 1.9. The proof uses something called the “Sandwich
Theorem” (sounds yummy!). Again, you won’t need to know how to do this proof.
OK. So we know that we’ve already lost most of you to tearing through the book to find page 63 and the proof
of this idea. But if you can wait (oh, the agony!), we’ll also show why this idea is true when we learn
L’Hopital’s Theorem in Unit 2.
Lesson 8: The Chain Rule (Derivatives of Composite Functions)
In this lesson you will learn:
1.)
How to use the Chain Rule to take derivatives of composite functions
2.)
How to combine the chain, power, product, and quotient rules to take derivatives of a huge new
class of functions
3.)
How to evaluate derivatives of functions at a given point without even knowing what the
functions are
4.)
A bit of new notation
How to use the Chain Rule
From previous lessons we know how to find derivatives of polynomial functions (use the Power Rule),
quotients of polynomial functions (use the Quotient Rule), products of polynomial functions (use the Product
Rule), and trig functions (use the 6 trig derivatives from lesson 7).
In this lesson, we expand our skills of finding derivatives by learning how to derive composite functions. Let’s
start off with an example. Remember that composite functions are really functions inside of other functions.
For example, sin (3x + 4) is a composite function. The “inside function” is (3x + 4). The main (or “outside”)
function is sin x.
Example A: If y = sin (x 2 ), then y’ = cos (x 2 )  2x.
In this case, sine is the “outside function” in the composition and x 2 is the “inside function.” We sort of
derived the “outside” first and then the “inside” last and multiplied the two derivatives. Here’s another
example.
(The “outer” is cos x and the derivative of the “outer” is –sin x.
The “inner” is x 3 and the derivative of the “inner” is 3x 2 .)
3
y’ = -sin(x )  3x 2


Example B: y = cos(x 3 ).
deriv. of outer
deriv. of inner
If this “inner-outer”, intuitive, example-based way of finding derivatives works for you, stay with it. If not,
here’s a more “symbolic” way of looking at it. It is known as The Chain Rule.
If y = f(g(x)), then y’ = f’(g(x))  g’(x)
Here are several examples.
Example C: y = sin (x 2 )
f(x) = sin x
g(x) = x 2
f’(x) = cos x
g’(x) = 2x
y’ = f’(g(x))  g’(x)
y’ = cos (x 2 )  2x
Example D: y = (2x + 5) 10
f(x) = x 10
f’(x) = 10x 9
g(x) = 2x + 5 g’(x) = 2
y’ = f’(g(x))  g’(x)
y’ = 10(2x + 5) 9  2 = 20(2x + 5) 9
Example E: y = sin 4 (x)
f(x) = x 4
g(x) = sin x
f’(x) = 4x 3
g’(x) = cos x
y’ = f’(g(x))  g’(x)
y’ = 4(sin x) 3  cos x = 4(sin 3 x)(cos x)
Sometimes our composite functions are “nested.” In other words there are 3 different levels of functions.
Example F:
y = sin 4 (3x 2 )
This is the same as y = (sin(3x 2 )) 4
Here, the outermost function is (x 4 ), the middle function is (sin x), and the innermost function is 3x 2 .
This could be written as y = f(g(h(x))) where f(x) = x 4 , g(x) = sin x, and h(x) = 3x 2 .
The chain rule now leads us to declare the following:
y’ = f’(g(h(x)))  g’(h(x))  h’(x)
So, to finish:
y = sin 4 (3x 2 )
f(x) = x 4
g(x) = sin x
h(x) = 3x 2
f’(x) = 4x 3
g’(x) = cos x
h’(x) = 6x
y’= 4(sin(3x 2 )) 3  cos(3x 2 )  6x
Isn’t this fun? I suppose it is if you like symbol-laden math problems with long answers. If you liked that, wait
until you see what’s coming…
Product/Chain Rule Problems
Example G: y = cos x  sin 2 (x)
It’s a product rule so use:
u = cos x
v = sin 2 x
u’ = -sin x
v’ = ?
To find v’, use the chain rule on y = sin 2 x.
f(x) = x 2
f’(x) = 2x
g(x) = sin x g’(x) = cos x
So, y’= f’(g(x))  g’(x) = 2(sin x)  cos x.
We now have:
y = cos x  sin 2 (x)
This is the v’ you need above.
u = cos x
u’ = -sin x
y’ = v’u  u’v
= 2(sin x)  cos x  cos x + -sin x  sin 2 x




v’
u
u’
v
= 2(sin x)(cos 2 x) + (-sin 3 x)
v = sin 2 x
v’ = 2(sin x)  cos x
Some New Notation
The whole y = f(g(x))  y’ = f’(g(x))  g’(x) is a pretty complicated expression. It can be simplified this way:
Rewrite y = f(g(x)) as y = f(u) with u = g(x). Now y = f(u) so y’ = f’(u)  u’.
Example H:
y = cos (x 4 )  y = cos (u) where u = x 4 and u’ = 4x 3 .
If y = cos (u), then y’ = -sin(u)  4x 3 = -sin(x 4 )  4x 3 .
More Notation Notes
dy
dy
= 2x. This expression “
” is the same as y’—it’s the derivative
dx
dx
of y—but it’s pronounced “dee why dee ex” and means “the derivative of function y with respect to variable x”.
If y = x 2 we say y’ = 2x. But we also say
dy
notation yet because all of our functions have been of the form y = f(x). Now, however,
dx
dy
we have a need to call in the
notation because of this u = g(x) idea above. If we have a composite function
dx
y = f(g(x)) and we let g(x) = u, we get
y = f(u) with u = g(x).
We haven’t used
We now can say if y = f(u) and u = g(x), then
dy dy du

=
dx du dx
Example I:
y = cos (x 4 ).
Let x 4 = u
y = cos u
dy dy du


dx du dx
dy
du
  sin u
 4x 3
du
dx


since y = cos u since u = x 4
dy
so,
= -sin u  4x 3 = -sin (x 4 )  4x 3
dx
Example J:
Find
dy
if y = (x 2 1) 5
Let u = x 2 + 1
dx
dy
 5u 4
y = u 5 , so
du
dy dy du

Finally,
=
= 5u 4  2x = 5(x 2 + 1) 4  2x
dx du dx
du
= 2x
dx
I guess there’s one thing to realize here. Everything we’ve done in the lesson revolves around the central idea
introduced at the beginning of the unit. You’ve seen this idea presented via different examples using different
notation. IT IS NOT THE NOTATION THAT IS IMPORTANT! If you understand the central idea behind the
chain rule, you can use whatever notation you want to express that idea.
Derivatives Without Functions:
Sometimes the book will ask you to find the derivative of a function at a point without telling you the function!
What could they possibly be thinking?!
Hey! Never fear! We’ll work though this. Just make sure you understand what’s happening along the way!
OK. Finding f’(3) if f(x) = x 2 isn’t too bad…
f(x) = x 2 ,
f’(x) = 2x.
f’(3) = 23 = 6
But to say find f’(3) and not tell you what f(x) is…
We can handle this. You’ll have a table of values that will help you get the answer.
Example K: Given:
x
2
3
f(x)
4
5
g(x)
2
3
f’(x)
-1
-2
g’(x)
-3
-4
Here are some problems that require you to use this table.
1.)
2.)
3.)
Find f(3)…duh…5
Find f’(2)…duh…-1
Find the derivative of f(x)  g(x) at x = 3.
O.K. This is tougher. First, write g’(x)  f(x) + f’(x)  g(x)

the product rule on f(x)  g(x)
At x = 3, we get: g’(3)  f(3) + f’(3)  g(3)
-4  5 + -2  3 = -20 + -6 = -26
4.)
Using the table of values we get:
Find the derivative of f(g(x)) at x = 3.
Write f’(g(x))  g’(x)
This is the Chain Rule on f(g(x)).
Now, f’(g(3))  g’(3)
x=3
Now, f’(3)  -4
since g(3) = 3 and g’(3) = -4
Finally, -2  -4 = 8.
since f’(3) = -2
With the power, product, quotient, and chain rules under our belts, we can derive any product, quotient or
composition of polynomial and trig functions! (Well, almost!) And by the time you’re done with the
assignment, you’ll feel like you’ve found the derivative of all of them. But there’s more to come.
For example, how do we derive x 2 y = 2yx + x 2 ?
Or, y = log x?
Or y = 4 ( x 3) ?
Time will tell.
Lesson 9: Implicit Differentiation
Implicit Functions:
The function y = sin (x) is explicitly defined because it’s possible to get y alone in the equation. In fact, most
functions we’ve dealt with so far this year are explicit functions.
The function xy + x2 = sin (x) is an implicitly defined function because we can’t get y alone. (Most implicit
functions aren’t even functions because they fail the vertical line test, but we’ll call them functions anyway
because… well, just because!)
Working with Implicit Functions:
dy du

Remember the
notation you learned in the last lesson. It’s going to come in handy when working with
du dx
implicit function as well. That’s because we use the chain rule when we derive implicit functions.
Some “facts” to get us going…
d ( x2 )
= 2x
dx
d ( y)
The derivative of y (with respect to x) is just y’. In other words,
= y’
dx
The derivative of x2 (with respect to x) is 2x. We can write this:
The derivative of y2 (with respect to x) is really a chain rule problem.
We have
d ( y2 )
dy 2 dy
=
= 2y  y’  The derivative of y with respect to x

dx
dy dx

The derivative of y2 with respect to y
Here are some additional examples…
Example A:
Example B:
Example C:
d (sin y )
= cos(y)  y’
dx
d (3 y )
= 3  y’
dx
d (cos 2 y )
= 2cos y  -sin y  y’
dx
Example D:
d ( x2 y2 )
= 2x  y2 + 2y  y’  x2
dx
 


u’  v + v’  u
This is just the product rule with u = x2 and v = y2
Example E:
d ( xy)
= 1  y + x  y’
dx
   
u’ v + u  v’
Product rule again!
If you accept the examples above, these steps will always find the derivative of an implicitly defined function:
1. Take the derivative of each term (with respect to x)
2. Isolate y’
Example F: Find y’ for the following:
y2 + cos y = x2 + y
1. Derive each term
2. Get all y’ terms on the left-hand side
3. Factor out y’ from all of the terms
2y  y’ + -sin y  y’ = 2x + y’
2y  y’ + -sin y  y’ – y’ = 2x
y’ (2y – sin y – 1) = 2x
2x
y’ =
2 y  sin y  1
4. Divide
Example G: Find y’ for the following:
1. Derive each term (see Ex. E above)
2. Get all y’ terms on the left-hand side
3. Factor out y’
4. Divide
y + 2x = xy
y’ + 2 = 1  y + x  y’
y’ – xy’ = y – 2
y’ (1 – x) = y – 2
y2
y’ =
1 x
Evaluating Implicit Derivatives:
It makes sense to ask, “What’s the slope of y = x2 at x = 10?” Since y’ = 2x, the answer is 2(10) = 20. (Note
that y = x2 is an explicit function.)
It makes less sense to ask, “What’s the slope of y + 2x = xy at x = 10?” (Note that this is an implicit function.)
y2
y2
as we found in Example G above. At x = 10, y’ =
1 x
1  10
which doesn’t mean too much because the y-variable is still unknown. To resolve this problem, when you are
finding the slope of an implicit function at some point, you need to know the (x, y)-coordinate pair of that point.
Here’s why. The slope of y + 2x = xy is y’ =
Example H: Find the slope of y2 = 2x – 6 + xy at (3, 0).
1. Derive
2y  y’ = 2 + 0 + 1  y + x  y’
2. Isolate
2y  y’ – xy’ = 2 + y
y’ (2y – x) = 2 + y
2 y
y’ =
2y  x
20
2
3. Evaluate
y’ =
=
3
2(0)  3
Vertical Tangent Lines:
You guessed it. Just like explicit functions can have vertical tangent lines (can you think of one right now?!), so
can implicitly defined functions.
Example: Find the points at which y2 = 2x – 6 + xy has a vertical tangent line.
Well, a vertical tangent line means the slope is infinite. And, an infinite slope means there is an infinite
2 y
derivative. The derivative of this function (from Example H above) is y’ =
and this derivative will be
2y  x
infinite if the denominator = 0. Setting 2y – x = 0 and solving for x gives x = 2y.
Use this in the original equation…
y2 = 2x – 6 + xy becomes
y2 = 2(2y) – 6 + (2y)(y) or
y2 = 4y – 6 + 2y2 or
0 = y2 + 4y – 6
Using your TI-83 (TI-84), you can solve for y. You should get y = -5.163 or 1.163.
Now use x = 2y to find the corresponding x-values. You should get x = -10.326 for the first one, and x = 2.326
for the second.
This means at (-10.326, -5.163) and at (2.326, 1.163) the function has vertical tangent lines.
Higher Order Derivatives:
You already know how to find the derivative of many functions. For example, you know that for y = x 3 the
derivative is y’ = 3x2. But y’ = 3x2 also has a derivative. It is y” = 6x. We pronounce that “y-double-prime”
and call it the second derivative of y = x3. There’s no need to stop at the second derivative either. In this case,
y’’’ = 6, y(4) = 0, y(5) = 0 and so on to infinity. (Note the change in notation after y’’’— the third derivative.)
From this example it’s easy to see that algebraic functions (polynomials) have a limited number of distinct
derivatives—at some point the power rule runs out of gas and you get y(n) = 0. Trig functions are very different,
however. If y = sin (x), y’ = cos (x), y’’ = -sin (x), y’’’ = -cos (x), and y (4) = sin (x), starting the infinite
derivative loop over again.
Composite functions get rather messy in a hurry. Consider the function, y = sin (x7). Here, y’ = cos (x7)  7x6.
For y” you need to use a joint chain-product rule. This would produce y” = cos (x7)  42x5 + -sin (x7)  7x6  7x6.
To find y’’’, you would need to run a joint product-chain rule for each term. Yulk!! 
Here’s one last note on notation. The second derivative of y can be written as
d2y
.
d ( x2 )
Second Derivatives of Implicit Functions:
If y2 = x2 + sin y, finding y’ isn’t tough…
2y  y’ = 2x + cos y  y’
2y  y’ – cos y  y’ = 2x
y’ (2y – cos y) = 2x
2x
y’ =
Whoop! There it is!! 
2 y  cos y
Finding y”, on the other hand, is a bit trickier. Start by deriving the first derivative. 
Using the quotient rule with u = 2x, and v = 2y – cos y,
(2 y  cos y )  2  2 x(2 y ' sin y  y ' )
y” =
(2 y  cos y ) 2
Now we need to eliminate all of the y’ terms from the second derivative equation. How?
2x
By plugging
in for y’. (We found that above, remember?)
2 y  cos y
 

x
(2 y  cos y )  2  2 x 2
  sin
 2 y  cos y 

y’’ =
(2 y  cos y ) 2


x
y
 
 2 y  cos y  
Yup!! That’s it! Pretty simple, huh? You may simplify this at your own risk! 
Lesson 10: Derivatives of Inverse Functions
Here’s the central, shining fact about the lesson:
If f(x) and g(x) are inverses of one another, then, f’(a) =
1
, assuming of course that f(x) and g(x)
g ' ( f (a))
are both differentiable functions and that g’(f(a))  0).
Here’s why it’s true:
1.)
2.)
3.)
If f(x) and g(x) are inverses and if f(x) contains (a, b) and (c, d) then g(x) contains (b, a) and (d, c).
That’s because the definition of inverse functions involves reflecting across the line y = x…switching x
for y.
bd
Now, if f(x) contains (a, b) and (c, d), the slope of the line segment is m 1 =
. The corresponding
ac
1
ac
points for g(x) yield a slope m 2 =
. Clearly, m 1 =
.
bd
m2
Now, imagine a point (x 1 , y 1 ) on f(x) such that a < x 1 < c and b < y 1 < d. Imagine points (a, b) and (c,
d) sliding along f(x) towards x 1 …the slope of this ever-shortening segment approaches f’(x 1 ), right?
Likewise, for g(x), the slope is approaching g’(y 1 )…and the reciprocal relationship from point 2 above
1
persists: f’(x 1 ) =
.
g ' ( y1 )
If you understand all of this, great! If not, a picture may help. We’re going to work through the graphical
representation of this idea with you. If this perspective is what makes you understand the concept, be sure to
copy our completed graph below.
Once again the central idea of this unit is this fact:
Slope of tangent line to f(x) at x 1 = reciprocal of slope of tangent line to g(x) at y 1 .
Example: If f(x) = x 2 , its inverse is g(x) =
1
equation f’(a) =
is true.
g ' (b)
f’(x) = 2x
g’(x) =
1
2 x
a = 3, b = 9
x . Show that at the point (3, 9) on f(x) and (9, 3) on g(x) the
f’(a) = f’(3) = 2(3) = 6
Since f’(3) is the reciprocal of g’(3), then the relationship is true.
g’(b) =g’(9) =
1
2 9
=
1
.
6