The following content is
provided under a Creative
Commons license.
Your support will help
MIT OpenCourseWare
continue to offer high quality
educational resources for free.
To make a donation, or to
view additional material
from hundreds of MIT courses,
visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: Right now,
we're finishing up
with the first
unit, and I'd like
to continue in this
lecture, lecture seven,
with some final remarks
about exponents.
So what I'd like to do
is just review something
that I did quickly
last time, and make
a few philosophical
remarks about it.
I think that the steps involved
were maybe a little tricky,
and so I'd like to go
through it one more time.
Remember, that we were
talking about this number a_k,
which is (1 + 1/k)^k.
And what we showed was
that the limit as k
goes to infinity of a_k was e.
So the first thing
that I'd like to do
is just explain the proof
a little bit more clearly
than I did last time
with fewer symbols,
or at least with this
abbreviation of the symbol
here, to show you
what we actually did.
So I'll just remind you
of what we did last time,
and the first
observation was to check,
rather than the limit
of this function,
but to take the log first.
And this is
typically what's done
when you have an exponential,
when you have an exponent.
And what we found was
that the limit here
was 1 as k goes to infinity.
So last time, this
is what we did.
And I just wanted to
be careful and show you
exactly what the next step is.
If you exponentiate this fact;
you take e to this power,
that's going to tend to
e^1, which is just e.
And then, we just observe
that this is the same as a_k.
So the basic ingredient
here is that e^ln a = a.
That's because the log
function is the inverse
of the exponential function.
Yes, question?
STUDENT: [INAUDIBLE]
PROFESSOR: So the question
was, wouldn't the log of this
be 0 because a_k
is tending to 1.
But a_k isn't tending to 1.
Who said it was?
If you take the logarithm,
which is what we did last time,
logarithm of a_k is
indeed k * ln(1 + 1/k).
That does not tend to 0.
This part of it tends to 0, and
this part tends to infinity.
And they balance each
other, 0 times infinity.
We don't really know yet
from this expression,
in fact we did some cleverness
with limits and derivatives,
to figure out this limit.
It was a very subtle thing.
It turned out to be 1.
All right?
Now, the thing that
I'd like to say
- I'm sorry I'm going
to erase this aside here
- but you need to go
back to your notes
and remember that this
is what we did last time.
Because I want to
have room for the next
comment that I want to make on
this little blackboard here.
What we just derived
was this property here,
but I made a claim
yesterday, and I just
want to emphasize it
again so that we realized
what it is that we're doing.
I looked at this backwards.
One way you can think of this
is we're evaluating this limit
and getting an answer.
But all equalities can
be read both directions.
And we can write
it the other way:
e equals the limit, as k goes
to infinity, of this expression
here.
So that's just the same thing.
And if we read it
backwards, what we're saying
is that this limit
is a formula for e.
So this is very
typical of mathematics.
You want to always reverse
your perspective all the time.
Equations work both
ways, and in this case,
we have two different
things here.
This e was what we
defined as the base,
which when you graph e^x,
you get slope 1 at 0.
And then it turns out to be
equal to this limit, which
we can calculate numerically.
If you do this on your
calculators, you, of course,
will have a way of
programming in this number
and evaluating it for each k.
And you'll have another button
available to evaluate this one.
So another way of
saying it is it
that there's a relationship
between these two things.
And all of calculus is a matter
of getting these relationships.
So we can look at these things
in several different ways.
And indeed, that's
what we're going
to be doing at least
at the end of today
in talking about derivatives.
A lot of times when we
talk about derivatives,
we're trying to look at them
from several perspectives
at once.
Okay, so I have to keep
on going with exponents,
because I have one loose end.
One loose end that
I did not cover yet.
There's one very important
formula that's left,
and it's the derivative
of the powers.
We actually didn't
do this - well we
did it for rational numbers r.
So this is the formula here.
But now we're going to check
this for all real numbers, r.
So including all the
irrational ones as well.
This is also good
practice for using base e
and using logarithmic
differentiation.
So let me do this
by our two methods
that we can use to handle
exponential type problems.
So method one was base e.
So if I just rewrite
this base e again,
that's just this
formula over here.
x^r = (e^ln x)^r,
which is e^r ln x.
Okay, so now I can
differentiate this.
So I get that d/dx (x^r),
now I'm going to use prime
notation, because I don't want
to keep on writing that d/dx
here; (e^(r ln x))'.
And now, what I can do is
I can use the chain rule.
The chain rule says
that it's the derivative
of this times the
derivative of the function.
So the derivative of the
exponential is just itself.
And the derivative
of this guy here,
well I'll write it out
once, is (r ln x)'.
So what's that equal to?
Well, e^(r ln x) is is just x^r.
And this derivative here is--
Well the derivative of r is 0.
This is a constant.
It just factors out.
And ln x now has derivative--
What's the derivative of ln x?
1/x, so this is going
to be times r/x.
And now, we rewrite it in the
customary form, which is r,
we put the r in front, x^(r-1).
Okay?
So I just derived
the formula for you.
And it didn't matter whether
r was rational or irrational,
it's the same proof.
Okay so now I have to show you
how method two works as well.
So let's do method
two, which we call
logarithmic differentiation.
And so here I'll use a
symbol, say u, for x^r,
and I'll take its logarithm.
That's r ln x.
And now I differentiate it.
I'll leave that in
the middle, because I
want to remember
the key property
of logarithmic differentiation.
But first I'll differentiate it.
Later on, what I'm going to
use is that this is the same
as u'/u.
This is one way of evaluating
a logarithmic derivative.
And then the other
is to differentiate
the explicit function
that we have over here.
And that is just,
as we said, r/x.
So now, I multiply through, and
I get u' = ur/x which is just
x^r r/x, which is just
what we did before.
It's r x^(r-1).
Again, you can now
see by comparing
these two pieces of arithmetic
that they're basically
the same.
Pretty much every time
you convert to base e
or you do logarithmic
differentiation,
it'll amount to the
same thing, provided
you don't get mixed up.
You generally have to
introduce a new symbol here.
On the other hand, you're
dealing with exponents there.
It's worth it to know
both points of view.
All right, so now I want to
make one last remark before we
finish with exponents.
And, I'll try to sell this
to you in a lot of ways
as the course goes on,
but one thing that I
want to try to emphasize is that
the natural logarithm really
is natural.
So, I claim that the
natural log is natural.
And the example that we're going
to use for this illustration
is economics.
Okay?
So let me explain to why the
natural log is the one that's
natural for economics.
If you are imagining the
price of a stock that you own
goes down by a dollar, that's a
totally meaningless statement.
It depends on a lot of things.
In particular, it depends on
whether the original price
was a dollar or 100 dollars.
So there's not much meaning
to these absolute numbers.
It's always the
ratios that matter.
So, for example, I just
looked up an hour ago,
the London Exchange closed,
and it was down 27.9,
which as I said, is
pretty meaningless
unless you know what the
actual total of this index is.
It turns out it was 6,432.
So the change in
the price, divided
by the price, which in
this case is 27.9 / 6,432,
is what matters.
And, in this case, it
happens to be .43%.
All right?
That's what happened today.
And similarly, if you take
the infinitesimal of this,
people think of days as being
relatively small increments
when you're
investing in a stock,
you would be interested in
the infinitesimal sense,
you would be interested in p'/p.
The derivative of
p divided by p.
That's just (ln p)'.
So this is the - let me
put a little box around it
- the formula of
logarithmic differentiation.
But let me just emphasize that
it has an actual significance,
and it's the one that's used
by economists and people who
are modeling prices of
things all the time.
They never use absolute prices
when there are large swings.
They always use the
log of the price.
And there's no point in using
log base 10, or log base 2.
Those give you junk.
They give you an
extra factor of log 2.
It's the natural log that's
the obvious one to use.
It's completely
straightforward that this
is a simpler expression
than using log base 10
and having a factor of
natural log of 10 there,
which would just
mess everything up.
All right, so this is
just one illustration.
Anything that has
to do with ratios
is going to
encounter logarithms.
All right, so that's
pretty much it.
That's all I want to
say for now anyway.
There's lots more
to say, but we'll
be saying it when we do
applications of derivatives
in the second unit.
So now, what I'd like to
do is to start a review.
I'm just going to run through
what we did in this unit.
I'll tell you
approximately what I
expect from you on the test
that's coming up tomorrow.
And, well, so let's
get started with that.
So this is a review of Unit One.
And I'm just going to put on
the board all of the things
that you need to think about,
anyway, keep in your head.
And there are what are
called general formulas
for derivatives.
And then there are
the specific ones.
And let me just remind you
what the general formulas are.
There's what you
do to differentiate
a sum, a multiple of a
function, the product
rule, the quotient rule.
Those are several
general formulas.
And then there's
one more, which is
the chain rule, which
I'm going to say just
a little bit more about.
So the derivative of a
function of a function
is the derivative of the
function times the derivative
of the other function.
So here, I've
abbreviated u is u(x).
Right, so this is one of
two ways of writing it.
The other way is also one
that you can keep in mind
and you might find
easier to remember.
It's probably a good idea
to remember both formulas.
And then the last type
of general formula
that we did was implicit
differentiation.
Okay?
So when you do implicit
differentiation,
you have an equation
and you don't
try to solve for the
unknown function.
You just put it in its simplest
form and you differentiate.
So, we actually did this,
in particular, for inverses.
That was a very, very key
method for calculating
the inverses of functions.
And it's also true that
logarithmic differentiation
is of this type.
This is a transformation.
We're differentiating
something else.
We're transforming the equation
by taking its logarithm
and then differentiating.
Okay, so there are a number of
different ways this is applied.
It can also be applied,
anyway, these are two of them.
So maybe in parenthesis.
These are just examples.
All right.
I'll try to give examples of
at least a few of these rules
later.
So now, the specific
functions that you know how
to differentiate: well you know
how to differentiate now x^r
thanks to what I just did.
We have the sine and
the cosine functions,
which you're
responsible for knowing
what their derivatives are.
And then other trig functions
like tan and secant.
We generally don't bother
with cosecants and cotangents,
because everything can be
expressed in terms of these
anyway.
Actually, you can really
express everything
in terms of sines and cosines.
But what you'll
find is that it's
much more convenient to remember
the derivatives of these
as well.
So memorize all of these.
All right, and then
we had e^x and ln x.
And we had the inverses
of the trig functions.
These were the two that we did:
the arctangent and the arcsine.
So those are the ones
you're responsible for.
You should have enough time,
anyway, to work out anything
else, if you know these.
All right, so
basically the idea is
you have a bunch of
special formulas.
You have a bunch of
general formulas.
You put them
together, and you can
generate pretty much anything.
Okay, so let's do a few
examples before going on
with the review.
Okay, so I do want to do a few
examples in sort of increasing
level of difficulty
in how you would
combine these things together.
So first of all,
you should remember
that if you differentiate the
secant function, that's just
- oh I just realized that
I wanted to say something
else before - so forget that.
We'll do that in a second.
I wanted to make
some general remarks.
So there's one rule that you
discussed in my absence, which
is the chain rule.
And I do want to make
just a couple of remarks
about the chain rule now to
remind you of what it is,
and also to present
some consequences.
So, a little bit of
extra on the chain rule.
The first thing that I want
say is that we didn't really
fully explain why it's true.
And I do want to just
explain it by example, okay?
So imagine that you have
a function which is, say,
10x + b.
All right?
So y = 10x + b.
Then obviously, y is changing
10 times as fast as b, right?
The issue is this number
here, dy/dx, is 10.
And now if x is a
function of something,
say t, shifted by some other
constant here, then dx/dt = 5.
Now all the chain rule is saying
is that if y is going 10 times
as fast as t, I'm sorry as
x, and x is going 5 times
as fast as t, then y is
going 50 times as fast as t.
And algebraically, all
this means is if I plug
in and substitute, which is
what the composition of the two
functions amounts to, 10(5t +
a) + b and I multiply it out,
I get 50t + 10a + b.
Now these terms don't matter.
The constant terms don't matter.
The rate is 50.
And so the consequence,
if we put them together,
is that dy/dt =
10*5, which is 50.
All right, so this
is in a nutshell
why the chain rule works.
And why these rates multiply.
The second thing that I wanted
to say about the chain rule
is that it has a few
consequences that
make some of the other
rules a little easier
to remember or
possibly to avoid.
The messiest rule
in my humble opinion
is the quotient rule, which is
kind of a nuisance to remember.
So let me just
remind you, if you
take just the reciprocal
of a function,
and you differentiate
it, there's
another way of looking at this.
And it's actually
the way that I use,
so I want to encourage you to
think about it this way too.
This is the same as (v^(-1))'.
.
And now instead of using
the quotient rule, which
we could've used, we can
use the chain rule here
with the power -1, which
works by the power law.
So what is this equal to?
This is equal to -v^(-2) v'.
So here, I've applied the
chain rule rather than
the quotient rule.
And similarly, suppose I wanted
to derive the full quotient
rule.
Well, now this may
or may not be easier.
But this is one way of
remembering what's going on.
If you convert it to uv^(-1)
and you differentiate that,
now I can use the
product rule on this.
Of course, I have to use
the chain rule and this rule
as well.
So what do I get?
I get u', the inverse,
plus u, and then I have
to differentiate the v inverse.
That's the formula
right up here.
That's -v^(-2) v'.
So that's one way of doing it.
This actually explains
the funny minus sign
when you differentiate
v in the formula.
The other formula, the
other way that we did it,
was by putting this over
a common denominator.
The common denominator was v^2.
This comes from this v v^(-2).
And then the second
term is -uv'.
And the first term, we have
to multiply by an extra factor
of v, because we have a
v^2 in the denominator.
So it's u'v. All right, so
this is the quotient rule as we
wrote it down in lecture,
and this is just another way
of remembering it or deriving
it without remembering it,
if you just remember the chain
rule and the product rule.
Okay, so you'll find
that in many contexts,
it's easier to do
one or the other.
Okay, so now I'm ready to
differentiate the secant
and a few such functions.
So we'll do some
examples here here.
So here's the secant
function, and I
want to use that formula up
there for the reciprocal.
This is the way I think of it.
This is the cosine
function to the power -1.
And so, the formula
here is just what?
It's just -(cos
x)^(-2) times -sin x.
So now this is usually written
in a different fashion,
so that's why I'm doing
this for a reason actually.
Which is although there are
several formulas for things,
with trig functions,
there are usually
five ways of writing something.
So I'm writing this
one down so that you
know what the standard
way of presenting it is.
So what happens here is
that we have two minus signs
cancelling.
And we get sin x / cos^2 x.
That's a perfectly
acceptable answer,
but there's a customary
way in which is written.
It's written (1 / cos
x) (sin x / cos x).
And then we get rid
of the denominators
by rewriting it in terms
of secant and tangent,
so sec x tan x.
So this is the form
that's generally
used when you see these
formulas written in textbooks.
And so you know, you
need to watch out,
because if you ever want to
use this kind of calculus,
you'll have not be put off by
all the secants and tangents.
All right, so getting
slightly more complicated,
how about if we
differentiate ln(sec x)?
If you differentiate
the natural log,
that's just going to
be (sec x)' / sec x.
And plugging in
the formula that we
had before, that's sec x tan
x / sec x, which is tan x.
So this one also has
a very nice form.
And you might say that this
is kind of an ugly function,
but the strange thing is that
the natural log was invented
before the exponential by
a guy named Napier, exactly
in order to evaluate
functions like this.
These are the functions that
people cared about a lot,
because they were
used in navigation.
You wanted to multiply
sines and cosines together
to do navigation.
And the multiplication he
encoded using a logarithm.
So these were invented
long before people even
knew about exponents.
And it was a surprise,
actually, that it
was connected to exponents.
So the natural log was
invented before the log base 10
and everything else, exactly
for this kind of purpose.
Anyway, so this is
a nice function,
which was very important,
so that your ships wouldn't
crash into the reef.
Okay, let's continue here.
So there's another
kind of function
that I want to discuss with you.
And these are the
kinds in which there's
a choice as to which of
these rules to apply.
And I'll just give a
couple of examples of that.
There usually is a
better and a worse way,
so let me illustrate that.
Okay, yet another example.
I hope you've seen
some of these before.
Say (x^10 + 8x)^6.
So it's a little bit more
complicated than what
we had before, because there
were several more symbols here.
So what should we
do at this point?
There's one choice which
I claim is a bad idea,
and that is to expand
this out to the 6th power.
That's a bad idea,
because it's very long.
And then your answer
will also be very long.
It will fill the entire
exam paper, for instance.
Yeah?
STUDENT: Can you
use the chain rule?
PROFESSOR: Chain rule.
That's it.
We use the chain rule.
So fortunately, this
is relatively easy
using the chain rule.
We just think of this box
as being the function.
And we take 6 times
this guy to the 5th,
times the derivative of this
guy, which is 10x^9 + 8.
And this is, filling
this in, it's x^10 + 8x.
And that's it.
That's all you need to do
differentiate things like this.
The chain rule is
very effective.
STUDENT: [INAUDIBLE]
PROFESSOR: That's
a good question.
So I'm not really willing
to answer too many questions
about what's going
to be on the exam.
But the question
that was just asked
is exactly the kind of question
I'm very happy to answer.
Okay, the question was,
in what form is-- what
form is an acceptable answer?
Now in real life, that is
a really serious question.
When you ask a
computer a question
and it gives you 500
million sheets of printout,
it's useless.
And you really care what
form answers are in,
and indeed, somebody
might really
care what this thing
to the 6th power is,
and then you would be forced
to discuss things in terms
of that other functional form.
For the purposes of this
exam, this is okay form.
And, in fact, any correct
form is an okay form.
I recommend strongly that you
not try to simplify things
unless we tell you to.
Sometimes it will be to your
advantage to simplify things.
Sometimes we'll say simplify.
It takes a good
deal of experience
to know when it's really worth
it to simplify expressions.
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR: Right, so
turning to this example.
The question is what
is this derivative?
And here's an answer.
That's the end of the problem.
This is a more customary form.
But this is answer is okay.
Same issue.
That's exactly the point.
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
do you have to show the work?
Do you have to show the work?
Well if I ask you
what is d/dx of sec x,
then if you wrote
down this answer
or you wrote down this
answer showing no work,
that would be acceptable.
If the question was derive
the formula for this
from the formula for the
derivative of the cosine
or something like that, then
it would not be acceptable.
You'd have to carry
out this arithmetic.
So, in other words,
typically this
will come up, for instance,
in various contexts.
You just basically have
to follow directions.
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR: The next
question is, are you
expected to be
able to prove what
the derivative of
the sine function is?
The short answer to that is yes.
But I will be getting
to that when I discuss
the rest of the material here.
We're almost there.
Okay, so let me just
finish these examples
with one last one.
And then we'll talk
about this question
of things like the derivative
of the sine function,
and deriving it.
So the last example that I'd
like to write down is the one
that I promised you
in the first lecture,
namely to differentiate
e^(x tan^(-1) x).
Basically you're supposed
to be able to differentiate
any function.
So this is the one that we
mentioned at the beginning.
So here it is.
Let's do it.
So what is it?
Well, it's just equal to
- I have to differentiate.
I have to use the
chain rule - it's
equal to the exponential
times the derivative
of this expression here.
That's the chain rule.
That's the first step.
And now I have to apply
the product rule here.
So I have e^(x tan^(-1) x).
And I differentiate the first
factor, so I get tan^(-1) x.
Add to it what happens
when I differentiate
the second factor,
leaving alone the x.
So that's x / (1+x^2).
And that's it.
That's the end of the problem.
It wasn't that hard.
Of course, it requires you
to remember all of the rules,
and a lot of formulas
underlying them.
So that's consistent with
what I just told you.
I told you that you
wanted to know this.
I told you that you needed
to know this product rule,
and that you needed to
know the chain rule.
And I guess there
was one more thing,
the derivative of e^x
came into play there.
So of these formulas,
we used four of them
in this one calculation.
Okay, so now what other things
did we talk about in Unit One?
So the main other thing
that we talked about
was the definition
of a derivative.
And also there
was sort of a goal
which was to get to the
meaning of the derivative.
So these are things - so we
had a couple of ways of looking
at it, or at least
a couple that I'm
going to emphasize right now.
But first, let me remind
you what the formula is.
The derivative is the limit
as delta x goes to 0 of (f(x +
delta x) - f(x)) / delta x.
So that's it, and
this is certainly
going to be a
central focus here.
And you want to be able
to recognize this formula
in a number of ways.
So, how do we use this?
Well one thing we did
was we calculated a bunch
of these rates of change.
In fact, more or less,
they're the ones which
are written right over here.
This list of functions here.
Now, which ones did we
start out with just straight
from the definition here?
Which of these things?
There were a whole
bunch of them.
So we started out
with a function 1/x.
We did x^n.
We did sine x.
We did cosine x.
Now there was a little
bit of subtlety with sine
x and cosine x.
We got them using
something else.
We didn't quite get
them all the way.
We got them using
the case x = 0.
We got them from the
derivative at x = 0,
we got the formulas for the
derivatives of sine and cosine.
But that was an argument
which involved plugging in sin
(x + delta x), and
running through.
So that's one example.
We also did a^x.
And that may be it.
Oh yeah, I think
that's about it.
That may be about it.
No.
It isn't.
Okay, so let me make a
connection here which you
probably haven't yet made, which
is that we did it for (u v)'.
And we also did it for (u / v)'.
So sorry, I shouldn't
write primes,
because that's not consistent
with the claim there.
I differentiated the product;
I differentiated the quotient
using the same delta x notation.
I guess I forgot that because I
wasn't there when it happened.
So look, these are the
ones that you do by this.
And, of course, you might have
to reduce them to other things.
These involve using
something else.
This one involves using the
slope of this function at 0,
just the way the sine
and the cosine did.
This one involves the slopes
of the individual functions, u
and v. And this one also
involves the individual--
So, in other words,
it doesn't get you
all the way through
to the end, but it's
expressed in terms of something
simpler in each of these cases.
And I could go on.
We didn't do these in class,
but you're certainly--
e^x is a perfectly okay
one on one of the exams.
We ask you for 1/x^2.
In other words, I'm
not claiming that it's
going to be one on this
list, but it certainly
can be any one of these.
But we're not going
to ask you to go
all the way through to the
beginning in these formulas.
There are also some fundamental
limits that I certainly
want you to know about.
And these you can
derive in reverse.
So I will describe that now.
So let me also emphasize
the following thing: I want
to read this backwards now.
This is the theme from the
very beginning of this lecture.
Namely, if you're
given the function f,
you can figure out its
derivative by this formula
here.
That is the formula for
this in terms of what's
on the right hand side.
On the other hand,
you can also use
the formula in that
direction, and if you
know the slope of something,
you can figure out
what the limit is.
For example, I'll use
the letter x here,
even though it's cheating.
Maybe I'll call it delta
x so it's clearer to you.
Maybe I'll call it u.
Suppose you look
at this limit here.
Well, I claim that you
should recognize that is
the derivative with respect to
u of the function e^u at u = 0,
which of course we know to be 1.
So this is reading this
formula in reverse.
It's recognizing that
one of these limits -
let me rewrite this again
here - one of these so-called
difference quotient
limits is a derivative.
And since we know a formula
for that derivative,
we can evaluate it.
And lastly, there's
one other type of thing
which I think you should know.
These are the ones you do
with difference quotients.
There are also other
formulas that you
want to be able to derive.
You want to be able
to derive formulas
by implicit differentiation.
In other words,
the basic idea is
to take whatever
equation you've got
and simplify it as
much as possible,
without insisting
that you solve for y.
That's not necessarily
the most appropriate way
to get the rate of change.
The much simpler
formula is sin y = x.
And that one is easier to
differentiate implicitly.
So I should say, do
this kind of thing.
So that's, if you like,
a typical derivation
that you might see.
And then there's one last type
of problem that you'll face,
and it's the other thing
that I claim we discussed.
And it goes all the way
back to the first lecture.
So the last thing that we'll be
talking about is tangent lines.
All right?
The geometric point of
view of a derivative.
And we'll be doing more
of this in next the unit.
So first of all,
you'll be expected
to be able to compute
the tangent line.
That's often fairly
straightforward.
And the second thing
is to graph y' ,
the derivative of a function.
And the third thing,
which I'm going
to throw in here,
because I regard it
in a sort of geometric
vein, although it's got
an analytical aspect to it.
So this is a picture.
This is a computation.
And if you combine
the two together,
you get something else.
And so this is to recognize
differentiable functions.
Alright, so how do you do this?
Well, we really only
have one way of doing it.
We're going to check the
left and right tangents.
They must be equal.
So again, this is
a property that you
should be familiar with
from some of your exercises.
And the idea is simply, that
if the tangent line exists,
it's the same from the
right and from the left.
Okay, now I'm going to
just do one example here
from this sort of
qualitative sketching skill
to give you an example here.
And what I'm going
to do is I'm going
to draw a graph of a
function like this.
And what I want to
do underneath is draw
the graph of the derivative.
So this is the
function y = f(x),
and here I'm going to draw
the graph of the function y =
f'(x) right underneath it.
So now, let's think about what
it's supposed to look like.
And the one step that you need
to make in order to do this,
is to draw a few tangent lines.
I'm just going to
draw one down here.
And I'm going to
draw one up here.
Now, the tangent
lines here - notice
that the slope of these
tangent lines are all positive.
So everything I
draw down here is
going to be above the x-axis.
Furthermore, as I go
further to the left,
they get steeper and steeper.
So they're getting
higher and higher.
So the function is
coming down like this.
It starts up there.
Maybe I'll draw it in green
to illustrate the graph here.
So that's this function here.
As we get farther out, it's
getting flatter and flatter.
So it's leveling off, but
above the axis like that.
So one of the
things to emphasize
is, you should not
expect the derivative
to look like the function.
It's totally different.
It's keeping track at each
point of its tangent line.
On the other hand, you should
get some kind of physical feel
for it, and we'll be practicing
this more in the next unit.
So let me give you an
example of a function which
does exactly this.
And it's the function y = ln x.
If you differentiate
it, you get y' = 1/x.
And this plot above is, roughly
speaking, the logarithm.
And this plot underneath
is the function 1/x.
We still have time
for one question.
And so, fire away.
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR: The
question is, can you
show how you derive the
inverse tangent of x.
So that's in a lecture.
I'm happy to do it right
now, but it's going
to take me a whole two minutes.
So, here's how you do
it. y = tan^(-1) x.
And now this is hopeless
to differentiate,
so I rewrite it as tan y = x.
And now I have to
differentiate that.
So when I
differentiate it, I get
the derivative of
tan y with respect
to x-- with respect to y.
That's 1 / (1 + y^2) times y'.
So this is a hard step.
That's the chain rule.
And on the left side I get 1.
So I'm doing this
super fast because we
have thirty seconds left.
But this is the hard
step right here.
And it needs for you
to know that d/dy tan
y is equal to one over-- Oh,
bad bad bad, secant squared.
I was ahead of myself so fast.
So here's the identity.
So you need have
known this in advance.
And that's the input
into this equation.
So now, what we have is
that y' = 1 / sec^2 y y,
which is the same
thing as cos^2 y.
Now, the last bit
of the problem is
to rewrite this in terms of x.
And that you have to do
with a right triangle.
If this is x and this
is 1, then the angle
is y, because the
tangent of y is x.
So this expresses the fact
that the tangent of y is x.
And then the hypotenuse is
the square root of 1 + x^2.
And so the cosine is
1 divided by that.
So this thing is 1 divided by
the square root of 1 + x^2,
the quantity squared.
So, and then the last little bit
here, since I'm racing along,
is that it's 1 / (1 + x^2),
which I incorrectly wrote over
here.
Okay, so good luck on the test.
See you tomorrow.
