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Professor: I am Haynes
Miller, I am substituting
for David Jerison today.
So you have a substitute
teacher today.
So I haven't been here
in this class with you
so I'm not completely
sure where you are.
I think you've just been
talking about differentiation
and you've got some examples
of differentiation like these
basic examples: the
derivative of x^n is nx^(n-1).
But I think maybe you've
spent some time computing
the derivative of the sine
function as well, recently.
And I think you have
some rules for extending
these calculations as well.
For instance, I think you
know that if you differentiate
a constant times a
function, what do you get?
Student: [INAUDIBLE].
Professor: The constant
comes outside like this.
Or I could write (cu)' = cu'.
That's this rule,
multiplying by a constant,
and I think you also know
about differentiating a sum.
Or I could write this
as (u + v)' = u' + v'.
So I'm going to be using
those but today I'll
talk about a collection
of other rules
about how to deal with
a product of functions,
a quotient of functions,
and, best of all,
composition of functions.
And then at the end,
I'll have something
to say about higher derivatives.
So that's the story for today.
That's the program.
So let's begin by talking
about the product rule.
So the product
rule tells you how
to differentiate a
product of functions,
and I'll just give you
the rule, first of all.
The rule is it's u'v + uv'.
It's a little bit funny.
Differentiating a
product gives you a sum.
But let's see how that works
out in a particular example.
For example, suppose
that I wanted
to differentiate the product.
Well, the product
of these two basic
examples that we
just talked about.
I'm going to use
the same variable
in both cases instead of
different ones like I did here.
So the derivative
of x^n times sin x.
So this is a new thing.
We couldn't do this without
using the product rule.
So the first function is x^n
and the second one is sin x.
And we're going to
apply this rule.
So u is x^n. u' is, according
to the rule, nx^(n-1).
And then I take v and write it
down the way it is, sine of x.
And then I do it the other way.
I take u the way
it is, that's x^n,
and multiply it by the
derivative of v, v'.
We just saw v' is cosine of x.
So that's it.
Obviously, you can
differentiate longer products,
products of more things
by doing it one at a time.
Let's see why this is true.
I want to try to show you
why the product rule holds.
So you have a standard way
of trying to understand this,
and it involves looking at
the change in the function
that you're interested
in differentiating.
So I should look at how
much the product uv changes
when x changes a little bit.
Well, so how do
compute the change?
Well, I write down the
value of the function
at some new value
of x, x + delta x.
Well, I better write
down the whole new value
of the function, and
the function is uv.
So the whole new
value looks like this.
It's u(x + delta x)
times v(x + delta x).
That's the new value.
But what's the change
in the product?
Well, I better subtract
off what the old value
was, which is u(x) v(x).
Okay, according to the
rule we're trying to prove,
I have to get u' involved.
So I want to involve the
change in u alone, by itself.
Let's just try that.
I see part of the formula for
the change in u right there.
Let's see if we can get
the rest of it in place.
So the change in x is
(u(x + delta x) - u(x).
That's the change in x
[Correction: change in u].
This part of it occurs up here,
multiplied by v(x + delta x),
so let's put that in too.
Now this equality sign
isn't very good right now.
I've got this
product here so far,
but I've introduced
something I don't like.
I've introduced u times
v(x delta x), right?
Minus that.
So the next thing I'm gonna do
is correct that little defect
by adding in u(x)
v(x + delta x).
Okay, now I cancelled off
what was wrong with this line.
But I'm still not
quite there, because I
haven't put this in yet.
So I better subtract off
uv, and then I'll be home.
But I'm going to do
that in a clever way,
because I noticed that
I already have a u here.
So I'm gonna take
this factor of u
and make it the
same as this factor.
So I get u(x) times this,
minus u(x) times that.
That's the same thing as
u times the difference.
So that was a
little bit strange,
but when you stand
back and look at it,
you can see multiplied out,
the middle terms cancel.
And you get the right answer.
Well I like that because
it's involved the change in u
and the change in v. So this
is equal to delta u times v(x +
delta x) minus u(x) times
the change in v. Well,
I'm almost there.
The next step in computing
the derivative is
take difference quotient,
divide this by delta x.
So, (delta (uv)) /
(delta x) is well,
I'll say (delta u / delta
x) times v(x + delta x).
Have I made a mistake here?
This plus magically became a
minus on the way down here,
so I better fix that.
Plus u times (delta
v) / (delta x).
This u is this u over here.
So I've just divided
this formula by delta x,
and now I can take the limit
as x goes to 0, so this
is as delta x goes to 0.
This becomes the definition
of the derivative,
and on this side, I
get du/dx times...
now what happens to this
quantity when delta x goes
to 0?
So this quantity is getting
closer and closer to x.
So what happens
to the value of v?
It becomes equal to x of v.
That uses continuity of v. So,
v(x + delta x) goes
to v(x) by continuity.
So this gives me times v,
and then I have u times,
and delta v / delta
x gives me dv/dx.
And that's the formula.
That's the formula
as I wrote it down
at the beginning over here.
The derivative of a product
is given by this sum.
Yeah?
Student: How did you
get from the first line
to the second of
the long equation?
Professor: From here to here?
Student: Yes.
Professor: So maybe it's easiest
to work backwards and verify
that what I wrote
down is correct here.
So, if you look there's a u
times v(x + delta x) there.
And there's also one here.
And they occur with
opposite signs.
So they cancel.
What's left is u(x + delta
x) v(x + delta x) - uv.
And that's just
what I started with.
Student: [INAUDIBLE]
They cancel right?
Professor: I cancelled out
this term and this term,
and what's left is the ends.
Any other questions?
Student: [INAUDIBLE].
Professor: Well, I just
calculated what delta uv is,
and now I'm gonna divide
that by delta x on my way
to computing the derivative.
And so I copied down the right
hand side and divided delta x.
I just decided to divide the
delta u by delta x and delta v
by delta x.
Good.
Anything else?
So we have the
product rule here.
The rule for differentiating
a product of two functions.
This is making us stronger.
There are many
more functions you
can find derivatives of now.
How about quotients?
Let's find out how
to differentiate
a quotient of two functions.
Well again, I'll write down what
the answer is and then we'll
try to verify it.
So there's a quotient.
Let me write this down.
There's a quotient
of two functions.
And here's the rule for it.
I always have to think about
this and hope that I get it
right. (u'v - uv') / v^2.
This may be the craziest rule
you'll see in this course,
but there it is.
And I'll try to show you why
that's true and see an example.
Yeah there was a hand?
Student: [INAUDIBLE]
Professor: What letters look
the same? u and v look the same?
I'll try to make them
look more different.
The v's have points
on the bottom.
u's have little round
things on the bottom.
What's the new value of u?
The value of u at x + delta
x is u + delta u, right?
That's what delta u is.
It's the change in u when
x gets replaced by delta x
[Correction: x + delta x].
And the change in v, the
new value v, is v + delta v.
So this is the new value of u
divided by the new value of v.
That's the beginning.
And then I subtract off
the old values, which
are u minus v. This'll
be easier to work out
when I write it out this way.
So now, we'll cross
multiply, as I said.
So I get uv + (delta u)v minus,
now I cross multiply this way,
you get uv - u(delta v).
And I divide all this
by (v + delta v)u.
Okay, now the reason
I like to do it
this way is that you see the
cancellation happening here. uv
and uv occur twice and
so I can cancel them.
And I will, and I'll answer
these questions in a minute.
Audience: [INAUDIBLE].
Professor: Ooh,
that's a v. All right.
Good, anything else?
That's what all hands were.
Good.
All right, so I cancel these
and what I'm left with then
is delta u times v
minus u times delta v
and all this is over v
+ delta v times v. Okay,
there's the difference.
There's the change
in the quotient.
The change in this function
is given by this formula.
And now to compute
the derivative,
I want to divide by delta
x, and take the limit.
So let's write that down,
delta(u/v)/delta x is this
formula here divided by delta x.
And again, I'm going to put
the delta x under these delta u
and delta v. Okay?
I'm gonna put delta
x in the denominator,
but I can think of
that as dividing
into this factor
and this factor.
So this is (delta u/ delta
x)v - u(delta v/delta x).
And all that is divided
by the same denominator,
(v + delta v)v. Right?
Put the delta x up in
the numerator there.
Next up, take the limit
as delta x goes to 0.
I get, by definition,
the derivative of (u/v).
And on the right
hand side, well, this
is the derivative du/dx right?
Times v. See and then
u times, and here it's
the derivative dv/dx.
Now what about the denominator?
So when delta x goes to 0,
v stays the same, v stays
the same.
What happens to this delta v?
It goes to 0, again,
because v is continuous.
So again, delta v
goes to 0 with delta x
because they're continuous
and you just get v times v.
I think that's the formula
I wrote down over there.
(du/dx)v - u(dv/dx).
And all divided by the square
of the old denominator.
Well, that's it.
That's the quotient rule.
Weird formula.
Let's see an application.
Let's see an example.
So the example I'm going
to give is pretty simple.
I'm going to take the
numerator to be just 1.
So I'm gonna take u = 1.
So now I'm
differentiating 1 / v,
the reciprocal of a
function; 1 over a function.
Here's a copy of my rule.
What's du/ dx in that
case? u is a constant,
so that term is 0 in this rule.
I don't have to
worry about this.
I get a minus.
And then u is 1, and dv/dx.
Well, v is whatever v is.
I'll write dv/dx as v'.
And then I get a v^2
in the denominator.
So that's the rule.
I could write it as v^(-2) v'.
Minus v' divided by v^2.
That's the derivative of 1 / v.
How about sub-example of that?
I'm going to take the special
case where u = 1 again.
And v is a power of x.
And I'm gonna use the rule
that we developed earlier about
the derivative of x^n.
So what do I get here?
d/dx (1/x^n) is, I'm plugging
into this formula here with v =
x^n.
So I get minus, uh, v^-2.
If v = x^n, v^-2 is, by the
rule of exponents, x^(-2n).
And then v' is the derivative
of x^n, which is nx^(n-1).
Okay, so let's put
these together.
There's several
powers of x here.
I can put them together.
I get -n x to the -2n + n - 1.
One of these n's cancels.
And what I'm left
with is -n - 1.
So we've computed the
derivative of 1 / x^n,
which I could also
write as x^-n, right?
So I've computed the derivative
of negative powers of x.
And this is the
formula that I get.
If you think of this -n as a
unit, as a thing to itself,
it occurs here in the exponent.
It occurs here,
and it occurs here.
So how does that
compare with the formula
that we had up here?
The derivative of
a power of x is
that power times x to
one less than that power.
That's exactly the same as the
rule that I wrote down here.
But the power here happens
to be a negative number,
and the same negative number
shows up as a coefficient
and there in the exponent.
Yeah?
Student: [INAUDIBLE].
Professor: How did I do this?
Student: [INAUDIBLE].
Professor: Where did
that x^(-2n) come from?
So I'm applying this rule.
So the denominator in
the quotient rule is v^2.
And v was x^n, so the
denominator is x^(2n).
And I decided to
write it as x^(-2n).
So the green comments there...
What they say is that I
can enlarge this rule.
This exact same rule is true
for negative values of n,
as well as positive values of n.
So there's something
new in your list
of rules that you can apply,
of values of the derivative.
That standard rule is true for
negative as well as positive
exponents.
And that comes out
of a quotient rule.
Okay, so we've done two rules.
I've talked about the product
rule and the quotient rule.
What's next?
Let's see the chain rule.
So this is a composition rule.
So the kind of thing that
I have in mind, composition
of functions is
about substitution.
So the kind of function that I
have in mind is, for instance,
y = (sin t)^10.
That's a new one.
We haven't seen how to
differentiate that before, I
think.
This kind of power of a trig
function happens very often.
You've seen them happen,
as well, I'm sure, already.
And there's a little notational
switch that people use.
They'll write sin^10(t).
But remember that when
you write sin^10(t),
what you mean is
take the sine of t,
and then take the
10th power of that.
It's the meaning of sin^10(t).
So the method of dealing
with this kind of composition
of functions is to use
new variable names.
What I mean is, I can
think of this (sin t)^10.
I can think of it it
as a two step process.
First of all, I
compute the sine of t.
And let's call the result x.
There's the new variable name.
And then, I express
y in terms of x.
So y says take this and
raise it to the tenth power.
In other words, y = x^10.
And then you plug x
= sin(t) into that,
and you get the formula for
what y is in terms of t.
So it's good practice to
introduce new letters when
they're convenient, and
this is one place where
it's very convenient.
So let's find a rule
for differentiating
a composition, a
function that can
be expressed by
doing one function
and then applying
another function.
And here's the rule.
Well, maybe I'll actually
derive this rule first,
and then you'll see what it is.
In fact, the rule is
very simple to derive.
So this is a proof first, and
then we'll write down the rule.
I'm interested in delta y /
delta t. y is a function of x.
x is a function of t.
And I'm interested in how
y changes with respect
to t, with respect to
the original variable t.
Well, because of that
intermediate variable,
I can write this as (delta y /
delta x) (delta x / delta t).
It cancels, right?
The delta x cancels.
The change in that immediate
variable cancels out.
This is just basic algebra.
But what happens when I
let delta t get small?
Well this give me dy/dt.
On the right-hand side,
I get (dy/dx) (dx/dt).
So students will often
remember this rule.
This is the rule, by saying
that you can cancel out
for the dx's.
And that's not so
far from the truth.
That's a good way
to think of it.
In other words, this is
the so-called chain rule.
And it says that differentiation
of a composition is a product.
It's just the product
of the two derivatives.
So that's how you differentiate
a composite of two functions.
And let's just do an example.
Let's do this example.
Let's see how that comes out.
So let's differentiate,
what did I say?
(sin t)^10.
Okay, there's an inside function
and an outside function.
The inside function is
x as a function of t.
This is the inside function, and
this is the outside function.
So the rule says, first
of all let's differentiate
the outside function.
Take dy/dx.
Differentiate it with
respect to that variable x.
The outside function
is the 10th power.
What's its derivative?
So I get 10x^9.
In this account, I'm using
this newly introduced variable
named x.
So the derivative of the
outside function is 10x^9.
And then here's the
inside function,
and the next thing I want
to do is differentiate it.
So what's dx/dt, d/dt (sin
t), the derivative of sine t?
All right, that's cosine t.
That's what the
chain rule gives you.
This is correct, but
since we were the ones
to introduce this
notation x here,
that wasn't given to us in
the original problem here.
The last step in
this process should
be to put back,
to substitute back
in what x is in terms of t.
So x = sin t.
So that tells me that I get
10(sin(t))^9, that's x^9,
times the cos(t).
Or the same thing
is sin^9(t)cos(t).
So there's an application
of the chain rule.
You know, people often
wonder where the name chain
rule comes from.
I was just wondering
about that myself.
So is it because
it chains you down?
Is it like a chain fence?
I decided what it is.
It's because by
using it, you burst
the chains of differentiation,
and you can differentiate
many more functions using it.
So when you want to
think of the chain rule,
just think of that chain there.
It lets you burst free.
Let me give you another
application of the chain rule.
Ready for this one?
So I'd like to
differentiate the sin(10t).
Again, this is the
composite of two functions.
What's the inside function?
Okay, so I think I'll introduce
this new notation. x = 10t,
and the outside
function is the sine.
So y = sin x.
So now the chain
rule says dy/dt is...
Okay, let's see.
I take the derivative
of the outside function,
and what's that?
Sine prime and we can
substitute because we
know what sine prime is.
So I get cosine of whatever,
x, and then times what?
Now I differentiate the inside
function, which is just 10.
So I could write this
as 10cos of what?
10t, x is 10t.
Now, once you get used to
this, this middle variable,
you don't have to
give a name for it.
You can just to think
about it in your mind
without actually writing
it down, d/dt (sin(10t)).
I'll just do it again
without introducing
this middle variable explicitly.
Think about it.
I first of all differentiate
the outside function,
and I get cosine.
But I don't change the thing
that I'm plugging into it.
It's still x that I'm
plugging into it. x is 10t.
So let's just write 10t and
not worry about the name
of that extra variable.
If it confuses you,
introduce the new variable.
And do it carefully
and slowly like this.
But, quite quickly,
I think you'll
get to be able to keep
that step in your mind.
I'm not quite done yet.
I haven't differentiated
the inside function,
the derivative of 10t = 10.
So you get, again,
the same result.
A little shortcut that
you'll get used to.
Really and truly, once
you have the chain rule,
the world is yours to conquer.
It puts you in a very,
very powerful position.
Okay, well let's see.
What have I covered today?
I've talked about product rule,
quotient rule, composition.
I should tell you something
about higher derivatives,
as well.
So let's do that.
This is a simple story.
Higher is kind of
a strange word.
It just means differentiate
over and over again.
All right, so let's see.
If we have a function
u or u(x), please
allow me to just write
it as briefly as u.
Well, this is a sort
of notational thing.
I can differentiate
it and get u'.
That's a new function.
Like if you started
with the sine, that's
gonna be the cosine.
A new function, so I can
differentiate it again.
And the notation for the
differentiating of it again,
is u prime prime.
So u'' is just u'
differentiated again.
For example, if u is the sine
of x, so u' is the cosine of x.
Has Professor Jerison
talked about what
the derivative of cosine is?
What is it?
Ha, okay so u'' is -sin x.
Let me go on.
What do you suppose u''' means?
I guess it's the
derivative of u''.
It's called the
third derivative.
And u'' is called the
second derivative.
And it's u''
differentiated again.
So to compute u''' in this
example, what do I do?
I differentiate that again.
There's a constant term,
-1, constant factor.
That comes out.
The derivative of sine is what?
Okay, so u''' = -cos x.
Let's do it again.
Now after a while, you get
tired of writing these things.
And so maybe I'll use
the notation u^(4).
That's the fourth derivative.
That's u''''.
Or it's (u''') differentiated
again, the fourth derivative.
And what is that
in this example?
Okay, the cosine has
derivative minus the sine,
like you told me.
And that minus sign
cancels with that sign,
and all together, I get sin x.
That's pretty bizarre.
When I differentiate the
function sine of x four times,
I get back to the
sine of x again.
That's the way it is.
Now this notation, prime
prime prime prime, and things
like that.
There are different
variants of that notation.
For example, that's
another notation.
Well, you've used the
notation du/dx before. u'
could also be denoted du/dx.
I think we've
already here, today,
used this way of
rewriting du/dx.
I think when I was talking about
d/dt(uv) and so on, I pulled
that d/dt outside and
put whatever function
you're differentiating
over to the right.
So that's just a
notational switch.
It looks good.
It looks like good
algebra doesn't it?
But what it's doing is regarding
this notation as an operator.
It's something you apply to a
function to get a new function.
I apply it to the sine function,
and I get the cosine function.
I apply it to x^2, and I get 2x.
This thing here, that symbol,
represents an operator,
which you apply to a function.
And the operator says, take the
function and differentiate it.
So further notation
that people often use,
is they give a different
name to that operator.
And they'll write
capital D for it.
So this is just using capital
D for the symbol d/dx.
So in terms of that
notation, let's see.
Let's write down what higher
derivatives look like.
So let's see.
That's what u' is.
How about u''?
Let's write that in terms
of the d/dx notation.
Well I'm supposed to
differentiate u' right?
So that's d/dx applied
to the function du/dx.
Differentiate the derivative.
That's what I've done.
Or I could write that as d/dx
applied to d/dx applied to u.
Just pulling that u outside.
So I'm doing d/dx twice.
I'm doing that operator twice.
I could write that as
(d/dx)^2 applied to u.
Differentiate twice, and
do it to the function u.
Or, I can write it as,
now this is a strange one.
I could also write
as-- like that.
It's getting stranger
and stranger, isn't it?
This is definitely just a
kind of abuse of notation.
But people will go even
further and write d squared
u divided by dx squared.
So this is the strangest one.
This identity quality
is the strangest one,
because you may think that
you're taking d of the quantity
x^2.
But that's not what's intended.
This is not d(x^2).
What's intended is the
quantity dx squared.
In this notation,
which is very common,
what's intended
by the denominator
is the quantity dx squared.
It's part of this second
differentiation operator.
So I've written a bunch
of equalities down here,
and the only content
to them is that these
are all different notations
for the same thing.
You'll see this
notation very commonly.
So for instance the
third derivative
is d cubed u divided
by dx cubed, and so on.
Sorry?
Student: [INAUDIBLE].
Professor: Yes, absolutely.
Or an equally good notation is
to write the operator capital
D, done three times, to u.
Absolutely.
So I guess I should also write
over here when I was talking
about d^2, the
second derivative,
another notation is do the
operator capital D twice.
Let's see an example of
how this can be applied.
I'll answer this question.
Student: [INAUDIBLE].
Professor: Yeah,
so the question is
whether the fourth
derivative always gives you
the original function back,
like what happened here.
No.
That's very, very special
to sines and cosines.
All right?
And, in fact, let's
see an example of that.
I'll do a calculation.
Let's calculate the
nth derivative of x^n.
Okay, n is a number,
like 1, 2, 3, 4.
Here we go.
Let's do this.
So, let's do this bit by bit.
What's the first
derivative of x^n?
So everybody knows this.
I'm just using a new notation,
this capital D notation.
So it's nx^(n-1).
Now know, by the way, n could
be a negative number for that,
but for now, for
this application,
I wanna take n to be
1, 2, 3, and so on;
one of those numbers.
Okay, we did one derivative.
Let's compute the second
derivative of x^n.
Well there's this n
constant that comes out,
and then the exponent comes
down, and it gets reduced by 1.
All right?
Should I do one more?
D^3 (x^n) is n(n-1).
That's the constant from here.
Times that exponent,
n - 2, times 1 less, n
- 3 is the new exponent.
Well, I keep on going until
I come to a new blackboard.
Now, I think I'm
going to stop when
I get to the n minus
first derivative,
so we can see what's
likely to happen.
So when I took the
third derivative,
I had the n minus
third power of x.
And when I took the
second derivative,
I had the second power of x.
So, I think what'll
happen when I
have the n minus
first derivative
is I'll have the first
power of x left over.
The powers of x
keep coming down.
And what I've done it n - 1
times, I get the first power.
And then I get a big constant
out in front here times more
and more and more of these
smaller and smaller integers
that come down.
What's the last integer that
came down before I got x^1
here?
Well, let's see.
It's just 2, because this x^1
occurred as the derivative
of x^2.
And the coefficient
in front of that is 2.
So that's what you get.
The numbers n, n-1, and so
on down to 2, times x^1.
And now we can differentiate
one more time and calculate what
D^n x^n is.
So I get the same number, n
times n-1 and so on and so on,
times 2.
And then I guess
I'll say times 1.
Times, what's the derivative
of x^1= 1, so times 1.
Time 1, times 1.
Where this one means
the constant function 1.
Does anyone know what
this number is called?
That has a name.
It's called n factorial.
And it's written n
exclamation point.
And we just used an example
of mathematical induction.
So the end result is
D^n x^n is n!, constant.
Okay that's a neat fact.
Final question for the lecture
is what's D^(n + 1) applied
to x^n?
Ha.
Excellent.
It's the derivative
of a constant.
So it's 0.
Okay.
Thank you.
