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 view additional materials
from hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
HERBERT GROSS: Hi.
Today we do a somewhat computational bit.
And actually, the lecture for today
is not nearly as difficult, once you
get through the maze of symbolism,
as it is to apply the material.
In other words, we're going to devote the next two
units of our course to this particular topic, which
is known as the chain rule.
But we'll give one lecture to cover both units.
And again, the idea is that it's not so much
that the concept becomes more difficult as much as it
is that you must develop a certain amount
of dexterity keeping track of the various partial derivatives
and the like.
At any rate, maybe I think the best way
is to just barge into a hypothetical situation
and see what the situation really is.
The idea is essentially the following.
We're given some function, say, w.
w is a function of, say, the three independent variables
x, y, and z.
Now, for some reason or other, which
we won't worry about right now, it
turns out that x, y, and z are, in turn,
conveniently expressible in terms of the two variables
r and s.
In fact, if you want a physical interpretation of this,
you can think of if x, y, and z are functions
of the two independent variables r and s,
that means that we have two degrees of freedom.
So we may think of this as parametrically representing
the equation of a surface.
And what we're talking about here
is w being a function of something in space and asking,
what does w look like when you restrict your space
to a particular surface?
I mean, that's just a geometrical interpretation
that one could talk about.
But the idea is the following.
After all, if w depends on x, y, and z, and x, y,
and z each depend on r and s, in particular then,
it's clear that w itself is some function of r and s,
where, again, I use the usual notation
of using a g here rather than the f
up here to indicate that the relationship between r and s
which specifies w, may very well be
a different algebraic relationship than that which
relates x, y, and z to give w.
But the point that we have in mind is the following.
Given that w is a function of x, y,
and z, given that x, y, and z are functions of r and s,
hence w is a function of r and s.
The question that we ask in calculus of several variables
is, first of all, if we can be sure that these were all
continuously differentiable functions,
can we be sure that w will be a continuously differentiable
function of r and s?
That's the first question.
And the second question is, OK, assuming
that the answer to the first question is in the affirmative,
that w is a continuously differentiable function of r
and s, how could we compute, for example, the partial of w
with respect to r, knowing all of the so-called obvious
other partial derivatives?
What do I mean by that?
Well, what I mean is if you were to look just at this equation,
just looking at this equation, what
are the obvious partial derivatives to take?
You say, well, we'll take the partial of w
with respect to x, the partial of w with respect to y,
and the partial of w with respect to z.
And if you were to look, say, at this equation,
the natural thing to ask is, what is the partial
of x with respect to r?
What is the partial of x with respect to s?
Et cetera.
In other words, what we're saying is,
in this particular problem, we would like to figure out how,
for example, to compute the partial of w with respect
to r, knowing that we have at our disposal
the partials of w with respect to x, y, and z;
the partials of x, y, and z with respect to r; et cetera;
meaning we also have the partials of x, y,
and z with respect to theta.
Before I go any further, notice, by the way,
that if I left out that phrase that we were talking
about in our last lecture, "continuously differentiable,"
notice that all of this would make sense,
provided that the derivatives existed.
There was no place here do I make any statement
that the partials have to not only exist but be continuous.
I never say that at all.
This is what the problem is.
I would like to use the chain rule.
Do you see why it's called the chain rule?
w is a function of x, y, and z.
x, y, and z are each functions of r and s.
Now, what I claim is that not only is it possible
to do this but the recipe for doing
this is a very, very suggestive thing, one which is very, very
easy to remember, once you see how it's put together.
If you don't see how it's put together,
the thing is just a mess-- namely,
the claim is that the partial of w with respect to r
is the partial-- I'll just read it to you-- the partial of w
with respect to x times the partial of x with respect to r,
plus the partial of w with respect
to y times the partial of y with respect to r,
plus the partial of w with respect
to z times the partial of z with respect to r.
And as I say, if you try to memorize that,
it's a very, very nasty business.
But let's look at this in three separate pieces.
In a way, can you sense that this
is nothing more than the change in w with respect
to r due to the change in x alone?
In other words, you're taking here what?
The change in w due to x and multiplying that
by the change in x with respect to r.
So this is the contribution of the change in w with respect
to r due to x alone.
On the other hand, this is the partial of w with respect
to r due to the change in y alone.
And this is the partial of w with respect
to r due to the change in z alone.
And since x, y, and z are independent,
the change in x, the change in y, and the change in z
are also independent variables.
Consequently, it seems reasonable to assume
that to find the total change of w with respect to r,
we just add up all of the partial contributions.
Namely, we take the partial of w with respect
to r due to x alone, add on to that the partial of w
with respect to r due to y alone,
add on to that the partial of w with respect
to r due to z alone, and that that sum should
be the total change in w with respect r,
treating s as a constant.
And by the way, let me point out a pitfall with this notation.
We're so used to using fractional notation here.
Have you noticed that if you're not careful here,
you're almost tempted to cancel--
I don't want to write this, because you'll think that it's
the right way of doing it.
But see if we say, let's cancel the partials with respect
to x here, let's cancel the partials with respect
to y here, and let's cancel the partials with respect
to z here.
By the way, if you did that, notice what you would get
is the contradiction that the partial of w with respect to r
is equal to the partial of w with respect
to r, plus the partial of w with respect
to r, plus the partial of w with respect to r.
In other words, it seems that you
would get that the partial of w with respect to r
is always three times itself, which
is, I hope, a glaring enough contradiction so I don't
have to go into any more detail about the contradiction part.
Notice again, though, why I have made such a fetish
over labeling the variables.
Notice that when you're taking the partial of w
with respect to x, you're assuming that y and z
are the variables that are being held constant.
And when you're taking the partial of x with respect to r,
it's s that you're assuming is being held constant.
And as soon as you look at these subscripts here,
somehow or other that should put you on your guard
to be careful about crossing out because, after all,
the changes are being made with respect
to different sets of variables.
At any rate, this is the statement.
And my other claim is that the proof follows immediately
from the main, key theorem that we stressed last time, even
though we didn't prove it.
But we've had ample exercises using this.
Namely, notice that we have already
seen that if w does happen to be a continuously differentiable
function of x, y, and z, then delta w is the partial of w
with respect to x times delta x, plus the partial of w
with respect to y times delta y, plus the partial of w
with respect to z times delta z, plus an error term.
And what is that error term?
It's k_1 delta x, plus k_2 delta y, plus k_3 delta
z, where k_1, k_2, and k_3 all approach
0 as delta x, delta y, and delta z approach 0.
Now again, the key step in all of this
is that this amount here I could always call delta w tan,
or as Professor Thomas calls it for more than two variables,
delta w sub lin, l-i-n, meaning that this is a linear equation.
Remember-- I've made an abbreviation here--
these partials are assumed to be evaluated at a particular point
that we're interested in.
But the idea is, granted that I can always call this delta w
tan, to say that the error has this small a magnitude depends
on the fact that w is a continuously differentiable
function of x, y, and z.
That's why the theory is so important.
What happens in real life is that most examples
you encounter in real-life engineering, the functions
that you're dealing with are continuously differentiable.
So it seems like we're making a big issue over nothing.
I should point out that on the frontiers of knowledge,
enough situations occur where the functions that we're
dealing with are not continuously differentiable,
that some horrible mistakes can be made by assuming that you
can replace delta w by this, without any significant error.
But as long as this is the case, we can do this.
And now notice, what does the partial of w with respect to r
mean?
It means you take delta w divided by delta r.
And let me just do that here.
I'll just divide every term by delta r.
And now, what do I have to do next to get the partial?
I have to take the limit as delta r approaches 0.
Now the interesting point is as delta approaches 0,
holding s fixed, this term obviously becomes
the partial of x with respect to r, by definition.
This term becomes the partial of y with respect to r,
by definition.
And this term becomes a partial of z with respect to r,
by definition.
By the way, notice that even though delta x, delta y,
and delta z are all going to 0 as delta r goes to 0,
you can not immediately conclude that these terms drop out.
Because after all, delta r is also approaching 0.
So delta x over delta r is that 0 over 0 form.
In fact, that's precisely the partial
of x with respect to r term that we're talking about.
The beauty is what?
That as delta x, delta y, and delta z
approach 0, each of the k's approach 0.
You see, the reason that the error term becomes negligible,
becomes 0 in the limit, isn't because delta x, delta y,
and delta z are becoming small.
Because these small numbers are being divided
by another small number.
It's because the k_1, k_2, and k_3 are getting small.
At any rate, putting this all together,
notice that now, in a manner completely
analogous to our part-one treatment of the chain rule,
except that we're now dealing with several variables,
these three terms drop out.
And these three terms become the claim that we made before.
In other words, this is how the partial of w with respect to r
is computed.
And again, the theory is the easiest part of this.
That's the easy part.
The hard part is getting familiarity
with how to work with this.
And I think the best way to get some familiarity
for working with this is to pick particularly simple problems
for the lecture, problems where it's
so easy to do the problem both ways
that no hangup can possibly occur.
Let's take a very simple example.
Let's suppose that w equals x squared
plus y squared plus z squared.
Suppose we also know that x is r plus s, y is r minus s,
and z happens to be 2r.
In this particular case, notice that we
would find the partial of w with respect to r very conveniently
by direct substitution.
Namely, we simply replace x by r plus s,
we replace y by r minus s, we replace z by 2r.
And then w simply becomes this expression here,
which when we collect terms, becomes
6 r squared plus 2 s squared.
And again, the arithmetic there is simple enough
so I'm not even going to bother worrying about how
we justify these steps.
At which stage, to take the partial of w with respect to r,
holding s constant, this is simply what?
12r.
Because s is being treated as a constant,
its derivative with respect to r is 0.
You see, in an example like this,
one would not really be tempted to use the chain rule.
The chain rule is used in many cases
not just for convenience, but in cases of great theory
where you're only given that w is some function of x, y,
and z, and you're not told explicitly
what the function is.
You're just given f of x, y, z.
In the case where the function is given explicitly,
it's sometimes very easy to substitute directly.
At any rate, what the chain rule says is roughly this.
They say, lookit.
From this equation, you could immediately
say the partial of w with respect to x is 2x,
the partial of w with respect to y is 2y,
the partial of w with respect to z is 2z.
From this equation, you could immediately
say that the partial of x with respect to r is 1,
the partial of x with respect to s is 1,
the partial of y with respect to r is 1,
the partial of y with respect to s is minus 1,
the partial of z with respect to r is 2,
and the partial of z with respect to s is 0.
In particular, summarizing our results, we have these here.
Now, what the chain rule says is what?
To find the partial of w with respect to r,
you just take the partial of w with respect
to x times the partial of x with respect to r,
plus the partial of w with respect
to y times the partial of y with respect to r,
plus the partial of w with respect
to z times the partial of z with respect to r.
And if we do that in this case, we simply get what?
2x plus 2y plus 4z.
Now again, I picked, deliberately,
a very simple problem here.
Remember, by definition, x is r plus s, y is r minus s,
and z happens to be 2r.
And now you can see very quickly here that when I substitute in,
I get what?
2r plus 2r is 4r, plus 8r is 12r, and 2s minus 2s is 0.
The partial of w with respect to r
is also 12r, also meaning what?
We found that same answer before.
At least that's how the chain rule works.
And again, we have to remember that the chain rule does not
depend on the number of variables,
even though this may start to look a little bit sticky.
Let's word it as follows.
Suppose w happens to be a continuously differentiable
function of the n independent variables x_1 up to x_n.
See, that's what this parenthetical remark means.
I'm saying that not only do the partials of f with respect
to x_1 up to x_n exist at a given point,
but they are continuous there.
And why do I want that in there?
So I can say that my error term is never
any greater than that k_1 delta x_1, plus k_2 delta x_2,
plus, et cetera, k_n delta x_n, where the k's go
to 0 as the delta x's go to 0.
I'm going to spare you the details of proofs.
But I just want you to keep seeing
why these things are necessary.
At any rate, let's suppose now that each of the n variables
x_1 up to x_n turn out to be functions of the m variables.
n and m could conceivably be equal.
But m could even be more than n.
It can be less than n.
There's no reason why they have to be equal.
All we're saying is, speaking in the most general terms,
suppose each of the n x's is a continuously differentiable
function of the m independent variables y sub 1 up
to y sub m.
In fact, that's what this "et cetera" means here.
The "et cetera" refers to the parenthetical remark here.
I mean that not only are the x's functions of y_1 up to y_m,
but they're continuously differentiable functions.
Now obviously, what we're saying is that if w can be expressed
in terms of the x's, the x's can be expressed in terms
of the y's, obviously then, w can be expressed in terms
of the y's.
In other words, w is some function of y_1 up to y_m.
Now the question that comes up is that just looking at this,
I can talk about the partial of w with respect to y_1,
the partial of w with respect to y_2,
the partial of w with respect to y_3, et cetera, all the way up
to the partial of w with respect to y sub m.
And the question is, lookit.
From the original form of w, it was easy to talk about
the partials of w with respect to the x's.
From how the x's are given in terms of the y's, it's easy
to talk about the derivatives of the x's with respect
to the y's.
And so the question is, how do you
find the partial of w with respect to, say, y sub
1, given all of these other partial derivatives?
And the answer, again, is something that you just
have to get used to.
The proof goes through for n and m the same way
as it did for the lower-dimensional case.
And the intuitive interpretation is the same.
Namely, to find the partial of w with respect to y_1,
we simply see how much w changed with respect
to y_1 due to the change in x_1 alone,
add on to that the change in w with respect
to y_1 due to the change in x_2 alone, et cetera,
add on to that, finally, the change in w with respect to y
sub 1 due to the change in x sub n alone.
In other words, again, if you think
of this in terms of cancellation,
if you cross these things out, don't think of adding them,
but think of them as what?
Giving you the individual components
that tell you how the partial of w with respect to y_1
is made up.
By the way, there is one parenthetical remark
that I haven't written on the board
that I would like to make at this time.
In Professor Thomas's text, he has
elected to introduce matrix algebra prior
to this particular chapter.
It again turns out that one does not
need matrices to talk about the chain rule
but that if one had matrix notation,
the matrix notation is particularly
convenient for summarizing the chain rule.
I have elected to hold off on matrix algebra
till the near future because it comes up
in a much better motivated way, I
think, in terms of these linear approximations.
But the point is if, as you're reading the text,
you see the matrix notation, and you are not
familiar with the matrices, forget it.
All the matrix is, is a shortcut notation for saying this.
And if I want a shortcut notation here,
I don't need matrices for saying this.
I can say this in terms of our sigma notation.
Notice that one other way of writing
this thing very compactly that may be more suggestive
is the following.
Notice that I'm adding up n terms here.
Each term consists of two factors, each of which
looks like a fraction.
The numerator of the first fraction
is always a partial of w.
The denominator of the second fraction
is always the partial y_1.
And it appears that the denominator of the first
and the numerator of the second always have the same subscript,
but they seem to vary consecutively from 1 to n.
And that's precisely where the sigma notation comes in handy.
Why don't we just write, therefore,
that this is the sum, partial of w with respect
to x sub k, plus the partial of x sub k
with respect to y_1, as the subscript
k ranges through all integral values from 1 to n?
In other words, notice that in this particular form,
we have simply rewritten this thing compactly.
But if you look at this and look at this,
I think it's very suggestive to see how the chain rule works.
You see, here's your partial of w with respect to y_1.
And these are what?
The contributions due to each of the changes of the n variables.
This is the change due to the x sub k variable.
And you add these all up because they're independent variables,
as k goes from 1 to n.
And I write "et cetera" here simply to point out
that I could have computed the partial of w
with respect to y_2 instead of y sub 1.
By the way, the recipe would've looked exactly the same,
except that if there was a 2 here, there
would have been a 2 here.
If there were a 3 here, they would've been a 3 here.
If there were an m here, there would have been an m here.
OK, now lookit.
At this particular stage of our lecture
today, this could end with the idea
that for the unit that's now assigned,
this is as far as you have to go.
In other words, for the exercises
that I've given you in this particular unit,
we do nothing higher than using the chain rule
for first-order derivatives.
The point is that in many applications in real life,
we must take higher-order derivatives.
In other words, there are many differential equations,
partial differential equations, where we must work
with higher-order derivatives.
And for that reason, it becomes very important, sometimes,
to be able to take a second derivative
or a third derivative or a fourth derivative
by means of the chain rule.
Now the interesting point is that the theory that we've used
so far doesn't change at all.
What does happen is that the average student,
in learning this material for the first time,
gets swamped by the notation.
Consequently, what I want to do is
to give you the lecture on this material at the same time
that I'm lecturing on first-order derivatives, simply
because the continuity follows smoother this way,
so that you see what the whole overall picture is, then
to make sure that you cement these things down.
The next unit after this will give
you drill on taking higher-order derivatives.
What this may mean is that many of you
may prefer to watch this half of the film
a second time, after you've already
tried working some of the problems
with higher-order derivatives, if you're still
confused by this.
But at any rate, let's take a look
at a hypothetical situation.
Since we're so used to polar coordinates,
let's talk in terms of polar coordinates.
Suppose w happens to be a continuously differentiable
function of x and y.
x and y, in turn, are continuously differentiable
functions of the polar coordinates r and theta.
In fact, they're x equals r cosine theta,
y equals r sine theta.
Now lookit.
If all I want to do is find the partial of w with respect to r,
I can do that by the ordinary chain rule.
Namely, it's the partial of w with respect
to x times the partial of x with respect to r,
plus the partial of w with respect
to y times the partial of y with respect to r.
Now, knowing what x looks like explicitly
in terms of r and theta and what y looks like explicitly
in terms of r and theta, I can certainly
compute the partials of x and y with respect
to r, holding theta constant.
In particular, the partial of x with respect to r
is simply cosine theta.
And the partial of y with respect to r
is simply sine theta.
So the partial of w with respect to r
is partial of w with respect to x times
cosine theta, plus partial of w with respect to y times sine
theta.
And by the way, notice that I cannot simplify these terms.
I cannot simplify these terms in general,
because all I'm given is that w is some function of x and y.
I don't know what w looks like explicitly in terms of x and y.
So all I can do is talk about the partials of w with respect
to x, partial of w respect to y, without worrying
any more about this, with the understanding that if I knew
what w looked like explicitly in terms of x and y,
I could work out what this thing was.
Now, here's the key point.
That's why I've accentuated it.
In the same way that w is a function
of both x and y, so also are the partials of w with respect
to x and the partials of w with respect to y.
In other words, even though this looks
like this emphasizes the x, notice
that when you take the derivative
of a function of both x and y with respect to x, in general,
the resulting function will again
be a function of both x and y.
And so what we're saying is that if the partials of w
with respect to x and the partials of w with respect to y
also happen to be continuously differentiable functions of x
and y, we could, if we wished, use the chain rule again.
In other words, suppose in the particular problem
that I was dealing with, it wasn't
enough to know the partial of w with respect to r.
Suppose, for example, I wanted the second partial of w
with respect to r.
Well obviously, that simply means what?
Take the partial of this with respect to r.
In other words, the second partial of w with respect to r
is just the partial of the partial of w with respect to r,
with respect r.
I'm just going to differentiate this thing with respect to r.
In other words, writing this out more succinctly for you,
the second partial derivative of w with respect to r
is the partial with respect to r of cosine theta
partial of w with respect to x, plus sine theta partial
of w with respect to y.
Now, here's the key point.
When we differentiate here, we're
assuming that theta is being held constant.
Isn't that right?
So consequently, when I'm differentiating with respect
to r, cosine theta is a constant.
I can skip over that, see, and differentiate
what's left with respect to r.
In other words, that's what?
It's the partial of w with respect to x differentiated
with respect to r.
See, I'm using the ordinary rule for the derivative of a sum.
Now, the derivative of sine theta--
see, sine theta is a constant with respect to r.
So the derivative of this term is just
sine theta times the derivative of the partial
of w with respect to y, with respect to r, written this way.
Now, the key point is that both of these functions
here, both of these are functions of x and y.
x and y, in turn, are functions of r and theta.
So in other words, to differentiate
this thing with respect to r, I must use the chain rule again.
Now, because this may seem difficult for you,
all I'm really saying is, lookit.
If this term here looks messy, since we
know that the partial of w with respect to x
is some function of x and y, let's call that h of x, y.
Then all we're saying is that the partial of the partial of w
with respect to x, with respect to r,
is just the partial of h with respect to r.
But to find the partial of h with respect to r,
we know how to use the chain rule there.
It's just what?
It's the partial of h with respect
to x times the partial of x with respect to r,
plus the partial of h with respect
to y times the partial of y with respect to r.
Of course, if we now remember what h is-- see,
h is the partial of w with respect to x.
So if I differentiate again with respect to x,
I get the second partial of w with respect to x.
We've already seen that the partial of x with respect to r
is cosine theta, so I have this term.
The partial of h with respect to y really says what?
Differentiate the partial of w with respect
to x, with respect to y.
And the usual way of abbreviating that
is like this, which, again, is explained in the reading
material.
And we now multiply that by the partial of y with respect
to r, which happens to be sine theta.
Now, look at this a few times in your spare time,
if it's bothering you.
It is not really that difficult. It
is messy notation in the sense that you're not
used to notation that's quite that messy.
That's why it's messy.
Once you get used to it, it is not
any tougher than the chain rule for one independent variable.
In fact, to take the partial of the partial of w
with respect to y, with respect to r, I'll do that in one step
without using a substitution.
All I'm saying is that this function
depends on both x and y.
So to see what its derivative is with respect to r,
I'll see what the contribution of its derivative with respect
to r is due to just x alone.
Then I'll see what contribution of its derivative
with respect to r is due to just y alone.
And by the way, when I say it that way,
notice how quick it is to write this thing down.
I differentiate this with respect
to x multiplied by the partial of x with respect to r.
Add on to that the partial of this with respect to y.
Multiply that by the partial of y with respect to r.
If I do this, notice now I have what?
I have the partial with respect to y.
And I differentiate that with respect to x.
That's written this way.
And by the way, notice that this is the reverse order
of what we did over here.
Namely, in one case, we first differentiated with respect
to x and then with respect to y.
In the other case, we differentiated first
with respect to y and then with respect to x.
So that actually, conceptually there is a difference.
That's why we write these things differently.
It does, again, turn out that in most cases,
the answer that you get-- thank goodness--
doesn't depend on the order in which you
perform the derivatives.
But this is not at all self-evident,
even though you'd like to believe that it is.
But we'll talk about that more in the exercises and the like.
But all I'm saying now is that if we put everything together
of what we've had before, we can obtain,
in this particular case, that the second partial of w
with respect to r is this somewhat messy but nonetheless
straightforward expression.
And see, I've circled these things
to sort of tell you that if it is permissible to interchange
the order of differentiation, we could combine these two terms.
On the other hand, if you couldn't interchange the order,
this would be a rather dangerous thing
to do over here because these might be different answers.
As I say again, if you have enough continuity,
it turns out that these two factors are the same.
But that's not the important issue here.
The important issue here is that I
can keep using the chain rule to take higher-order derivatives.
And even though the notation is messier,
this happened when we dealt with functions of a single variable.
Remember when we used the chain rule
to find dy/dx when y and x were given, say, as functions of t?
We could also use the chain rule to find the second derivative
of y with respect to x.
But we had to be a little bit more careful of the computation
because certain factors crept in that we had to keep track of.
At any rate, again, to illustrate this idea
rather than to keep droning on about it,
let me take a particularly simple computational problem
to check this thing on.
In other words, what I'm going to do
is take this messy formula over here and apply it to a case
where the arithmetic happens to be very, very simple.
I'm going to rig this very, very nicely.
I'm going to let f of x, y just be x squared plus y squared,
in this case.
Let w be x squared plus y squared.
In polar coordinates, notice that x squared plus y squared
is just r squared.
So w is just r squared.
What is the partial of w with respect to r, then?
The partial of w with respect to r is 2r.
And if I now differentiate that with respect to r,
the second partial of w with respect to r is 2.
Obviously, one would not use the chain rule in real life
to find the answer to this particular problem.
We've chosen this problem simply to emphasize how
the chain rule would work here.
At any rate, going back here, notice
that it's very simple to see from this equation
that the partial of w with respect to x is 2x.
Therefore, the second partial of w with respect to x is 2.
The partial of w with respect to y is 2y.
Therefore, the second partial of w with respect to y is also 2.
The partial of w with respect to x is a function of x alone,
in this case.
Consequently, the derivative with respect to y will be 0.
Similarly, the partial of w with respect to y is a function of y
alone.
Consequently, when I differentiate
that with respect to x, meaning I'm holding y constant,
that derivative will also be 0.
And the interesting point now is if I take these values
and substitute those into this equation, what happens?
Look.
The second partial of w with respect to x is just 2.
The second partial of w with respect to y is just 2.
The mixed partials are both 0, regardless of which order
you did them in.
That's what we saw over here.
So consequently, according to this recipe,
the second partial of w with respect to r
is 2 cosine squared theta plus 0 plus 2 sine
squared theta plus 0, where the reason I've
written these 0's in is simply so
that when you're looking at your notes later,
that traces the analog of these terms over here.
At any rate, notice now that if I add these up,
2 cosine squared theta plus 2 sine
squared theta, since sine squared
theta plus cosine squared theta is 1,
this sum is just-- I'll write that in white chalk
just so we don't accentuate it.
Let it just be part of the answer.
This is 2 plus 0, which is 2.
And this certainly does check with the result
that we got the so-called easier way.
And again, I don't want to leave you with the idea
that the second way was just the hard way of doing
the same problem that we did easily the first way.
I picked a simple example so you can see how this works.
I'm going to have a multitude of exercises
for you to do in the next unit, simply so that you'll pick up
the kind of know-how that will allow you to change variables
using the chain rule, with a minimum degree of difficulty.
In fact, hopefully, I would like to feel
by the time we're through with the next two units,
you will be doing this almost as second nature.
Well, we have other topics to consider
in terms of our linear approximations and the like.
We'll talk about that more as the course unfolds.
For the time being, I would like you to concentrate simply
on mastering the chain rule.
And so until we meet next time, good bye.
Funding for the publication of this video
was provided by the Gabriella and Paul Rosenbaum Foundation.
Help OCW continue to provide free and open access to MIT
courses by making a donation at ocw.mit.edu/donate.