DAVID JORDAN: Hello, and
welcome back to recitation.
Today the problem I'd
like to work with you
is about computing
partial derivatives
and the total differential.
So we have a function z which
is x squared plus y squared.
So it depends on the
two variables x and y.
Now the variables
x and y themselves
depend on two auxiliary
variables, u and v.
So that's the
setup that we have.
So in part a, we just want to
compute the total differential
dz in terms of dx and dy.
So u and v aren't going
to enter into the picture.
And then in part b,
we're going to compute
the partial derivative
partial z partial u
in two different ways.
First, we're going to compute
it using the chain rule.
And then we're going to compute
it using total differentials.
And so we'll substitute
in some of the work
that we had in a
to solve that part.
So why don't you pause the video
now and work on the problem.
We'll check back and
we'll do it together.
Hi, and welcome back.
Let's get started.
So first, computing
a is not so bad.
So we just need
to first remember,
what does it mean to compute
the total differential?
So the total
differential dz is just
the partial derivative of z
in the x-direction dx plus
z in the y-direction dy.
OK?
So now, looking at our
formula here for z,
we have-- so the
partial derivative of z
in the x-direction is
2x, so this is 2x dx.
And the partial derivative
of z in the y is 2y,
so we have 2y dy.
OK, and that's all
we have to do for a.
Now for b, we want to compute
this partial derivative
in two different ways.
First, using the chain rule.
So let's remember what
the chain rule says.
So whenever I think
about the chain rule,
I like to draw this
dependency graph.
OK?
And this is just a
way for me to organize
how the different variables
depend on one another.
So at the top, we have z.
And z is a function of
x and y, but x is itself
a function of both u
and v, and y is also
a function of u and v.
So z depends on x and y,
and x and y each jointly
depend on u and v.
So it's a little
bit complicated,
the relationships here.
So now, what the
chain rule says is
that if we take a partial
derivative-- partial z
partial u-- we have to go
through our dependency graph.
Every way that we
can get from z to u,
we get a term in our summation
for each one of those.
So for instance, z
goes to x goes to u.
So that means that we
have partial z partial x,
partial x partial u.
And then we can also go
z goes to y goes to u.
And that will give us partial z
partial y, partial y partial u.
And now these partials
are ones that we can just
compute from our formulas.
So for instance,
partial z partial x
is 2x, which we computed.
Now partial x partial u, we have
to remember that x is defined
as u squared minus v squared.
And so partial x
partial u, that's 2u.
Partial z partial y, again,
is this 2y that we computed.
And partial y partial u is v.
This v is just because u was
u*v, and we take a partial
in the u-direction.
OK.
So altogether this
is 4u*x plus 2v*y,
and that's our
partial derivative.
So notice that, you know,
x is a function of u and v.
So if I really wanted
to, I could substitute
for x its formula for
u and v, but that's not
really necessary.
You know, what's interesting
about these problems is
how the differentials
depend on one another,
and I'm perfectly happy
with an answer that
has mixed variables like this.
That's fine.
So now, let's go
over here and let's
see if we can get the
same answer by using
total differentials.
Now, I have to say
that the chain rule
that we used on the
previous problem,
it's the quickest way to
do these sorts of things.
I like to do total differentials
if I have some time
to actually explore the problem
and get comfortable with it.
I prefer to use total
differentials because I think
it's a little bit clearer.
Somehow, this chain
rule it's just, to me,
it's just a prescription,
it's not an explanation.
So why don't we compute
some total differentials.
So we already saw-- let
me just repeat over here.
We already saw that dz
is 2x dx plus 2y dy.
OK.
Now, we want to
use the fact that x
is itself a function of
u and v. So that's what
we need to do now.
So that tells us that dx is 2u
du minus 2v dv in the same way.
And dy.
So remember, y was u*v.
So taking d of u*v,
we get v du plus u dv.
OK?
So now, so what we've
done is we've just
listed out all of the
total differentials.
And the nice thing about
this is once you've
done these computations,
now it's just substitution.
So what we really want
to know is how does z
depend on u and v. And
so all we need to do
is substitute in our
formulas for dx here.
So this tells us
that dz is-- OK,
so we have 2x-- instead of
dx, we just plug in here--
so we have 2u du minus 2v dv.
So that was this term.
And now we have
plus 2y-- and now
we just plug in this--
so v du plus u dv.
You see?
It's just substitution.
So then now, we just
expand everything out.
And so we get-- OK,
so let's collect
all the things involving du.
So if we collect all the
things involving du, we have--
2 times 2 times x times
u-- 4x*u plus 2y*v.
This whole quantity times du.
And then if we collect the
terms in dv, we have 2y*u.
So that's coming from here, and
then we have a minus 4x*v. OK?
And now what that tells
us is that-- so let's
just remember that
one definition
of the partial derivative
partial z partial u
is this coefficient.
So if I go over here, if we
write the total differential
dz, we can write that as
partial z partial u du
plus partial z partial v dv.
Right?
Well, look.
What we have here
on these two sides
is essentially the
same expression.
So that means if
we want to compute
partial z partial u, then that's
just equal to this coefficient
here.
So we get that partial z
partial u is 4x*u plus 2--
that should be v.
One of those is an x.
Let's see.
So where did this come from.
Yeah, one of those
is an x, sorry--
SPEAKER 1: It's a y.
DAVID JORDAN: --is a y.
2v*y, OK.
Now just as a sanity
check, why don't we
go back to the
middle of the board,
and we'll see that we
got the same thing.
So 4x*u plus 2v*y, that's what
we concluded for partial z
partial u.
And then going back to the
middle of the board, that's we
found again.
So let's just go over
the two different methods
and compare them.
So if I'm in a rush to
do a computation-- maybe
I'm taking an
exam-- I definitely
think it's the quickest
to just compute,
to figure out what the
dependency of the variable is,
and I use this dependency graph.
And then I just trace
all the paths from z
to the independent variable
u that I'm interested in.
And then I multiply all
the partial derivatives
that correspond to each edge
and I get an expression.
Now if I have more
time, then I really
prefer to use the method
of total differentials
that we did on the third board.
I like it, because once you
do some simple calculus,
and then after that it's
just, it's basic algebra.
I find that I'm
less likely to make
a mistake doing that method.
But as you saw, it
involves computing
a lot more derivatives
that we didn't actually
use in the final answer.
For instance, when we
computed total differentials,
we got an expression
for partial z partial v
at the end of the
day, even though we
weren't asked to do that.
So it's lengthier, but I
think more conceptually
straightforward.
So I think I'll
leave it at that.
