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DAVID JORDAN: Hello, and welcome
back to recitation.

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Today the problem I'd like
to work with you is about

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computing partial derivatives
and the total differential.

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So we have a function z which
is x squared plus y squared.

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So it depends on the two
variables x and y.

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Now the variables x and y
themselves depend on two

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auxiliary variables, u and v.
So that's the setup that we

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have.

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So in part a, we just want to
compute the total differential

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dz in terms of dx and dy.

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So u and v aren't going to
enter into the picture.

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And then in part b, we're going
to compute the partial

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derivative partial z partial
u in two different ways.

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First, we're going to compute
it using the chain rule.

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And then we're going to
compute it using total

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differentials.

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And so we'll substitute in some
of the work that we had

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in a to solve that part.

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So why don't you pause
the video now

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and work on the problem.

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We'll check back and we'll
do it together.

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Hi, and welcome back.

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Let's get started.

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So first, computing
a is not so bad.

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So we just need to first
remember, what does it mean to

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compute the total
differential?

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So the total differential dz is
just the partial derivative

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of z in the x-direction dx plus
z in the y-direction dy.

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OK?

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So now, looking at our
formula here for z.

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We have, so, the partial
derivative of z in the

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x-direction is 2x,
so this is 2x dx.

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And the partial derivative
of z in the y is 2y,

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so we have 2y dy.

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OK, and that's all we
have to do for a.

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Now for b, we want to compute
this partial derivative in two

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different ways.

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First, using the chain rule.

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So let's remember what
the chain rule says.

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So whenever I think about the
chain rule, I like to draw

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this dependency graph.

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OK?

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And this is just a way for me to
organize how the different

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variables depend
on one another.

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So at the top, we have z.

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And z is a function of x and y,
but x is itself a function

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of both u and v, and y is also
a function of u and v. So z

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depends on x and y, and x and y
each jointly depend on u and

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v. So it's a little bit
complicated, the

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relationships here.

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So now, what the chain rule
says is that if we take a

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partial derivative--

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partial z partial u--

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we have to go through our
dependency graph.

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Every way that we can get from
z to u, we get a term in our

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summation for each
one of those.

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So for instance, z goes
to x goes to u.

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So that means that we have
partial z, partial x, partial

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x, partial u.

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And then we can also go
z goes to y goes to u.

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And that will give us partial
z, partial y,

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partial y, partial u.

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And now these partials are ones
that we can just compute

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from our formulas.

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So for instance, partial
z partial x is

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2x, which we computed.

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Now partial x, partial u, we
have to remember that x is

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defined as u squared
minus v squared.

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And so partial x, partial
u, that's 2u.

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Partial z, partial y, again, is
this 2y that we computed.

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And partial y, partial u is v.
This v is just because u was

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uv, and we take a partial
in the u-direction.

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OK.

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So altogether this is 4ux plus
2vy, and that's our partial

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derivative.

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So notice that, you know, x is a
function of u and v. So if I

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really wanted to, I could
substitute for x its formula

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for u and v, but that's
not really necessary.

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You know, Wwat's interesting
about these problems is how

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the differentials depend on one
another, and I'm perfectly

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happy with an answer that has
mixed variables like this.

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That's fine.

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So now, let's go over here and
let's see if we can get the

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same answer by using total
differentials.

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Now, I have to say that the
chain rule that we used on the

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previous problem, it's the
quickest way to do

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these sorts of things.

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I like to do total differentials
if I have some

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time to actually explore
the problem and get

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comfortable with it.

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I prefer to use total
differentials because I think

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it's a little bit clearer.

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Somehow, this chain rule it's
just, to me, it's just a

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prescription, it's not
an explanation.

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So why don't we compute some
total differentials.

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So we already saw-- let me
just repeat over here.

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We already saw that dz
is 2x dx plus 2y dy.

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OK.

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Now, we want to use the fact
that x is itself a function of

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u and v. So that's what
we need to do now.

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So that tells us that
dx is 2u du minus 2v

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dv in the same way.

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And dy.

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So remember, y was uv. So taking
d of uv, we get v du

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plus u dv. OK?

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So now, so what we've done is
we've just listed out all of

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the total differentials.

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And the nice thing about this
is once you've done these

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computations, now it's
just substitution.

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So what we really want to know
is how does z depend on u and

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v. And so all we need to
do is substitute in our

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formulas for dx here.

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So this tells us that dz
is, OK, so we have 2x--

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instead of dx, we just plug in
here-- so we have 2u du minus

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2v dv. So that was this term.

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And now we have plus 2y--

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and now we just plug in this--

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so v du plus u dv. You see?

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It's just substitution.

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So then now, we just expand
everything out.

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And so we get--

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OK, so let's collect all the
things involving du.

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So if we collect all the things
involving du we have--

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2 times 2 times x times u--

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4xu plus 2yv. This whole
quantity times du.

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And then if we collect the
terms in dv, we have 2yu.

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So that's coming from here,
and then we have

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a minus 4xv. OK?

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And now what that tells
us is that--

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so let's just remember that one
definition of the partial

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derivative partial z partial
u is this coefficient.

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So if I go over here, if we
write the total differential

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dz, we can write that as partial
z, partial u du plus

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partial z, partial
v dv. Right?

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Well, look.

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What we have here on these two
sides is essentially the same

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expression.

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So that means if we want to
compute partial z partial u,

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then that's just equal to
this coefficient here.

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So we get that partial z partial
u is 4xu plus 2--

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that should be v. One
of those is an x.

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Let's see.

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So where did this come from.

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Yeah, one of those
is an x, sorry--

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SPEAKER 1: It's a y.

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DAVID JORDAN: --is a y.

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2vy, OK.

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Now just as a sanity check, why
don't we go back to the

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middle of the board, and
we'll see that we

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got the same thing.

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So 4xu plus 2vy, that's
what we concluded for

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partial z partial u.

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And then going back to the
middle of the board, that's we

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found again.

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So let's just go over
the two different

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methods and compare them.

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So if I'm in a rush to
do a computation--

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maybe I'm taking an exam--

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I definitely think it's the
quickest to just compute, to

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figure out what the dependency
of the variable is, and I use

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this dependency graph.

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And then I just trace all
the paths from z to the

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independent variable u that
I'm interested in.

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And then I multiply all the
partial derivatives that

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correspond to each edge and
I get an expression.

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Now if I have more time, then
I really prefer to use the

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method of total differentials
that we did

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on the third board.

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I like it, because once you do
some simple calculus, and then

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after that it's just,
it's basic algebra.

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I find that I'm less
likely to make a

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mistake doing that method.

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But as you saw, it involves
computing a lot more

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derivatives that we didn't
actually use

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in the final answer.

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For instance, when we computed
total differentials, we got an

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expression for partial z partial
v at the end of the

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day, even though we weren't
asked to do that.

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So it's lengthier, but I
think more conceptually

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straightforward.

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So I think I'll leave
it at that.

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