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DAVID JORDAN: Hello, and
welcome back to recitation.

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So today, the problem
I'd like to work with you

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is about taking
partial derivatives

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in the presence of constraints.

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So this is a pretty
subtle business.

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So take your time when
you work these problems.

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So what we have is we
have this function w,

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and it's a function of four
variables: x, y, z, and t.

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OK?

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But it's not really a function
of these four variables

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because we have a constraint.

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So we want to
study how w changes

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as we vary the parameters,
except that we have imposed

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this constraint here.

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So that really we kind of
only have three variables,

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because we have four
variables and one constraint.

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So that's what partial
derivatives with constraints

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help us do.

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So let's explain
first the notation.

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OK?

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So it says partial w
partial z, and then we

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have the subscripts x and y.

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So what's important
about this notation

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is not what you see as
much as what you don't see.

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What you don't see
is the variable t.

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OK?

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So what this notation
means is, as always,

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the denominator in our
derivative expression--

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partial z here-- that means
that we want to vary z.

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And we want to see how
w changes as we vary z.

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And the x and y here mean that
we want to keep x and y fixed.

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So if we didn't
have a constraint,

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this x and y here
would be superfluous.

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Because by partial
derivative, we always

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mean to keep the other
unlisted variables unchanged.

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However, the fact that
t is missing here,

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it means that-- so if you
think about it-- if we vary z,

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and we keep x and y fixed,
then t also is varying.

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Right?

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Because we have this
constraint here.

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And so it wouldn't
make sense for me

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to ask you to compute the
partial derivative of w

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in z varying x, y, and t
because-- excuse me-- keeping

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x, y, and t fixed,
because then there

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would be no room for z to vary.

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OK?

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So this notation means that z
is going to be allowed to vary,

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but it's going to vary
in a way that we're just

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going to ignore.

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So you will see how this
works out in the problem.

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So what we're
really interested in

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is making sure that x and y
stay fixed and that z varies.

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And then we're going to need
to-- when we do some algebra,

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we're going to need to
get rid of any mention

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of the variable t.

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OK.

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So the first way that we're
going to work this out

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is using total differentials.

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And I like to use
total differentials

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when I'm on new ground
because-- they're not

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the most computationally
effective,

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because they involve computing
all the derivatives that we

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might possibly need in sight.

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So they're not the most
efficient computationally.

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But if you go ahead and compute
the total differentials,

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then all the other
computations that you have

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to do are just substitution.

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So it really just
becomes linear algebra,

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and that's what I like about it.

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In part b, we'll see a shortcut
using implicit differentiation

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and the chain rule.

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And this is going to
be a little bit tricky.

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So we have these
two equations, we

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need to turn them both into
differential equations.

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And so we'll do that
using a combination

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of implicit differentiation
and the chain rule.

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So I'll let you pause
the video and get

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started on these problems.

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And you can check back and
we'll work it out together.

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OK, welcome back.

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So let's start by doing a,
let's start with problem a.

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So we have total differentials
is the suggested way

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to attack this.

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So why don't we
just start computing

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the total differentials
that we know.

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So we have two equations.

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w in relation to the other
variables and the constraint

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equation.

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And what we first
want to do is just

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take the total differential
of both of those equations

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to get started.

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So we can take the first one
and it tells us that dw is equal

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to-- OK, so we have 3 x
squared y dx, plus x cubed dy,

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minus 2z*t dz,
minus z squared dt.

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OK.

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Now right away, we can
simplify this equation.

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So this is the
total differential,

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but we have to remember
that in the setting we're

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interested in, x and
y are held fixed.

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And so holding x and y fixed
means that the differentials dx

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and dy are both set to 0.

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So that lets us rewrite this
first differential equation is

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just dw equals minus 2z*t
dz minus z squared dt.

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So that's our first
equation that we get.

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Let me just check with
my notes to make sure.

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That's right.

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OK.

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And so now, we have
the constraint equation

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from the original
statement of the problem.

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And we need to take
its differential.

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So on the one hand,
we get x dy plus y dx.

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That's the total differential
of the left-hand side.

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And then on the right-hand
side, we have t dz plus z dt.

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OK?

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And now we notice that now the
left-hand side of this equation

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is just 0 for the same reason.

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dy and dx are being held fixed.

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So the relation that
we end up getting

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is we get that dt is equal
to minus t over z dz by just

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doing straightforward algebra.

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OK.

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So, with that in
hand now, we can-- so

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remember I mentioned
in the beginning

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that our goal was-- so
from the very beginning,

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we knew that if we varied z,
because of our constraint,

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we're going to be
forced to be varying t.

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And that's exactly what this
equation says, doesn't it?

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We got this by just taking the
differential of the constraint.

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And it says if you
vary z, you have

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to vary t in an
appropriate way, and that's

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what this coefficient tells us.

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So what we're
really interested in

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is how does w vary
in terms of z here.

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And so we want to get
rid of this dt here.

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And in fact, we can by
using the constraint.

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So combining this equation
with this equation,

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we get that dw here
is equal to-- OK,

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so we have minus 2z*t dz.

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And then we have minus-- OK--
z squared times another minus

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times t over z, so this
all becomes a plus z*t dz.

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So all I did is I plugged in
for dt using our formula here.

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And so this altogether is
equal to just minus z*t dz.

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And that tells us that the
partial derivative that we're

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after is just this
coefficient, right?

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The partial derivative
is just defined

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to be the coefficient
of the differential

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once you work everything out.

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And so this is minus z*t.

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OK, so that's a.

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So now let's see if we can
use some tricks to make

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the computation a bit shorter.

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So the tricks that
we're going to use

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are implicit differentiation
and the chain rule.

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So at the end of the
day-- excuse me--

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we're interested in
partial w partial z.

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And what we're going to
do is use the chain rule

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to just take a straightforward
partial derivative

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of our original expression.

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So remember, w was x
cubed y minus z t squared.

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And so let's just take
a partial derivative

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of that in the z-direction.

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So the partial derivative in
the z-direction of x cubed y

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is just 0.

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So that will go away.

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And so we only have minus--
we have a 2z*t component.

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That's just because the partial
derivative of z squared is 2z.

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And then we have another term
which is minus z squared,

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and now we need to take
the partial derivative of t

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in the z-direction.

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So, you know, often times when
we take partial derivatives

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of one variable in
terms of the other,

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it's common to think that
the partial derivative of one

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variable in terms of
the other is just 0.

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Because usually our
variables are independent.

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They don't vary in
terms of one another.

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But this is exactly
a situation where

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t does vary depending on z,
and so we had to include that

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into our notation.

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OK.

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So now this is
almost what we want,

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except we have this
mystery component here.

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And of course,
there's only one way

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we can solve this mystery, which
is the same way we solved it

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in part a.

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We have to use the constraint.

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So let's take partial z of
our constraint equation.

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And remember, our constraint
equation was x*y equals z*t.

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OK.

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So if we take the partial
derivative of this equation--

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so if I take the partial
derivative of x and y

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in terms of z, then I do
get 0, because x and y are

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genuinely independent from z.

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It's only t that depends on z.

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So on this side we get 0.

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Now, on the other side I just
need to use the product rule.

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So I get t, plus z
partial t partial z.

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OK?

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So we can rewrite this as saying
that partial t partial z is

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minus t over z.

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OK?

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Now, you might notice that,
you know, this is formally

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very similar to what we did
in part a, and of course,

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that's no surprise.

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When we are manipulating
using implicit differentiation

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and the chain rule,
it's just a compact way

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of doing what we were doing
with the total differentials.

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I mean, to me, the chain
rule is a computation

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which you could prove by
doing the corresponding thing

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with total differentials.

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And so we get this same
coefficient negative t over z,

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which you recall that
we got in part a.

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OK.

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So now we have, once again
we have this, two equations,

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and we just can do substitution.

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So we get that partial w partial
z is equal to minus 2z*t.

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And now again, we get
minus another minus,

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and z here cancels the z
squared, so we get plus z*t.

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And so we get minus z*t.

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OK, and finally, if we
remember our assumptions,

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our assumptions were that x
and y were independent of z.

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That was our notation.

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And we use that assumption
at this step right here.

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So in fact, we don't just have
the partial derivative of w

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with respect to z.

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We need to specify that
we held x and y fixed.

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OK.

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So just to review
again, if we look now

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at what we did in part b, you
know, the meat of the argument

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was the exact same as
what we did in part a.

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The meat of the
argument was right here.

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We took some derivative and
then this was an unknown.

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The definition of
w doesn't know how

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t and z depend on one another.

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That you can only find by
looking at the constraint.

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And so we just went
through the problem

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and we took derivatives
of the constraint,

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and that gave us an equation
that we were looking for.

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Now if we go back now
to part a over here.

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So as you can see, there's a lot
more work involved in part a.

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On the other hand, to me it
was more straightforward.

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We just had to compute
the total differentials

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and then do some linear
algebra with cancellations.

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And somehow, when you
do total differentials,

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you just compute everything
that could possibly come up,

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and then you just
substitute it in.

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And indeed, we got
the same answer:

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partial w partial z
as being minus z*t.

248
00:15:12,690 --> 00:15:14,615
OK, and I think I'll stop there.