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DAVID JORDAN: Hello, and
welcome back to recitation.

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So in this problem, we're
considering a function

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f of three variables, f of x,
y, z, and it's differentiable.

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And we're not told
a formula for f.

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We just know that it's
differentiable at this point

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P, which is 1, minus
1, 1, and we're

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told that the gradient
of f at that point

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is this particular vector 2i
plus j minus 3k, at that point

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P. So all we understand
about f is how

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it looks around the point P.

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Now, we also have this
relation between the variables,

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so x, y and z aren't unrelated.

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They're related
by this constraint

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that z is x squared
plus y plus 1.

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So with this
information, we want

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to compute the gradient
of a new function g,

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and the new function g is a
function of two variables,

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and this function g is obtained
from f by just plugging

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in our relation for y.

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So we can use our
constraint to solve for y,

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and then this function g is just
f with that constraint applied.

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And what we really
want to do is we

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want to find the gradient
of g at the point (1, 1).

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So notice that when
g is equal to 1,

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1, that means that--
sorry, when the input of g

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is 1, 1, that means the
input of f is P. OK?

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So why don't you pause the video
and work this out for yourself.

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Check back with me and
we'll work it out together.

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OK, welcome back.

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So let's get started.

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So this problem may
not look like it's

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a problem about partial
derivatives with constraints,

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but that's what it's really
going to boil down to,

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which is to say that when
we want to compute-- so

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when we want to answer
this question by computing

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the gradient, the first thing
we're going to want to do

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is compute the partial
derivative of g

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and its variable x.

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And from the way we
set up the problem,

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that's just the
same as computing

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the partial derivative
of f with respect

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to x and keeping z constant.

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Now, remember, when we
do partial derivatives

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with constraints, what's
important about the notation

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is what's missing.

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The variable y is
missing, and that's

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because we are going to use the
constraint to get rid of it,

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and that's exactly
how we define g,

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so this is the key observation.

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So computing the
gradient of g is just

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going to be computing
these partial derivatives

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with constraints.

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So we'll do that in a moment,
and I'll also just write

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that the partial derivative
of g in the z-direction

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is partial f partial z,
now keeping x constrained.

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All right, so we need to compute
these partial derivatives

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with constraints.

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And so you remember
how this goes.

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The way that I
prefer to do this is

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to compute the
total differentials.

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So let's compute over here.

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The total differential df
is-- the total differential

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of f is just the partials
of f in the x-direction, f

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in the y-direction, f in the
z-direction, and each of these

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is multiplied by the
corresponding differential.

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And we don't know f, so we can't
compute the partial derivatives

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of it in general, but we do
know these partial derivatives

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at this point.

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And so in the problem,
we were given that this

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is 2dx plus dy minus 3dz.

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So this is just using the
fact that the gradient of f we

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were given is 2, 1 minus 3, OK?

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So that's the total
differential of f,

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and now we have this constraint.

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And remember, when we do
these partial derivatives

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with constraints, the trick
is to take the differential

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of the constraint.

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So we had this equation z
equals x squared plus y plus 1,

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and what we need to do
is take its differential.

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So we have dz is 2x dx plus dy.

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So that's our constraint.

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Now here we have
this variable x,

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but we're not varying
x in this problem.

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We're only focused on the point
P, and at the point P, x is 1.

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So, in fact, dz is
just 2dx plus dy, OK?

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So now what we need to do is we
need to combine the constraint

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equation and the
total differential

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for f into one
equation, and so this

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is just linear algebra now.

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So I'll just come over here.

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So we can rewrite our total
differential for the constraint

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as saying that dy is
equal to dz minus 2dx,

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and then we can plug that back
into our total differential

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for f.

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And so we get that df
is equal to 2dx plus--

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now I plug in dy here--
so dz minus 2dx, and then

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finally, minus 3dz.

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So altogether, I
get a minus 2dz,

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because this and this cancel.

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OK.

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We get a minus 2dz.

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So what does that tell
us about the gradient?

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So remember that
the differential now

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of g and its variables
is partial g partial x

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dx plus partial g partial z dz.

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And remember, the trick
about partial derivatives

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and their relation to
total differentials

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is that the partial derivative
is just this coefficient.

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So the fact that we
computed df and we

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found that it was minus
2dz, that tells us

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that dg-- so the thing
that we need to use

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is that g, remember, is just
the specialization of f.

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So dg is equal to
df in this case,

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because g is just a special
case of f with constraints.

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So when we computed
df here, we were

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imposing exactly the constraints
that we used to define dg,

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and so what this tells
us is that we can just

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compare the coefficients
here and so our gradient is

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0, minus 2.

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So the 0 is because there
is no dependence anymore

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on x at this point.

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There wasn't a dx term.

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There could have been
and there wasn't.

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And the minus 2 is because
that's the dependence on z.

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OK, so this is a
bit complicated,

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so why don't we
review what we did.

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So going back over to
the problem statement,

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we first needed to
realize that saying that g

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was a special case of f where
we use our constraints to solve

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for y, that's what put us
in the context of problems

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with constraints.

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So the fact that
the dependence on y

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was so simple that we could
just use the constraint.

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OK.

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So then, what that
allowed us to do

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is it allowed us to say that
the partial derivative of g

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in the x-direction is just
the partial derivative

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in the x-direction
subject to the constraint,

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and that's what we did here.

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And so then, the next
steps that we did

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are the same steps
that we would always

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do to compute partial
derivatives with constraints,

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and so we finally got
a relationship that

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said that, subject
to these constraints,

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df is equal to minus 2dz.

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And the point is that
g is just the function

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f with those
constraints imposed,

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and so dg and df are the same,
and so, in particular, dg

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is minus 2dz.

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And then, we remember
that you can always

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read off partial derivatives
from the total differential.

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They're just the coefficients.

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And so in the end, we got that
the gradient was 0 minus 2.

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And I think I'll
leave it at that.