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DAVID JORDAN: Hello, and
welcome back to recitation.

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The problem I'd like
to work with you

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now is simply to compute
some partial derivatives,

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using the definitions we
learned today in lecture.

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So first we're going to
compute the partial derivative

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in the x-direction
of this function x

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y squared plus x squared y.

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Then we're going to compute its
derivative in the y-direction,

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and then finally we're going to
evaluate the partial derivative

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in the x-direction at a
particular point (1, 2).

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That's the first problem.

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And in the second
problem, we're going

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to compute second
partial derivatives.

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Now these we just
compute by taking

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the derivative of
the derivative,

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just as we do in
one-variable calculus.

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So why don't you work on
these, pause the tape,

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and I'll check back
in a moment, and we'll

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see how I solve these.

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OK, welcome back.

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Let's get started.

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So we have x squared
y-- excuse me,

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x y squared plus x squared y.

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That's our f.

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So when we take the partial
derivative in the x-direction--

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remember, this just
means that we treat y

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as if it were a
constant, and we just

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take an ordinary derivative
in the x-direction

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as we would do in
one-variable calculus.

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So the derivative of
this in the x-direction

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is just y squared, because we
only differentiate the x here.

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Similarly here, the
derivative of x squared is 2x,

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and y just comes along for the
ride as if it were a constant.

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For the partial derivative
in the y-direction,

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we do the same thing,
except now, x is a constant,

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and we're taking an ordinary
derivative in the y-direction.

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So we have 2xy plus x squared.

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And then the final
thing that we need to do

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is we want to evaluate partial
f, partial x at the point

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(1, 2).

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And so all that
means is that we have

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to plug in x equals
1 and y equals 2

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into our previous computation,
and so we get 2 squared

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plus 2 times 1 times 2.

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So altogether, we get 8.

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So that's computing
partial derivatives.

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Now let's move on and compute
the second partial derivatives.

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So, for instance,
we want to compute

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the second partial derivative
both times in the x-direction.

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So all this means is that when
we took the first partial,

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we got a function of
x and y, and now we

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just need to take its partial.

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So we just need to take
the derivative of this

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again in the x-direction.

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So now, the derivative
of y squared--

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be careful-- the derivative of
y squared in the x-direction

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is just zero, because y is
a constant relative to x.

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And so, then altogether,
we just get 2y.

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When we take the derivative
to this x, we just get 1.

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So that's our second partial
derivative in the x-direction.

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And now you can also
take mixed partials.

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So here, we take
a derivative of f.

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First we take the derivative
in the y-direction

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and then we take a derivative
of that in the x-direction.

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So we can look at our derivative
here, partial f, partial y,

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and we need to take its
partial in the x-direction.

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And so we get 2y plus 2x.

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Now let's see what happens
if we switch the order here

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and we take, instead,
the partial derivative

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in the opposite order.

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So now let's go back to
our partial derivative of f

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in the x-direction and let's
take its derivative now

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in the y-direction.

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So the first term
there, y squared,

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gives us a 2y and the
second term gives us a 2x.

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I want to just note
that these are equal.

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In fact, the mixed
partial derivatives,

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whether you take them in the
xy order or the yx order,

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for the sorts of functions that
we're going to be considering

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in this class-- for instance,
all polynomial functions

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and all differentiable
functions of several variables--

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these mixed partials
are going to be equal.

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In your textbook,
there are some examples

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of sort of
pathological functions

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where these are not
equal, but certainly

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for any polynomial
functions, these

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are always going to be equal.

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And I think I'll
leave it at that.