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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 now is simply to

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compute some partial derivatives
using the

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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, this
function xy squared

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plus x squared y.

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Then we're going to compute
its derivative in the

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y-direction, and then finally
we're going to evaluate the

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partial derivative in the
x-direction at a particular

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point 1, 2 for the first
problem, and in the second

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problem, we're going to
compute second partial

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derivatives.

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Now these we just compute by
taking the derivative of the

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derivative, just as we do in
one-variable calculus.

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

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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--

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excuse me, xy 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

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x-direction--

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remember, this just means that
we treat y as if it were a

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constant, and we just take an
ordinary derivative in the

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x-direction as we would do
in one-variable calculus.

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So the derivative of this in
the x-direction is just y

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squared, because we only
differentiate the x here.

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Similarly here, the derivative
of x squared is 2x, and y just

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

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For the partial derivative in
the y-direction, we do the

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same thing, except now, x is a
constant, and we're taking an

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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 is we want to

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evaluate partial f, partial
x at the point 1, 2.

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And so all that means is that we
have to plug in x equals 1

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and y equals 2 into our previous
computation, and so

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we get 2 squared 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

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derivatives.

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So, for instance, we want to
compute the second partial

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derivative both times
in the x-direction.

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

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

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need to take its partial.

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So we just need to take the
derivative of this again in

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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 one.

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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 and then we

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take a derivative of that
in the x-direction.

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

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partial y, and we need to
take its partial in the

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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 and

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we take, instead, the partial
derivative in

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the opposite order.

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

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

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y-direction.

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So the first term there, y
squared, gives us a 2y and the

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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, whether you take

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them in the xy order or the
yx order for the sorts of

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functions that we're going to be
considering in this class,

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for instance, all polynomial
functions and all

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differentiable functions of
several variables, these mixed

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partials are going
to be equal.

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In your textbook, there are
some examples of sort of

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pathological functions where
these are not equal, but

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certainly for any polynomial
functions, these are always

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going to be equal.

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And I think I'll leave
it at that.

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