DAVID JORDAN: Hello, and welcome back to recitation.
The problem I'd like to work with you
now is simply to compute some partial derivatives,
using the definitions we learned today in lecture.
So first we're going to compute the partial derivative
in the x-direction of this function x
y squared plus x squared y.
Then we're going to compute its derivative in the y-direction,
and then finally we're going to evaluate the partial derivative
in the x-direction at a particular point (1, 2).
That's the first problem.
And in the second problem, we're going
to compute second partial derivatives.
Now these we just compute by taking
the derivative of the derivative,
just as we do in one-variable calculus.
So why don't you work on these, pause the tape,
and I'll check back in a moment, and we'll
see how I solve these.
OK, welcome back.
Let's get started.
So we have x squared y-- excuse me,
x y squared plus x squared y.
That's our f.
So when we take the partial derivative in the x-direction--
remember, this just means that we treat y
as if it were a constant, and we just
take an ordinary derivative in the x-direction
as we would do in one-variable calculus.
So the derivative of this in the x-direction
is just y squared, because we only differentiate the x here.
Similarly here, the derivative of x squared is 2x,
and y just comes along for the ride as if it were a constant.
For the partial derivative in the y-direction,
we do the same thing, except now, x is a constant,
and we're taking an ordinary derivative in the y-direction.
So we have 2xy plus x squared.
And then the final thing that we need to do
is we want to evaluate partial f, partial x at the point
(1, 2).
And so all that means is that we have
to plug in x equals 1 and y equals 2
into our previous computation, and so we get 2 squared
plus 2 times 1 times 2.
So altogether, we get 8.
So that's computing partial derivatives.
Now let's move on and compute the second partial derivatives.
So, for instance, we want to compute
the second partial derivative both times in the x-direction.
So all this means is that when we took the first partial,
we got a function of x and y, and now we
just need to take its partial.
So we just need to take the derivative of this
again in the x-direction.
So now, the derivative of y squared--
be careful-- the derivative of y squared in the x-direction
is just zero, because y is a constant relative to x.
And so, then altogether, we just get 2y.
When we take the derivative to this x, we just get 1.
So that's our second partial derivative in the x-direction.
And now you can also take mixed partials.
So here, we take a derivative of f.
First we take the derivative in the y-direction
and then we take a derivative of that in the x-direction.
So we can look at our derivative here, partial f, partial y,
and we need to take its partial in the x-direction.
And so we get 2y plus 2x.
Now let's see what happens if we switch the order here
and we take, instead, the partial derivative
in the opposite order.
So now let's go back to our partial derivative of f
in the x-direction and let's take its derivative now
in the y-direction.
So the first term there, y squared,
gives us a 2y and the second term gives us a 2x.
I want to just note that these are equal.
In fact, the mixed partial derivatives,
whether you take them in the xy order or the yx order,
for the sorts of functions that we're going to be considering
in this class-- for instance, all polynomial functions
and all differentiable functions of several variables--
these mixed partials are going to be equal.
In your textbook, there are some examples
of sort of pathological functions
where these are not equal, but certainly
for any polynomial functions, these
are always going to be equal.
And I think I'll leave it at that.