DAVID JORDAN: Hello, and welcome back to recitation.
Today the problem I'd like to work with you
is about computing partial derivatives
and the total differential.
So we have a function z which is x squared plus y squared.
So it depends on the two variables x and y.
Now the variables x and y themselves
depend on two auxiliary variables, u and v.
So that's the setup that we have.
So in part a, we just want to compute the total differential
dz in terms of dx and dy.
So u and v aren't going to enter into the picture.
And then in part b, we're going to compute
the partial derivative partial z partial u
in two different ways.
First, we're going to compute it using the chain rule.
And then we're going to compute it using total differentials.
And so we'll substitute in some of the work
that we had in a to solve that part.
So why don't you pause the video now and work on the problem.
We'll check back and we'll do it together.
Hi, and welcome back.
Let's get started.
So first, computing a is not so bad.
So we just need to first remember,
what does it mean to compute the total differential?
So the total differential dz is just
the partial derivative of z in the x-direction dx plus
z in the y-direction dy.
OK?
So now, looking at our formula here for z,
we have-- so the partial derivative of z
in the x-direction is 2x, so this is 2x dx.
And the partial derivative of z in the y is 2y,
so we have 2y dy.
OK, and that's all we have to do for a.
Now for b, we want to compute this partial derivative
in two different ways.
First, using the chain rule.
So let's remember what the chain rule says.
So whenever I think about the chain rule,
I like to draw this dependency graph.
OK?
And this is just a way for me to organize
how the different variables depend on one another.
So at the top, we have z.
And z is a function of x and y, but x is itself
a function of both u and v, and y is also
a function of u and v. So z depends on x and y,
and x and y each jointly depend on u and v.
So it's a little bit complicated,
the relationships here.
So now, what the chain rule says is
that if we take a partial derivative-- partial z
partial u-- we have to go through our dependency graph.
Every way that we can get from z to u,
we get a term in our summation for each one of those.
So for instance, z goes to x goes to u.
So that means that we have partial z partial x,
partial x partial u.
And then we can also go z goes to y goes to u.
And that will give us partial z partial y, partial y partial u.
And now these partials are ones that we can just
compute from our formulas.
So for instance, partial z partial x
is 2x, which we computed.
Now partial x partial u, we have to remember that x is defined
as u squared minus v squared.
And so partial x partial u, that's 2u.
Partial z partial y, again, is this 2y that we computed.
And partial y partial u is v. This v is just because u was
u*v, and we take a partial in the u-direction.
OK.
So altogether this is 4u*x plus 2v*y,
and that's our partial derivative.
So notice that, you know, x is a function of u and v.
So if I really wanted to, I could substitute
for x its formula for u and v, but that's not
really necessary.
You know, what's interesting about these problems is
how the differentials depend on one another,
and I'm perfectly happy with an answer that
has mixed variables like this.
That's fine.
So now, let's go over here and let's
see if we can get the same answer by using
total differentials.
Now, I have to say that the chain rule
that we used on the previous problem,
it's the quickest way to do these sorts of things.
I like to do total differentials if I have some time
to actually explore the problem and get comfortable with it.
I prefer to use total differentials because I think
it's a little bit clearer.
Somehow, this chain rule it's just, to me,
it's just a prescription, it's not an explanation.
So why don't we compute some total differentials.
So we already saw-- let me just repeat over here.
We already saw that dz is 2x dx plus 2y dy.
OK.
Now, we want to use the fact that x
is itself a function of u and v. So that's what
we need to do now.
So that tells us that dx is 2u du minus 2v dv in the same way.
And dy.
So remember, y was u*v. So taking d of u*v,
we get v du plus u dv.
OK?
So now, so what we've done is we've just
listed out all of the total differentials.
And the nice thing about this is once you've
done these computations, now it's just substitution.
So what we really want to know is how does z
depend on u and v. And so all we need to do
is substitute in our formulas for dx here.
So this tells us that dz is-- OK,
so we have 2x-- instead of dx, we just plug in here--
so we have 2u du minus 2v dv.
So that was this term.
And now we have plus 2y-- and now
we just plug in this-- so v du plus u dv.
You see?
It's just substitution.
So then now, we just expand everything out.
And so we get-- OK, so let's collect
all the things involving du.
So if we collect all the things involving du, we have--
2 times 2 times x times u-- 4x*u plus 2y*v.
This whole quantity times du.
And then if we collect the terms in dv, we have 2y*u.
So that's coming from here, and then we have a minus 4x*v. OK?
And now what that tells us is that-- so let's
just remember that one definition
of the partial derivative partial z partial u
is this coefficient.
So if I go over here, if we write the total differential
dz, we can write that as partial z partial u du
plus partial z partial v dv.
Right?
Well, look.
What we have here on these two sides
is essentially the same expression.
So that means if we want to compute
partial z partial u, then that's just equal to this coefficient
here.
So we get that partial z partial u is 4x*u plus 2--
that should be v. One of those is an x.
Let's see.
So where did this come from.
Yeah, one of those is an x, sorry--
SPEAKER 1: It's a y.
DAVID JORDAN: --is a y.
2v*y, OK.
Now just as a sanity check, why don't we
go back to the middle of the board,
and we'll see that we got the same thing.
So 4x*u plus 2v*y, that's what we concluded for partial z
partial u.
And then going back to the middle of the board, that's we
found again.
So let's just go over the two different methods
and compare them.
So if I'm in a rush to do a computation-- maybe
I'm taking an exam-- I definitely
think it's the quickest to just compute,
to figure out what the dependency of the variable is,
and I use this dependency graph.
And then I just trace all the paths from z
to the independent variable u that I'm interested in.
And then I multiply all the partial derivatives
that correspond to each edge and I get an expression.
Now if I have more time, then I really
prefer to use the method of total differentials
that we did on the third board.
I like it, because once you do some simple calculus,
and then after that it's just, it's basic algebra.
I find that I'm less likely to make
a mistake doing that method.
But as you saw, it involves computing
a lot more derivatives that we didn't actually
use in the final answer.
For instance, when we computed total differentials,
we got an expression for partial z partial v
at the end of the day, even though we
weren't asked to do that.
So it's lengthier, but I think more conceptually
straightforward.
So I think I'll leave it at that.