Let's try to get a better intuition of the chain rule,
and in the process, we'll get a better intuition of how
it applies to implicit differentiation or vice versa.
So let's say-- and I'm going to get out of the world of x's and
y's and f of x's for a second, just so that you see that those
are just letters and there's nothing special about them.
And I think sometimes it helps develop the intuition.
So let's say that a is equal to b squared and that b is
equal to the sine of c.
And my question to you is what is the derivative
of a with respect to c?
We have a defined in terms of b, so it's not like we can
just take the derivative.
I mean, how do we do that?
Well, that's where the chain rule comes in.
And the chain rule just tells us that the derivative of a
with respect to c is equal to the derivative of a with
respect to b times the derivative of b
with respect to c.
And you shouldn't really-- you don't even have
to memorize this.
We're just multiplying two fractions where the denominator
in the first one is the same as the numerator in
the second one.
So if you were to multiply fractions, they would
cancel out, right?
Or if you were to actually multiply these fractions
out, you'd get da times db on the numerator.
In the denominator, you would get-- da-- sorry, db times dc
in the numerator, and then these would cancel out, and
then you're back with da over dc.
So there's nothing fancy, really, about the chain rule
when you actually view them for what they are.
They're fractions, but the values in the numerator and the
denominator, they are these differentials, these infinitely
small changes in these variables.
But anyway, the chain rule tells us: change of a with
respect to c is equal to the change of a with respect to
b times the change of b with respect to c.
And these are pretty easy to calculate based on
the information that I've just given you.
What is the change of a with respect to b?
The derivative of a with respect to b?
Well, that's pretty straightforward, right?
The derivative a with respect to b is equal to 2b.
So this is 2b.
And what's the derivative of b with respect to c?
Derivative of b with respect to c, well, that's
equal to cosine of c.
And so we're done.
The derivative of a with respect to c is equal to this,
which is this, 2b, the derivative of a with respect to
b times the derivative of b with respect to c.
So times cosine of c.
And then if you don't like it in this format, if you don't
like the derivative of a with respect to c being dependent or
being a function of both b and c, you can just substitute for
b because b is also a function of c.
So let's do that.
So the derivative of a with respect to c is equal to 2b.
But what's b?
b is just sine of c.
So 2-- just substitute sine of c for b-- 2 sine of c, and then
we have the remainder, right?
This is this, and then times cosine of c,
and that's just that.
And we're done.
And I want to now try to make the connection between this and
the kind of-- I don't want to say-- well, you know, kind of
the plain English way I talked about doing that
chain rule before.
So all of the chain rule problems that I think I've done
so far would have had a explicitly defined in terms of
c, but you're saying, oh, but then, you know, I said outside
and inside and et cetera.
And we can do that.
We can make a explicitly defined in terms of c,
and how do we do that?
We could substitute before we even take the derivative.
So we can substitute this for here, and we get what?
a is equal to b squared, b is sine of c, so it's
sine of c squared.
And now we could take the derivative of a with respect
to c using the chain rule.
And this is the more traditional way that I showed
it, but maybe it gave you more intuition.
It probably gave you more intuition of how to actually do
the problem, but it probably gave you less intuition of how
it applies to the chain rule.
So hopefully, this will connect everything together.
So the derivative of a with respect to c is equal to-- I
think sometimes I said that I like to take the derivative of
the inside and then multiply that times the derivative of
the outside, but we could do it in either order.
So let's take the derivative of the outside.
So what's the derivative of the outside?
sine of c squared, well, it's 2 sine of c times the
derivative of the inside.
The derivative of the inside is cosine of c.
The phone is ringing.
I'm not going to answer it.
And so we ended up with the exact same result, and it makes
sense, because when I say the derivative of the outside, when
I take the derivative of sine c squared and I get 2 sine of c,
I'm really just taking the derivative of a with
respect to b, right?
Because the derivative of a with respect to b
is 2b or 2 sine of c.
And then when I take the derivative of the inside, I'm
really taking the derivative of b with respect to c.
So hopefully, that gives you a little intuition.
So now let's take that intuition and tackle some
implicit differentiation problems.
And if you think about it, this kind of was an implicit
differentiation problem.
So let's say I want to take the derivative with respect to x
of y squared x to the third.
And some of you all might know how to mechanically do this.
Well, I won't go into the mechanics because the
whole point of this video is the intuition.
And maybe you have the intuition in which case you
might not have to watch this video, but let me give
you the intuition.
And I think the best way to give the intuition
is to actually do some variable substitution.
Let's define a is equal to y squared and b is equal to
x to the third, right?
Then this is the same thing, and I just made this stuff--
you know, this could have been any letters.
This is the same thing as the derivative with respect
to x of a times b, right?
And what does the chain rule tell us?
Well, actually, we could just do the product rule here, so
let's do the product rule.
So we're taking the derivative with respect to x of a times b.
So let's do that.
This is just the product rule so it's the derivative of a
with respect to x times b plus the derivative of b with
respect to x times a.
That's just the product rule.
But what's the derivative of a with respect to x and what's
the derivative of b with respect to x?
So the derivative of a with respect to x-- and this is
where we get into the chain rule.
The derivative of a with respect to x is equal to the
derivative of a with respect to y times the derivative of y
with respect to x, right?
This is just the chain rule.
These dy in the denominator and the numerator,
they'll cancel out.
So this is just the chain rule.
And what is this?
What's the derivative of a with respect to y?
Well, a is equal to y squared, so that is equal to 2y times
the derivative of y with respect to x.
Well, we haven't explicitly defined y in terms of x, so
we're just going to have to leave it like that.
So times dy/dx.
So this, right here, the derivative of a with respect
to x, is just that.
And what's the derivative of b with respect to x?
Well, here b is explicitly defined in terms of x, so this
is pretty straightforward.
There's no chain rule application here, so that is
just going to be 3x squared.
And so what's our final answer?
The final answer is the derivative with respect to x of
y squared x to the third is equal to derivative of a with
respect to x, or you could view that as the derivative of y
squared with respect to x, which is 2y dy/dx, right?
You could say this is the derivative of a with respect to
x or the derivative of y squared with respect to x, and
this is just from the chain rule, right?
This was the chain rule.
2y times dy/dx and then times b.
Well, what's b? b is x to the third plus the derivative of b
with respect to x, well, that's 3x squared, times a.
a is just y squared.
So hopefully, that gave you the intuition and made it a little
bit clearer of why implicit differentiation-- you're really
just applying the chain rule.
And then, you know, just the way I've been doing in all the
problems without going through all of this mess just to show
you that it isn't that hard of a problem, that you could
just do it straight up.
You could just say, OK, this is the product rule.
It's the derivative of the first expression times the
second expression plus the derivative of the second
expression times the first.
What's the derivative of this first expression y squared?
Well, it's the derivative of that expression with respect
to y times the derivative of y with respect to x.
And it's only because we're taking the derivative with
respect to x times the second expression, x to the third,
plus the derivative of the second expression with respect
to x, 3x squared, times the first expression, y squared.
Well, anyway, I'm at my time limit.
Hopefully, that helped a little bit and didn't confuse things,
and let me know if you want me to do even more videos on this.
See you soon.