Welcome to the video on implicit differentiation.
Let's just explain the difference between implicit
and explicit first.
So if I had a function that was, let's say, y is equal to
x squared plus 2x plus 3.
In this situation, y, the variable y, is defined
explicitly in terms of x.
How do I know that?
Well, if you give me an x, I can just input it into this
expression right there, and I explicitly get y.
Right?
It just pops out of the equation.
But on the other hand, what if I had something like this.
What if I had x squared plus y squared is equal to 100.
Well now, if I give you an x, you still can figure out a y,
but it's just not as easy.
Right?
If I said, x is equal to, I don't know, x is equal to 9,
you would put 9 in here, you would get 81, plus y squared is
equal to 100, you'd subtract that, you'd get y squared
is equal to, what, 19?
Those numbers are weird.
And then you'd get y is equal to the plus or
minus square root of 19.
So you still could solve for y, but this expression right
here, or this equation, defines y implicitly.
And that's a different, you know, when people use the
word explicit or implicit in everyday language,
it's the same thing.
When you say, he explicitly said this,
it means, I am hungry.
I explicitly stated that I'm hungry.
But you could implicitly state that you're hungry.
You could say, oh well, you know, I haven't eaten for a
long time, and you know, boy, it would be nice to maybe eat
something, and that would be kind of an implicit
expression of your hunger.
Because someone would have to kind of do a little bit of work
to come to the conclusion that you're hungry, instead
of you telling them.
So with that out of the way, let's actually do
some differentiation.
So everything you've learned about differentiation right
now has been explicit differentiation.
So example.
For this function, we know how to take the explicit
derivative here.
Well, it's an explicitly defined function, so we
just take the derivative.
We apply the derivative operator.
We take the rate of change of both sides of this equation
with respect to x.
You could do it like this.
We're taking the derivative operator, I know this is
getting messy, we're taking the derivative operator on that
side, and we're also taking the derivative operator
on this side.
I didn't plan to write that there that's
why-- so it happens.
On the left side, the rate of change at y changes
with respect to x, well we just write that.
The rate of change of y, for every small change in
y, what is change in x.
That equals-- and then, we know how to do this
type of derivative.
This is just a simple quadratic.
So it equals 2x plus 2 plus 0.
So there.
We just did some explicit differentiation.
So how do we do the same thing here?
Well, we're going to do the same thing.
We're going to apply the derivative operator to both
sides of this equation.
So let me do that, I'll do that in a different color.
Let me erase all of this, too, because this is kind
of obvious things that you might already know.
Let me erase all of that.
OK, so I'll take the derivative operator to both sides
of this equation.
So this is no different than what we did up here.
And as you learned in algebra, anything you do to one side of
the equation, you have to do to the other side of the equation.
Otherwise it would no longer be an equation.
Take the derivative operator-- because we're taking the
derivative with respect to x.
We could've taken it with respect to y.
x squared plus y squared is equal to, have to take the
derivative of both sides, right?
Derivative of 100 with respect to x, right?
So this equals the derivative with respect to x, let
me put it in brackets.
I was going to do parentheses, but I think brackets.
Of x squared, plus the derivative with respect
to x of y squared.
And what's the derivative with respect to x of 100?
For every change in x, how much does this change?
Well, it doesn't change at all.
It's going to stay 100, right?
So this is 0.
The derivative of a constant, really with respect to any
changing variable, is going to be 0.
So that was just 0.
And what's the derivative of x squared with respect to x?
Well, we did that up here, right?
When we took this derivative, we just took the derivative
of each of these terms, the derivative of x squared with
respect to x, that's just 2x.
So we get 2x plus-- now this term here is where you might
get a little bit confused.
I am taking the derivative of y squared with respect to x.
So let me take you [? on the side. ?]
You know you shouldn't do this every time that you do one of
these implicit differentiation problems, but I'm doing this
because I want you to understand what we're doing.
So let's say I just defined g as equal to y squared, right?
So let's say this was g.
And I want to take the derivative of g
with respect to x.
So I apply the derivative operator on both sides of
this, so let me do that.
So the derivative, let me do a d of x on that side,
d of x on that side.
That's a d, there you go.
So what we get is the derivative of g with respect to
x is equal to the derivative with respect to x of y squared.
And so far, we haven't really done much.
I've just rewrote things.
But what does the chain rule tell us?
The chain rule tells us that the derivative, let me
stay in the same color.
I'm always switching colors.
The derivative of any variable with respect to x, or with
respect to another variable, is equal to the derivative of g
with respect to y times the derivative of y
with respect to x.
And if you don't have the full intuition of the chain rule
yet, there's a couple of ways to do it.
You know, all of these dg's and dy's that you see throughout
integral and differential calculus, all these really mean
are really, really, really, really, small change in g, a
really, really, really, small change in y, a really,
really, small change in x.
So you can almost view these, if you multiply this out, this
really, really small change in y is the same as this really,
really small change in y.
So these are essentially fractions.
So you could cancel these out, and that's why you would get
the change in g with respect to x, or the derivative of
g with respect to x.
So this is what the chain rule tells us.
So given that this is what the chain rule tells us, how does
that help us figuring out this?
Well, we know that the derivative of g with respect to
x is equal to the derivative of g with respect to y.
But we already know that g is equal to y squared.
I just made that definition.
That's where we started from.
So what's the derivative of g with respect to y?
Well, that's just 2y, right, just using our derivative
rules, or our polynomial rules, which hopefully you believe
actually work, because there are several videos
where we prove it.
And then, times the derivative of y with respect to x.
Which we don't know, and that's actually kind of what we're
going to try to solve for.
So the derivative of g with respect to x is 2y times
the derivative of y with respect to x.
Well, g is just the same thing is y squared, right?
So the derivative of, this is g.
And normally when you do this, you don't have to go through
the whole problem of substituting for g.
I just wanted to give you that intuition.
So what's this derivative right here?
Well, it's going to be the derivative of this expression
with respect to y, from the chain rule, so that's 2y.
Times the derivative of y with respect to x is equal to 0.
When you do it mechanically, it's actually quite simple.
You just say, oh, well, let me pretend like y is like an x,
take a derivative, and I multiply it times
the derivative of y with respect to x.
Mechanically it's very simple, but I really want you to
understand that intuition that this is coming out
of the chain rule.
And actually, the chain rule itself is not
this bizarre concept.
It's really, you're just multiplying fractions.
So let me clear up some space here, and we can
finish this problem.
And then I'll do a problem where you have to do
implicit differentiation.
And even in this one that I just gave, it's really useful
to do implicit differentiation.
So let me, I could clear this guy out up here, clear that
out, clear all of this stuff out, there we go.
Clear all of that out.
Ready to do some more work.
So now we can solve this for the derivative of
y with respect to x.
So we get, let's subtract 2x from both sides.
We get 2y dy dx is equal to-- I'm just subtracting 2x
from both sides-- minus 2x.
And then what do we get?
We get dy over dx is equal to, divide both sides by 2y, the
2's cancel out, you get minus x over y.
And this is interesting, just as it is, right?
Because what was this?
This was the equation for a circle with radius 10, right?
You know, r squared.
And if you know your trigonometry and all of that,
this tells you that the slope of a circle, at any point, is
equal to the negative of x over the y.
Well, we know from trigonometry that the x is the cosine, and
the y is the sine, and so that tells you something interesting
about how the derivative is related to the tangent
and all of that.
But I won't go into that.
We're just learning implicit differentiation here.
But it's sometimes it's fun to just think about things.
So how's this different from everything else
we learned so far?
Well, here we have the derivative, the derivative is
actually a function of both the x and the y value.
And if we feel really uncomfortable having this y
here, if we don't like that, what we could do, is
we can substitute.
So let's solve for y here.
We get y squared is equal to 100, minus x squared, and then
you get why is equal to the plus or minus square root
of 100 minus x squared.
So you just substitute that in here.
Another way of writing the derivative of y with
respect to x, sometimes you'll see y prime.
I'm just switching notation, just so you're
familiar with it.
Is equal to minus x over the plus or minus square root
of 100 minus x squared.
And to some degree, since we have plus or minus down here,
this minus sign kind of doesn't matter, right?
Because when it's a minus here it becomes a plus, when it's
plus it's minuses, so you can kind of ignore this.
That's something interesting for you to think about as well.
Anyway.
And this is interesting, because our traditional way of
solving derivatives, we had never seen a plus
or minus here.
So that might be a little bit interesting to you.
So now let's do a problem that is actually a bit
of a trick problem.
It's not a trick problem but it's actually something, that I
think, I first saw it either when I was practicing for the
AP Calculus exam, or actually saw it on the AP Calculus exam.
So it's something that you might encounter.
But it's really neat.
Let's say I have y is equal to x to the x.
So first, you say, oh, this looks easy.
Maybe I could use the power rule, but no.
Chain rule, you have 2 x's here, and I mean, you might
want to just pause and think about this for a little bit,
because this actually isn't that straightforward
of a problem.
And even though right here, y is explicitly defined in terms
of x, right, if I give you x, if I say x is 3, well then y
is 3 to the third, y is 27.
So this is an explicit function.
I explicitly defined y in terms of x.
But it's very hard to differentiate explicitly.
And here's the trick.
Let's take the natural log, or actually you could take any
logarithm, but let's take the natural log on both
sides of this equation.
So you get the natural log of y is equal to the natural
log of x to the x.
And why did I do that?
Because now I can use what we know about
logarithm properties.
We know that log of a to the b is equal to b log of a.
And you can watch the exponent properties videos,
if you forgot that.
So let's rewrite this using this property.
And I'm going to switch colors.
So we get the natural log of y is equal to x times
the natural log of x.
And now we can implicitly differentiate this.
So let's take the derivative with respect to x of both
sides of this equation.
And I'll switch colors again.
d dx.
And I'm writing this just so you know what I'm doing.
is equal to d dx of x natural log of x.
So the derivative of this expression with respect to x is
equal to the derivative of this expression with respect to y,
times the derivative of y with respect to x.
So what's the derivative of this with respect to y?
What's 1/y?
1/y times the derivative, just the chain rule, the derivative
of y with respect to x.
dy dx is equal to-- and remember, just mechanically, if
you want to make sure you're doing it right, you just take
the derivative with respect to y, and you multiply
it times dy dy.
That's just the chain room.
Equals, and now we take the derivative of this.
Well, now we can just use the product rule.
The derivative of the first, right?
It's this times this.
So what's the derivative of x?
It's 1.
So it's 1 times the natural log of x plus the derivative of
the second term, 1/x, times the first term, times x.
Let me clear that out up there.
I'm already over time, but I've become a YouTube partner,
so my videos can be longer.
But I don't want to make them too long, because
you'll get bored.
All right.
So let's simplify this.
We get 1/y dy dx is equal to the natural log of x, plus,
well, that just cancels out and becomes 1, right?
And so we have the derivative of y with respect
to x is equal to y.
I'm just multiplying both sides of this equation times y.
y times the natural log of x plus 1.
Now, you might say you're done, but some people just don't like
it, the change of y with respect to x, they don't like
it being defined in terms of y.
So let's substitute.
We know that y is equal x to the x.
So that equals x to the x times the natural log of x plus one.
So another way of saying it is that y prime, or the change in
y with respect to x, or the slope at any point of this
curve, is equal to x to the x, natural log of x,
plus x to the x.
I just distributed the x to the x.
But that's pretty neat.
This is kind of a, you know, a five-star problem that you
might see in your books, and you might see it in a math
competition, and people will be really impressed.
And they'd be even more impressed if you kind of
recognized this pattern, and you're able to find the
antiderivative of it, and recognize that it
is x to the x.
Anyway, I've talked for too long.
I will see you in the next video.