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## Implicit differentiation

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# Implicit differentiation

AP Calc: FUN‑3 (EU), FUN‑3.D (LO), FUN‑3.D.1 (EK)

## Video transcript

So we've got the
equation x squared plus y squared is equal to 1. I guess we could call
it a relationship. And if we were to graph
all of the points x and y that satisfied
this relationship, we get a unit circle like this. And what I'm curious
about in this video is how we can figure out the
slope of the tangent line at any point of
this unit circle. And what immediately might be
jumping out in your brain is, well a circle defined this
way, this isn't a function. It's not y explicitly
defined as a function of x. For any x value we actually
have two possible y's that satisfy this
relationship right over here. So you might be tempted
to maybe split this up into two separate
functions of x. You could say y is equal to
the positive square root of 1 minus x squared. And you could say y is equal to
the negative square root of 1 minus x squared. Take the derivatives of
each of these separately. And you would be able to find
the derivative for any x, or the derivative of the
slope of the tangent line at any point. But what I want to
do in this video is literally leverage
the chain rule to take the
derivative implicitly. So that I don't have
to explicitly define y is a function of x either way. And the way we do
that is literally just apply the derivative operator
to both sides of this equation. And then apply what we
know about the chain rule. Because we are not
explicitly defining y as a function of x, and
explicitly getting y is equal to f prime
of x, they call this-- which is really just an
application of the chain rule-- we call it implicit
differentiation. And what I want you to keep
in the back of your mind the entire time
is that it's just an application of
the chain rule. So let's apply the derivative
operator to both sides of this. So it's the derivative
with respect to x of x squared plus y
squared, on the left hand side of our equation. And then that's going to
be equal to the derivative with respect to x on
the right hand side. I'm just doing the
same exact thing to both sides of this equation. Now if I take the derivative
of the sum of two terms, that's the same thing as taking
the sum of the derivative. So this is going to
be the same thing as the derivative with
respect to x of x squared, plus the derivative with
respect to x of y squared. I'm writing all my
orange stuff first. So let's see. This is going to be x squared,
this is going to be y squared. And then this is going to
be equal to the derivative with respect to x of a constant. This isn't changing
with respect to x. So we just get 0. Now this first term
right over here, we have done many, many,
many, many, many, many times. The derivative with
respect to x of x squared is just the power rule here. It's going to be 2 times
x to the first power. We could just say 2x. Now what's interesting is what
we're doing right over here. The derivative with
respect to x of y squared. And the realization here is
to just apply the chain rule. If we're taking the
derivative with respect to x of this
something, we just have to take the derivative--
let me make it clear-- we're just going to take the
derivative of our something. The derivative of y squared--
that's what we're taking, you can kind of view that as
a function-- with respect to y and then multiply that
times the derivative of y with respect to x. We're assuming that y does
change with respect to x. y is not some type of
a constant that we're writing just an abstract terms. So we're taking the
derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking
the derivative of y with respect to x. It might be a little bit
clearer if you kind of thought of it as the derivative
with respect to x of y, as a function of x. This might be, or y is
a function of x squared, which is essentially another way
of writing what we have here. This might be a
little bit clearer in terms of the chain rule. The derivative of
y is a function of x squared with
respect to y of x. So the derivative of
something squared with respect to that something, times the
derivative of that something, with respect to x. This is just the chain rule. I want to say it
over and over again. This is just the chain rule. So let's do that. What do we get on the
right hand side over here? And I'll write it
over here as well. This would be equal to the
derivative of y squared with respect to y, is just
going to be 2 times y. 2 times y, just an
application of the chain rule. And the derivative of
y with respect to x? Well, we don't
know what that is. So we're just going to leave
that as times the derivative of y with respect to x. So let's just write
this down over here. So we have is 2x plus the
derivative of something squared, with respect to
that something, is 2 times the something. In this case, the something
is y, so 2 times y. And then times the derivative
of y with respect to x. And this is all going
to be equal to 0. Now that was interesting. Now we have an equation
that has the derivative of a y with respect to x in it. And this is what we
essentially want to solve for. This is the slope of the
tangent line at any point. So all we have to do
at this point is solve for the derivative of
y with respect to x. Solve this equation. So let's do that. And actually just so we
can do this whole thing on the same page so we can see
where we started, let me copy and paste this up here. This is where we left off. And let's continue there. So let's say let's subtract
2x from both sides. So we're left with 2y
times the derivative of y, with respect to x, is equal
to-- we're subtracting 2x from both sides-- so it's
equal to negative 2x. And if we really want to
solve for the derivative of y with respect to x, we can
just divide both sides by 2y. And we're left with
the derivative of y with respect to x. Let's scroll down a little bit. The derivative of
y with respect to x is equal to, well
the 2s cancel out. We we're left with
negative x over y. So this is interesting. We didn't have to us
explicitly define y as a function of x here. But we got our derivative
in terms of an x and a y. Not just only in terms of an x. But what does this mean? Well, if we wanted
to find, let's say we wanted to find the
derivative at this point right over here. Which, if you're familiar
with the unit circle, so if this was a
45 degree angle, this would be the square
root of 2 over 2 comma the square root of 2 over 2. What is the slope of
the tangent line there? Well, we figured it out. It's going to be
negative x over y. So the slope of the
tangent line here is going to be
equal to negative x. So negative square root
of 2 over 2 over y. Over square root of 2 over 2,
which is equal to negative 1. And that looks just about right.

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