Second derivatives (implicit equations): find expression | AP Calculus AB | Khan Academy

- [Instructor] Let's say
that we're given the equation

that y squared minus x squared

is equal to four.

And our goal is to find
the second derivative of y

with respect to x, and we want to

find an expression for it
in terms of x's and y's.

So pause this video, and see
if you can work through this.

All right, now let's do it together.

Now, some of you might
have wanted to solve for y

and then use some traditional techniques.

But here, we have a y squared,

and so it might involve a
plus or a minus square root.

And so some of y'all
might have realized, hey,

we can do a little bit of
implicit differentiation,

which is really just an
application of the chain rule.

So let's do that.

Let's first find the first derivative

of y with respect to x.

And to do that, I'll
just take the derivative

with respect to x of both
sides of this equation.

And then what do we get?

Well, the derivative with
respect to x of y squared,

we're gonna use the chain rule here.

First, we can take the derivative

of y squared with respect to y,

which is going to be equal to two y,

and then that times the
derivative of y with respect to x.

Once again, this comes
straight out of the chain rule.

And then, from that, we will subtract,

what's the derivative of x
squared with respect to x?

Well, that's just going to be two x.

And then last, but not least,

what is the derivative of a
constant with respect to x?

Well, it doesn't change,

so it's just going to be equal to zero.

All right, now we can solve for

our first derivative
of y with respect to x.

Let's do that.

We can add two x to both sides,

and we would get two y
times the derivative of y

with respect to x is equal to

two x.

And now I can divide both
sides by two y, and I am going

to get

that the derivative of y

with respect to x is equal to x,

x over y.

Now, the next step is
let's take the derivative

of both sides of this with respect to x,

and then we can hopefully
find our second derivative

of y with respect to x.

And to help us there,
actually let me rewrite this.

And I always forget the quotient rule,

although it might be a useful
thing for you to remember.

But I could rewrite this as a product,

which will help me at least.

So I'm going to rewrite
this as the derivative of y

with respect to x is equal to

x times y to the negative one power,

y to the negative one power.

And now, if we want to
find the second derivative,

we apply the derivative operator

on both sides of this equation,

derivative with respect to x.

And our left-hand side is exactly

what we eventually wanted to get,

so the second derivative
of y with respect to x.

And what do we get here
on the right-hand side?

Well, we can apply the product rule.

So first, we can say the
derivative of x with respect to x,

well, that is just going to
be one times the other thing,

so times y to the negative one power,

y to the negative one power.

And then we have plus

x times the derivative
of y to the negative one.

So plus x,

what's the, times,

what's the derivative of y
to the negative one power?

Well, first, we can find the derivative

of y to the negative one
power with respect to y.

We'll just leverage the power rule there.

So that's going to be negative one

times y to the negative two power.

And then we would multiply that

times the derivative
of y with respect to x,

just an application of the chain rule,

times dy/dx.

And remember, we know what the derivative

of y with respect to x is.

We already solved for that.

It is x over y.

So this over here is going to be x over y.

And so now we just have to
simplify this expression.

This is going to be equal to,

and I'll try to do it part by part,

that part right over there is
just going to be a one over y.

And then all of this business,

let's see if I can simplify that.

This negative is going to
go out front, so minus,

and then I'm going to have
x times x in the numerator.

And then it's going to
be divided by y squared

and then divided by another y.

So it's going to be minus x squared

over y to the third,

over y to the third, or
another way to think about it,

x squared times y to the negative three.

And we are done.

We have just figured out
the second derivative of y

with respect to x

in terms of x's and y's.