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# Implicit differentiation (advanced example)

Sal finds dy/dx for e^(xy²)=x-y using implicit differentiation. Created by Sal Khan.

Video transcript

So let's take another
implicit derivative of the somewhat
crazy relationship. And I've graphed the
relationship here. As you can see, it is
actually quite bizarre. e to the x times y squared
is equal to x minus y. This is some at least in the
range that's depicted here, the x's and the y's that
satisfy this relationship. Well let's take the
derivative of both sides. So we'll apply our derivative
operator to both sides. And actually this
is a good chance to maybe explore some
different notation. We tend to be using
that notation, but oftentimes you'll see
a derivative operator that just looks like a big
capital D. So maybe we'll do that right over here. So let me make it clear. This is the equivalent
of saying d over dx. I'll just use this
big capital D operator so that you are familiar
with the notation. And instead of using dy
dx for the derivative of y with respect to x, I'm just
going to write that as y prime. So we get a little
bit of practice with different notation. So let's take the derivative
of this thing right over here. Well we're going to
apply the chain rule. Actually, we're going
to apply the chain rule multiple times here. The derivative of e to
the something with respect to that something
is going to be e to the something
times the derivative of that something
with respect to x. So times the derivative
of xy squared. So that's our left-hand side. We aren't done taking
the derivative yet. And on our right-hand side,
the derivative of x is just 1. And the derivative
with respect to x of y is just going to be minus-- or
I could write-- negative dy dx. But instead of writing dy dx,
I'm going to write y prime. As you can tell, I like this
notation and this notation more because it
makes it explicit that I'm taking the
derivative with respect to x. Here, we just have
to assume that we're taking the derivative
with respect to x. Here, we have to assume
that's the derivative of y with respect to x. But anyway let's stick with
this notation right over here. Actually, let me make
all of my y primes, all my derivatives of
y with respect to x, let me make them pink
so I keep track of them. So once again,
this is going to be equal to e to the xy squared
times the derivative of this. Well the derivative
of this, we can just use the product and actually
a little bit of the chain rule here. So the derivative of x is just
1 times the second function. So it's going to
be times y squared. And then to that,
we're going to add the product of the
first function which is this x times the derivative
of y squared with respect to x. Well that's going to be
the derivative of y squared with respect to y, which
is just going to be 2y times the derivative
of y with respect to x, which we are now
writing as y prime. And then that's going to be
equal to 1 minus y prime. And like we've
been doing, we now have to just solve for y prime. So let's distribute
this exponential, this e to the xy squared. And we get e, or maybe I
should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative
of y with respect to x, is equal to 1 minus the
derivative of y with respect to x. Now let's get all of our
y primes on one side. So let's add y
prime to both sides. So let's add-- and
just to be clear, I'm adding one y
prime to both sides. So let's add a y
prime to both sides. And let's subtract this
business from both sides. So let's subtract y
squared e to the xy squared subtracting from both sides. So we're going to subtract y
squared e to the xy squared. And we are left
with 2xye to the xy squared plus 1 times y prime. We had this many
y primes and then we add another 1y prime so we
have this many plus 1 y primes. That's going to be equal
to-- well, I purposely added y prime to both sides
and so we are left with 1 minus-- this is kind of a crazy
expression-- y squared times e to the xy squared. And now we just have to
divide both sides by this. And we're left with the
derivative of y with respect to x is equal to this, which
I will just copy and paste. Actually, let me
just rewrite it. Scroll down a little bit. It's equal to 1
minus y squared e to the xy squared
over this business. Let me get some more space. 2xye to the xy squared plus 1. And we're done. It was kind of crazy, but
fundamentally no different than what we've been doing
in the last few examples.