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# Implicit differentiation example

Video transcript

Let's get some more practice
doing implicit differentiation. So let's find the derivative
of y with respect to x. We're going to assume
that y is a function of x. So let's apply our
derivative operator to both sides of this equation. So let's apply our
derivative operator. And so first, on
the left hand side, we essentially are just going
to apply the chain rule. First we have the
derivative with respect to x of x minus y squared. So the chain rule
tells us this is going to be the derivative
of the something squared with respect to the
something, which is just going to be 2 times x
minus y to the first power. I won't write the
1 right over there. Times the derivative of the
something with respect to x. Well, the derivative of x
with respect to x is just 1, and the derivative
of y with respect to x, that's what
we're trying to solve. So it's going to
be 1 minus dy dx. Let me make it a
little bit clearer what I just did right over here. This right over here
is the derivative of x minus y squared with
respect to x minus y. And then this right over
here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right
hand side of this equation. This is going to be equal to
the derivative of x with respect to x is 1. The derivative of y
with respect to x. We're just going to write
that as the derivative of y with respect to x. And then finally, the
derivative with respect to x of a constant, that's
just going to be equal to 0. Now let's see if we can
solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I
can rewrite as 2x minus 2y. So let me do that so
I can save some space. This is 2x minus 2y If
I just distribute the 2. And now I can distribute
the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is
just going to be 2x minus 2y. And then 2x minus 2y
times negative dy dx, that's just going to be
negative 2x minus 2y. Or we could write that as
2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy
dx's in orange now. 1 plus dy dx. So now there's a
couple of things that we could attempt to do. We could subtract 2x
minus 2y from both sides. So let's do that. So let's subtract 2x
minus 2y from both sides. So over here, we're going
to subtract 2x minus 2y from that side. And then we could also subtract
a dy dx from both sides, so that all of our dy dx's
are on the left hand side, and all of our non dy dx's
are on the right hand side. So let's do that. So we're going to subtract a
dy dx on the right and a dy dx here on the left. And so what are we left with? Well, on the left hand
side, these cancel out. And we're left with 2y minus
2x dy dx minus 1 dy dx, or just minus a dy dx. Let me make it clear. We could write this
as a minus 1 dy dx. So this is we can
essentially just add these two coefficients. So this simplifies to 2y minus
2x minus 1 times the derivative of y with respect
to x, which is going to be equal to-- on this
side, this cancels out. We are left with 1
minus 2x plus 2y. So let me write it that way. Or we could write
this as-- so negative, negative 2y is
just a positive 2y. And then we have minus 2x. And then we add that 1, plus 1. And now to solve
for dy dx, we just have to divide both sides
by 2y minus 2x minus 1. And we are left
with-- we deserve a little bit of a drum
roll at this point. As you can see, the
hardest part was really the algebra to solve for dy dx. We get the derivative
of y with respect to x is equal to
2y minus 2x plus 1 over 2y minus 2x minus 1.