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## AP®︎/College Calculus AB

### Unit 2: Lesson 2

Defining the derivative of a function and using derivative notation- Formal definition of the derivative as a limit
- Formal and alternate form of the derivative
- Worked example: Derivative as a limit
- Worked example: Derivative from limit expression
- Derivative as a limit
- The derivative of x² at x=3 using the formal definition
- The derivative of x² at any point using the formal definition
- Finding tangent line equations using the formal definition of a limit

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# The derivative of x² at x=3 using the formal definition

AP.CALC:

CHA‑2 (EU)

, CHA‑2.B (LO)

, CHA‑2.B.2 (EK)

, CHA‑2.B.3 (EK)

, CHA‑2.B.4 (EK)

Sal finds the limit expression for the derivative of f(x)=x² at the point x=3 and evaluates it. Created by Sal Khan.

## Video transcript

In the last video we tried to
figure out the slope of a point or the slope of a
curve at a certain point. And the way we did, we said OK,
well let's find the slope between that point and then
another point that's not too far away from that point. And we got the slope
of the secant line. And it looks all fancy, but
this is just the y value of the point that's not too far away,
and this is just the y value point of the point in
question, so this is just your change in y. And then you divide that
by your change in x. So in the example we did,
h was the difference between our 2 x values. This distance was h. And that gave us the
slope of that line. We said hey, what if we take
the limit as this point right here gets closer
and closer to this point. If this point essentially
almost becomes this point, then our slope is going to be the
slope of our tangent line. And we define that as the
derivative of our function. We said that's equal
to f prime of x. So let's if we can apply this
in this video to maybe make things a little bit more
concrete in your head. So let me do one. First I'll do a particular
case where I want to find the slope at exactly some point. So let me draw my axes again. Let's draw some
axes right there. Let's say I have the curve--
this is the curve-- y is equal to x squared. So this is my y-axis, this is
my x-axis, and I want to know the slope at the point
x is equal to 3. When I say the slope you can
imagine a tangent line here. You can imagine a tangent line
that goes just like that, and it would just barely graze
the curve at that point. But what is the slope
of that tangent line? What is the slope of that
tangent line which is the same as the slope of the curve
right at that point. So to do it, I'm actually going
to do this exact technique that we did before, then we'll
generalize it so you don't have to do it every time for
a particular number. So let's take some
other point here. Let's call this 3 plus delta x. I'm changing the notation
because in some books you'll see an h, some books you'll see
a delta x, doesn't hurt to be exposed to both of them. So this is 3 plus delta x. So first of all what is
this point right here? This is a curve y is equal to
x squared, so f of x is 3 squared-- this is the point 9. This is the point
3,9 right here. And what is this
point right here? So if we were go all the way
up here, what is that point? Well here our x is
3 plus delta x. It's the same thing as
this one right here, as x naught plus h. I could have called this
3 plus h just as easily. So it's 3 plus
delta x up there. So what's the y
value going to be? Well whatever x value is, it's
on the curve, it's going to be that squared. So it's going to be the point
3 plus delta x squared. So let's figure out the
slope of this secant line. And let me zoom in a little
bit, because that might help. So if I zoom in on just
this part of the curve, it might look like that. And then I have one point
here, and then I have the other point is up here. That's the secant line. Just like that. This was the point over
here, the point 3,9. And then this point up here is
the point 3 plus delta x, so just some larger number than
3, and then it's going to be that number squared. So it's going to be 3
plus delta x squared. What is that? That's going to be 9. I'm just foiling this out,
or you do the distribute property twice. a plus b squared is a squared
plus 2 a b plus b squared, so it's going to be 9 plus two
times the product of these things. So plus 6 delta x, and then
plus delta x squared. That's the coordinate
of the second line. This looks complicated, but I
just took this x value and I squared it, because it's on the
line y is equal to x squared. So the slope of the secant line is going to be the
change in y divided by the change in x. So the change in y is just
going to be this guy's y value, which is 9 plus 6 delta
x plus delta x squared. That's this guy's y value,
minus this guy's y value. So minus 9. That's your change in y. And you want to divide
that by your change in x. Well what is your change in x? This is actually going to
be pretty convenient. This larger x value-- we
started with this point on the top, so we have to start with
this point on the bottom. So it's going to be
3 plus delta x. And then what's this x value? What is minus 3? That's his x value. So what does this simplify to? The numerator-- this 9
and that 9 cancel out, we get a 9 minus 9. And in the denominator
what happens? This 3 and minus 3 cancel out. So the change in x actually end
up becoming this delta x, which makes sense, because this delta
x is essentially how much more this guy is then that guy. So that should be the
change in x, delta x. So the slope of my secant line has simplified to 6 times
my change in x, plus my change in x squared, all of that
over my change in x. And now we can simplify
this even more. Let's divide the numerator
and the denominator by our change in x. And I'll switch colors just
to ease the monotony. So my slope of my
tangent of my secant line-- the one that goes
through both of these-- is going to be equal if you
divide the numerator and denominator this becomes 6. I'm just dividing numerator
and denominator by delta x plus six plus delta x. So that is the slope
of this secant line So slope is equal
to 6 plus delta x. That's this one right here. That's this reddish line that
I've drawn right there. So this number right here, if
the delta x was one, if these were the points 3 and 4,
then my slope would be 6 plus 1, because I'm picking a point
4 where the delta x here would have to be 1. So the slope would be 7. So we have a general formula
for no matter what my delta x is, I can find the slope
between 3 and 3 plus delta x. Between those two points. Now we wanted to find the
slope at exactly that point right there. So let's see what happens
when delta x get smaller and smaller. This is what delta
x is right now. It's this distance. But if delta x got a little
bit smaller, then the secant line would look like that. Got even smaller, the secant line would look like that,
it gets even smaller. Then we're getting pretty
close to the slope of the tangent line. The tangent line is this
thing right here that I want to find the slope of. Let's find a limit as our
delta x approaches 0. So the limit as delta
x approaches 0 of our slope of the secant line of 6 plus delta
x is equal to what? This is pretty straightforward. You can just set this equal
to 0 and it's equal to 6. So the slope of our tangent
line at the point x is equal to 3 right there is equal to 6. And another way we could write
this if we wrote that f of x is equal to x squared. We now know that the derivative
or the slope of the tangent line of this function at the
point 3-- I just only evaluated it at the point 3 right there--
that that is equal to 6. I haven't yet come up with a
general formula for the slope of this line at any point, and
I'm going to do that in the next video.