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>> This is the second video on derivatives, which is a continuation
of what I was doing on the first video.
I was graphing the curve of a parabola and narrowing in on a point on the parabola
and trying to find the tangent line and the slope of a tangent line at that point.
So make sure you watch the first video so that the beginning of this video,
where I narrow in on that graph makes sense.
Then I begin part two, which shows a more general curve and move on from there.
So let's narrow in on the picture here.
So we have this graph, Y = F of X = X squared - 3.
And we're trying to find the tangent line to this point.
So the goal is to find the tangent line at point 1, -2.
Okay, so we're trying to find that tangent line.
And so we drew the secant line through that point, in the ordered pair 0, -3,
and that secant line had a slope of 1.
And we could see it's going to be much steeper than that,
so then we put another point that's a little bit closer to 1, -2 on the curve, at one-half,
-2 and three-fourths, and we found out that that had a slope of one and a half.
So imagine if we just got closer and closer and closer.
We tried to point until we got super close to here.
Or even we can come up from here.
We might have checked out a point up here.
And we get closer and closer, and we try doing secant points --
I'm sorry, we try doing secant lines through these points, we're gonna get pretty close
to getting the correct slope, but maybe not exactly right on.
So that's what we're gonna focus on the next video is how can -- we can use limits --
the idea of the limits and the idea of the formula for slope to find the exact slope
of the tangent line at that point.
So go on to part two of derivatives.
This is part two of derivatives.
In the first part, we were trying to find a tangent line to the curve Y = X squared.
And we use the idea of secant lines, where we could take two points on the graph,
and if we drew a line between those two points, it would get close
to the slope of the tangent line.
So let's just do this in general.
Let's say we have some sort of curve here, like this is Y = F
of X. I'm not exactly sure what curve it is, just part of some curve.
And we've got some point, C, F of C. So in our previous example,
that might have been the ordered pair 1, -2,
if we were doing the graph of F of X = X squared - 3.
I'm trying to be a little bit more general here.
So let's say this ordered pair is just C, F of C. Now, let's take a point close to this,
so up here, what point would that be?
Well, let's say this little distance right here, to the right, so you're on the X-axis.
You're going H spaces to the right.
So this ordered pair right there would be C + H, right?
And the height is still the same, so it's F of C, okay?
I hope you're with me.
But if we moved up, since this is C + H, right?
That's the X coordinate.
Then the Y coordinate is F of C + H. So if I wanted to get the slope of the secant line here,
which is not the tangent line, but we're gonna say it's not too far off if this is very close,
then, I'm gonna use the formula for slope between these two points.
I'm gonna call this point one and point two, a point that's near it, very close to it.
So we're gonna pretend like H is a very small number.
So the slope of line L here, which is the secant line, would be --
all right, we're gonna take one of the ordered pairs and do the Y over the X,
so F of C + H over its X value, which is C + H. And then the other ordered pair,
we're gonna do the Y value over the X value, so that's F of C over C. So this is just Y2 -
Y1 over X2 - X1, just the formula for slope.
And if we simplify that just a little bit more, in the denominator, C - C,
you're gonna end up with just an H in the denominator.
So we have F of C + H - F of C. So that's the slope
of the tangent line -- I'm sorry, of the secant line.
All right, now, let's say we got a point a little bit closer.
Imagine I could pick a point closer to this original point 1 over here.
I can get closer and closer.
See, I'm picking some points closer, so of course,
my little H is getting smaller and smaller and smaller.
Right, I would only go over a little bit.
I'd have a different value for H in this case, and imagine you're getting as close as possible.
We can use the idea of limits to say what if you get so close that H is approaching 0,
so it's practically on top of that point?
You say that if H is approaching 0, that will be the exact slope of the tangent line.
And that's by definition.
All right, so that's what we're finally getting to.
Up until this point, I've been appealing to your intuition
to get a feel for the following concepts.
One, the tangent line to a point of a smooth curve, and two,
the slope of the line tangent to a curve at a given point.
I've been somewhat imprecise in my language,
but now we come to a formal definition for the slope of a tangent line.
Here's the definition, and it's quite a mouthful.
Definition of a tangent line with slope M. If a function,
F is defined on an open interval containing C --
this means it's a curve without any breaks basically.
C's just the X coordinate of the point on the curve.
And if the limit -- here's the limit -- the limit as H approaches 0 --
that's like the really tiny number -- of F of C + H - F of C over H equals M, if it exists --
the reason I have to say it exists, we have to make sure
that you're not gonna get something undefined -- then the line passing through the point C,
F of C, with slope M, is the tangent line to the graph of F at the point C, F of C. Let's go back
up to our original problem, F at Y = F of X = X squared - 3.
[Inaudible] what were trying to find.
We are trying to find the tangent line, right, at the point 1, -2,
so we needed to find the slope of the tangent line at 1, -2.
So at this point, our C is equal to 1.
So let's use this formula we just got.
So at this point, our point is 1, -2.
Remember, that is your C and your F of C, right?
So we're gonna figure this out now.
The limit as H approaches 0 of F of --
now remember, C is 1 and F of C is -2 from this ordered pair 1, -2.
So we've got F of 1 + H - F of 1 over H, right?
That just comes from the slope.
And what's F of 1 + H?
Well, we're gonna have to plug in 1+ H in for X into the function here, right?
So you've got to put in 1 + H for X.
So we've got 1 + H squared -- -3 - -- now, what's our F of 1?
Well, F of 1 is -2, right?
That's your 1, F of 1 is -2.
So in our particular instance, C is 1, so F of 1 was -2.
So this is - a -2 all over H. And now, you have to do same algebra.
See what we get here.
So we've got 1 + 2H + H squared when we square the 1 + H squared -
3 + 2 all over H. And what do we have here?
1 - 3 + 2, we're gonna end up with a numerator 2H + H squared, which you can factor an H out
or divide the top, each term, by H to get the limit as H approaches 0.
You're gonna get 2 + H after you factor out an H and cancel.
So you're gonna get 2 + H. So what does that give you when you plug in?
Can you directly substitute 0 for H?
Yes. That's gonna give you 2 + 0, which is 2.
So you know what?
The slope of that tangent line is 2.
Let's kind of look back up there one more time.
All right, I want you to go onto the third video, and we will finish discussing this.