[MUSIC] Here I've graphed some function. Notice what happens if I zoom in on one
little piece of the graph, right?
If I just focus on one part of the graph, well, that little bit of graph sort of
looks like part of a straight line. Another way to think about this is as a
limit of secant lines. By secant line, I mean, I'm going to pick
two points on the graph and I'm going to put a line, the secant line, through
those two points. But, I can do better.
By taking this point and moving it closer to a, this red line is going to be a
better approximation to the orange curve near the point a.
So, instead of putting it through those two points, let me put the secant line
through, say, these two points, this point at a and this point that's nearby.
And now, the line segment through those two points is a much better approximation
of the orange curve. And I can do better and better by taking
a limit, by putting those two points closer and closer together, I can get my
secant line to be a better and better approximation to the orange curve near
the point a. What we're really calculating here is a
limit. So here, I've got an input a and an input
a+h, and here are the corresponding points on the graph of the function.
I'm going to put a secant line through those points.
What I want to know is what's the slope of that secant line because I'm going to
take the limit as this a+h point is moved closer to a as h goes to zero in other
words, and that'll make this secant line move closer and closer to the tangent
line, to the curve.
Okay, so what's the slope of the secant line?
Well, the rise is f(a)+h-f(a). The run is h.
So, the slope of that secant line is this, f(a+h)-f(a)/h.
If I take the limit as h goes to 0, I get this: the limit as h goes to 0 of this
slope. We've seen this, this is the derivative
of the function at the point a. Let's find the equation of the tangent
line in a concrete example. Here's a graph of y=x^2.
Let's figure out the equation of the tangent line to that graph at the point
(3,9). So to do that, we're going to use the
derivative and I know that the derivative of x^2 is 2x.
So, the slope of the tangent line at the point 3 is 2*3=6,
and that's the slope of the tangent line. I also know a point that the tangent line
passes through. It should be passing through the point
(3,9), right? The point (3,9) is on the tangent line.
So, if I know the slope of the tangent line and I know a point on the line, I
can write down an equation for the line using point slope form.
So, y minus the y coordinate on the line is the slope time x minus the x
coordinate on the line. So, this is an equation for this red
tangent line.
[MUSIC]