[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]