Page 17 - Calc for the Clueless

Given points P 1—coordinates (x 1,y 1)—and P 2—coordinates (x 2,Y 2). Draw the line segment through P 1 parallel to
the x axis and the line segment through P 2 parallel to the y axis, meeting at point Q. Since everything on P 1Q has
same y value, the y coordinate of Q is y 1.

Everything on P 2Q has the same x value. The x value of Q is x 2. The coordinates of Q are (x 2,y 1).

Since everything on P 1Q has the same y value, the length of P 1Q = x 2 - x 1. Since everything on P 2Q has the same
x value, the length of P 2Q = y 2 - y 1. The slope

∆ = delta, another Greek letter.

Let's do the same thing for a general function y = f(x).

Let point P 1 be the point (x,y) = (x,f(x)). A little bit away from x is x + ∆x. (We drew it a lot away; other wise
you could not see it.) The corresponding y value is f(x + ∆x). So P 2 = (x + ∆x,f(x + ∆x)). As before, draw a line
through P 1 parallel to the x axis and a line through P 2 parallel to the y axis. The lines again meet at Q. As before,
Q has the same x value as P 2 and the same y value as P 1. Its coordinates are (x + ∆x,f(x)). Since all y values on
P 1Q are the same, the length of P 1Q = (x + ∆x) - x = ∆x. All x values on P 2Q are the same. The length of P 2Q =
f(x + ∆x) - f(x). The slope of the secant line

Now we do as before—let P 2 go to P 1. Algebraically this means to take the limit as ∆x goes to 0. We get the
slope of the tangent line L 2 at P 1. Our notation will be...

The slope of the tangent line

if it exists.