The Derivative at a Point

HELP

Problem: Given a function and a specific x-value x = c, compute the slope of the line tangent to at x = c.
We denote this slope by ′(c), and we say ′(c) is the derivative of at x = c.

Solution:
′(c) =
lim
h→0
(c + h) − (c)
h
With the applet above, we'll explore where this formula comes from. Note that in the applet you can move two points on the x-axis -- one with x-value c, and one with x-value c + h.

Explore

  1. Start with c = 2.5 and see how h changes as you move the other point. When is h positive? negative?
  2. Show the tangent line and the secant line, and see how the secant line approximates the tangent line when h is very small. What happens when h = 0? Why?
  3. How is the fraction
    (c + h) − (c)
    h
    related to the formula for the slope of a line? Fill in the blanks:
    (c + h) − (c)
    h
    is the slope of the line between points ______ and ______.
  4. How are
    (c + h) − (c)
    h
    and
    lim
    h→0
    (c + h) − (c)
    h
    different? In other words, what is the effect of sticking that limit out in front of the fraction?
  5. Hide the secant line. Notice that the graph is "pointy" at x = 1. What happens to the tangent line when c = 1?
  6. Show the secant line again, and keep c at 1. Does
    lim
    h→0
    (1 + h) − (1)
    h
    exist?
  7. Confirm by playing with the applet: is continuous at x = 1, but is not differentiable at x = 1.