The Derivative at a Point


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

[`f'(c) = lim_(h->0) (f(c + h) - f(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`.


  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 `(f(c + h) - f(c))/h` related to the formula for the slope of a line? Fill in the blanks: `(f(c + h) - f(c))/h` is the slope of the line between points ______ and ______.
  4. How are `(f(c + h) - f(c))/h` and `lim_(h->0) (f(c + h) - f(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) (f(1 + h) - f(1))/h` exist?
  7. Confirm by playing with the applet: `f` is continuous at `x = 1`, but `f` is not differentiable at `x = 1`.