3.5 The Derivative: Numerical and Graphical Viewpoints

This tutorial: Part A: Numerical Approach
Next tutorial: Part B: Graphical Viewpoint
Following tutorial: Part C: The Derivative Function

(This topic is also in Section 3.5 in Applied Calculus or Section 10.5 in Finite Mathematics and Applied Calculus))

 
Calculating the Instantaneous Rate of Change of a Function

In this tutorial, we continue with the topic of average rate of change over an interval discussed in the previous tutorial, but this time we look at shorter and shorter intervals.

You are based in Indonesia, and you monitor the value of the US Dollar on the foreign exchange market very closely during a rather active five-day period. Suppose you find that the value of one US Dollar can be well approximated by the function

where t is time in days. (t = 0 represents the value of the Dollar at noon on Monday.)

Let us begin by considering the average rate of change of the Dollar's value over various intervals. (To learn more about average rates of change, go to the preceding tutorial.)

Q What was the value of the Dollar at noon on Tuesday?

Q According to the graph, when was the value of the Dollar rising most rapidly?

Q Recall that the value of the Dollar was given by

Now compute the average rate of change of R over the intervals [1, 1+h] for h = 1,   0.1,   0.01,   0.001, and 0.0001 (see the last example in the previous tutorial). Note that there are several ways to do this:

R(t) =
a =     b =
Average Rate of Change:
   

Now complete the following table.

h getting smaller;     interval [1, 1+h] getting smaller →
h10.10.010.0010.0001
Ave. Rate of Change over [1, 1+h]
       

Q As the width h of the interval decreases to 0, the rates of change are getting closer and closer to exactly rupiahs per day.  

In terms of the value of the Dollar, this suggests that, as measured over a very small time interval around noon on Tuesday (t = 1), the Dollar was rising at a rate 300 rupiahs per day.

To put it another way, the Dollar was increasing at an instantaneous rate of 300 rupiahs per day at noon on Tuesday. And that is what much of calculus is concerned with: studying the instantaneous rate of change of a function.

The process of letting h get smaller and smaller is called taking the limit as h approaches 0. See the tutorials on limits to learn more about limits. Taking the limit of the average rates of change gives us the instantaneous rate of change. Here is the notation for this limit.

Instantaneous Rate of Change of f(x) at x = a: The Derivative

The instantaneous rate of change of f(x) at x = a is defined by taking the limit of the average rates of change of f over the intervals [a, a+h], as h approaches 0. We write:

    Instantaneous rate of change= lim
    h→0
    f(a+h) - f(a)

    h
    .
(We read this as "the limit as h approaches 0 of the difference quotient.".) We also call the instantaneous rate of change the derivative of f at x = a, which we write as f'(a) (read "f prime of a"). Thus,

    f'(a)= lim
    h→0
    f(a+h) - f(a)

    h

Units: The units of f' are units of f per unit of x.


Quick Example

If f(x) = 7,500 + 500x - 100x2, then the calculation you did above suggests (correctly) that

    f'(1)= lim
    h→0
    f(1+h) - f(1)

    h
    = 300 rupiahs per day

 

The cost (in dollars) of producing x dumbbell sets per day at the Taft Sports Company is calculated by its marketing staff to be given by the formula

Now make a table showing the values of the averge rate of change of C over the interval [100, 100+h] for h = 1, 0.1, 0.01, 0.001, and 0.0001. Use your table to estimate the instantaneous rate of change of cost that results from an increase in production level from the current level of 100 dumbbell sets.

Here again is that little utility that computes the average rate of change over any interval. Enter the technology formula for C(x) in the formula box below, and the values for the end-points a = 100, and b = 100+h using the various values of h.

C(x) =
a =     b =
Ave. Rate of Change:
   

Q Do we always need to make tables of difference quotients as above in order to calculate an approximate value for the derivative?
A We can usually approximate the value of the derivative by using a single, small, value of s. In the example above, the value h = 0.0001 would have given a pretty good approximation. The problems with using a fixed value of h are that (1) we do not get an exact answer, only an approximationof the derivative, and (2) how good an approximation it is depends on the function we're differentiating. With many of the functions you encounter, it is a good enough approximation.

Calculating a Quick Approximation of the Derivative

We can calculate an approximate value of f'(a) by using the formula

    f'(a)
    f(a+h) - f(a)

    h

with a small value of h (the value h = 0.0001 often works for this).

Alternative Formula: the "Balanced Difference Quotient"

The following alternative formula often gives a more accurate result, and is the one used in many calculators (the nDeriv function of the TI-83 does this; by default it uses h = 0.001, but this may be changed via an optional argument).

    f'(a)
    f(a+h) - f(a-h)

    2h


Example

Let f(x) = x2 - x-0.4 again, and use the balanced approximation with h = 0.0001 to estimate f'(5). You must round the answer to at least 4 decimal places!

    f'(5)

 

The following is similar to Example 3 in Section 3.5 of Applied Calculus.

If I throw a ball upward at a speed of 80 ft/s, its height t seconds later will be

How fast will the ball be rising exactly 2 seconds after I throw it (t = 2)?

You could now try some of the exercises in Section 3.5 in Applied Calculus or Section 10.5 in Finite Mathematics and Applied Calculus), but you will need the material in the next tutorial to answer the questions about graphs.

Alternatively, press "game version" on the sidebar to go to the game version of this tutorial (it has different examples to try and is a lot of fun!).

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Last Updated: September, 2007
Copyright © 2007 Stefan Waner