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 
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 specifically at rates of change over 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 fiveday period. Suppose you find that the value of one US Dollar can be well approximated by the function

The rupiah is the Indonesian currency
where t is time in days. (t = 0 represents the value of the Dollar at noon on Monday.)
Q What was the value of the Dollar at noon on ?
Q According to the graph, over which oneday period was the value of the Dollar most rapidly?
Starting Monday at noon  Starting Tuesday at noon  
Starting Wednesday at noon  Starting Thursday at noon  
Now that we have warmed up, let us begin by considering the average rate of change of the Dollar's value over various intervals. (If you are unsure about average rates of change, you should first go through the preceding tutorial.)
Q Recall that the value of the Dollar was given by
 Follow the method used in the previous tutorial.
 Follow one of the methods shown in the textbook (Section 3.5 in Applied Calculus or Section 10.5 in Finite Mathematics and Applied Calculus): Using a graphing calculator, set Y_{1} = . Then for, say, h = 0.01, you can compute the average rate of change using the formula
 (Y_{1}(1.01)Y_{1}(1))/0.01
 Use the following little utility that computes the average rate of change of any function of x or t over any interval. Enter the technology formula for R(t) in the formula box below, and the values for the endpoints a = 1, and b = 1 + h using the various values of h.
Now complete the following table.
We call this quantity the instantaneous rate of change of R(t) at t = 1 as opposed to the average rate of change. You can think of the instantaneous rate of change at t = 1 as the average rate of change over an extremely tiny interval [1, 1+h]. (We can make this statement more precise by using the language of limits see below). 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 obtained by taking the limit of the average rates of change of f over the intervals [a, a+h], as h approaches 0. We write:
f'(a) = \lim_{h\to 0}\frac{f(a+h)  f(a)}{h}
Units: The units of f' are units of f per unit of x. Quick Example If then the calculation you did above suggests (correctly) that

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 average rate of change of C over the interval [100, 100+h] for h = 1, 0.1, 0.01, 0.001, and 0.0001. Then 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 endpoints a = 100, and b = 100 + h using the various values of h.
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 approximation of 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
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 (also an average rate over a small interval) often gives a more accurate result, and is the one used in many calculators (the nDeriv function of the TI83 does this; by default it uses h = 0.001, but this may be changed via an optional argument).
Example Let , and use the balanced approximation with h = 0.0001 to estimate The answer must be accurate to at least 4 decimal places!

The following is similar to Example 3 in Section 3.5 of Applied Calculus.
If I throw a ball upward from the top of a tower of height ft at a speed of ft/s, its height t seconds later will be

feet
How fast will the ball be rising or falling exactly seconds after I throw it (t = )?
Cheaper that 'Help'  This will cost you! 
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 "next" button on the sidebar to go on to the next topic.