Let variable y be a function of variable x; we write y = f(x). The rate of change of y relative to x, when x changes from x_{1} to x_{2}, is r(x_{1}, x_{2}) = (f(x_{2})  f(x_{1}))/(x_{2 }  x_{1}).
Task.
Record gas prices at one service station for several consecutive weeks and create a table similar to the one given below.
Variables:
x fraction of the month (July 2011);
y price of gas in dollars per gallon at one gas station;
r rate of change measured in dollars per month, where one month equals four weeks.
Formulas.
For two consecutive weeks, the difference y_{2}  y_{1} represents the change during 1/4 of a month. Thus the rate of change in cents per month is r = (y_{2}  y_{1})/(1/4) = 4*(y_{2}  y_{1}).
x = 
y= 
y_{2}  y_{1} 
r= 
0 
3.779 




.05 
0.20 dollars/gallon/month 
1/4 
3.729 




.12 
0.48 dollars/gallon/month 
1/2 
3.609 




.01 
0.04 dollars/gallon/month 
3/4 
3.599 


***********************************************************************
Computation of rates of change with the TI83/84
For a given function f(x), and a list of values x_{1}, x_{2}, ... , x_{k}, compute the rates of change of y = f(x), and return them as a list r_{1}, r_{2}, ... , r_{(k1)}. Note that the list of rates has one less item than the list of values of x.
Define,
\Y1= f(X) (choose some function)
Enter
{x_{1},x_{2}, ...,x_{k}}→L1
ΔList(Y1(L1))/∆List(L1)→L2
Now L2 contains the required rates.
Experiment with variables x and y of your choice.
Let's look at two more practical examples.
To compute how many miles per gallon your car gets, you record the number of miles given on your odometer two times after filling the tank. These are the values y_{1} and y_{2}. The number of gallons you pumped the second time is your x_{2}  x_{1}. The quotient (y_{2}  y_{1})/(x_{2}  x_{1}) shows how many miles per gallon your car yielded during that period.
From the map of New Mexico you read that the distance from Las Cruces to Albuquerque is 223 miles, and the expected driving time is 3.5 hours. The ratio r = 223 miles/3.5 hours ≈ 63.7 mph indicates the average speed needed to make the trip in the quoted time.
In general when we have variables x and y =f(x), then in order to compute the rate of change of y relative to x, we need four numbers, x_{1}, y_{1} = f(x_{1}), x_{2}, and f(x_{2}) = y_{2}. The rate of change
(y_{2}  y_{1})/(x_{2}  x_{1}) is usually called the average rate.
Instantaneous rate of change.
When you are driving a car, your speedometer shows the speed at each moment, and this value can change very fast when you are braking or accelerating. It is not the average speed, which is the average rate of change, but the speed at a given moment, which is the instantaneous rate of change. Also a policeman uses a radar gun to measure the speed of a car at the moment that the measurement is made, and not its average speed over a prolonged period of time.
The mathematical concept of a derivative captures the intuitive concept of the instantaneous rate of change. Its introduction marked the creation of calculus and started the mathematical foundations of modern science. It was first used to describe instant velocity (velocity at a given moment) and instant acceleration, which is the rate velocity changes in time. One of the basic equations of physics discovered by Newton states that acceleration of an object is always proportional to the sum of all forces acting on the object.
Remark.
When we talk about velocity of an object, we take into consideration not only its speed but also the direction it is moving in space.
The mathematical concept of a derivative.
As before we have two variables, x and y = f(x). Here x is an "independent" variable, and the function f shows the manner in which y depends on x. But now we have only one value of x,
x = x_{0}, and the corresponding value y_{0} = f(x_{0}). We need to define the rate of change of y relative to x at x_{0}.
We introduce a new independent variable d, and consider x_{0} + d, and f(x_{0} +d). Now for any value of d we can look at the quotient (f(x_{0}+d)  f(x))/(x_{0}+d  x_{0}) = (f(x_{0}+d)  f(x_{0}))/d.
What happens when we choose different values of d closer and closer to 0? Both the denominator and the numerator of the quotient may get closer and closer to 0, but the quotient itself may "converge" to some number r_{0}. If this is the case, this number r_{0} is the rate of change of y = f(x) relative to x, at the value x = x_{0}. In mathematical jargon,
r_{0} is the value of the derivative of y = f(x)
relative to x, at x_{0}.
Task.
Choose some rather simple function f(x) and a value x_{0}. Compute the values of
(f(x_{0}+d)  f(x_{0}))/d, for d = 1/100, 1/200, 1/300, 1/400, ... , and observe what happens.
Example.
f(x) = x^{3}, x_{0} = 5.
Enter:
\Y1=X^{3}
From the home screen enter
0→N ENTER
N+100→N:1/N→D:(Y1
(5+D)Y1(5))/D repeat ENTER
Remarks.
The values of D can be negative.
When the values of D become very close to 0, for example, around 10^{8}, then the loss of precision for the TI83/84 may be so big that the results will be unreliable or simply wrong.
The derivatives exist only for some functions, and even then, not always for all values x_{0}.
There are other ways of computing a derivative, besides relying directly on its definition, that are much better.