What is the difference between average and instantaneous rate of change?

The key difference is whether you measure across an interval or at one input. An average rate of change compares the function’s outputs at two inputs and divides the output change by the input change. It is the slope of a secant line between two points on the graph. An instantaneous rate of change is the rate at a particular input; it is the derivative there, defined as the limit of the average-rate calculation as the interval shrinks toward zero. [S1]

In everyday terms, an average describes a stretch of the graph, while an instantaneous rate describes the rate at one input. You do not calculate it by choosing an interval of zero width: the limit of the average-rate quotient gives the point-specific rate.

Use f(x)=x². First find the average rate from x=1 to x=3. The outputs are f(1)=1 and f(3)=9, so the output change is 9−1=8 and the input change is 3−1=2. Divide: 8/2=4. Thus, the average rate on that interval is 4. [S1]

Now find the instantaneous rate at x=1 using the limiting quotient. Let h represent the change from 1 to a nearby input:

((f(1+h)−f(1))/h) = (((1+h)²−1)/h) = (2h+h²)/h = 2+h.

As h approaches zero, 2+h approaches 2. So the instantaneous rate at x=1 is 2. [S1]

The paired results are: average rate from 1 to 3 = 4; instantaneous rate at 1 = 2. They differ because the first summarizes the interval, while the second is the limiting rate at its starting input.

They answer different questions: the average rate summarizes the change between two endpoints, while the instantaneous rate is found by shrinking the interval toward the input of interest. For f(x)=x², the quotient from x=1 to x=1+h simplifies to 2+h. When h approaches zero, that quotient approaches 2, the instantaneous rate at x=1. [S1]

For comparison, from x=1 to x=1.1 the average rate is ((1.1)²−1²)/(1.1−1)=0.21/0.1=2.1. This is closer to 2 than the average rate over the wider interval from 1 to 3, which is 4. The smaller-interval calculation illustrates how the quotient approaches the point rate as its endpoint moves toward 1.

Name the quantity and show which input or interval you used. For an average rate, write the interval endpoints, substitute the function values, and divide the output change by the input change. For an instantaneous rate, state the input and evaluate the derivative there, or show the limit of the average-rate quotient if that is what the problem asks for.

For this example, a clear final response is: “The average rate of change of f(x)=x² from x=1 to x=3 is 4. The instantaneous rate of change at x=1 is 2.” These are different quantities, and this example shows that their values need not match. [S1]