Two Different Questions About a Quantity
A function can answer two related but different questions. First, how much of the quantity is there at a particular input? Second, how quickly is that quantity changing at that input? The function value answers the first question; the derivative value answers the second.
For example, a population model might give the number of people in a town at time \(t\). Evaluating the model at a particular time gives the population then. Differentiating the model and evaluating the derivative at that time gives the population’s instantaneous rate of change. These answers can both be useful, but they are not interchangeable.
What Each Value Represents
The notation \(f(a)\) means substitute \(a\) into the function rule and find the corresponding output. That output describes the quantity itself. If \(P(t)\) gives a population, for instance, \(P(10)\) gives the modeled population at time \(t=10\).
The notation \(f'(a)\) refers to the derivative function evaluated at \(a\). The derivative describes how the function’s output changes as its input changes. In context, \(P'(10)\) describes the population’s instantaneous rate of change at \(t=10\), not the population at that time.
Both expressions use the same input, \(a\), but they answer different questions. Evaluating \(f\) gives an amount; evaluating \(f'\) gives a rate. The result of one calculation cannot be substituted for the other just because both are evaluated at the same input.
Why the Derivative Represents a Rate
The definition of the derivative explains this distinction. For a function \(f\), the derivative at \(a\), when it exists, is the limit of the difference quotient as the input change approaches zero:
Here, \(f(a)\) is the output at the input \(a\). The numerator \(f(a+h)-f(a)\) is the change in output when the input moves from \(a\) to \(a+h\). Dividing by \(h\) gives the change in output per change in input for that interval. Taking the limit as \(h\) approaches zero gives the instantaneous rate at \(a\).
Evaluating \(f(a)\) tells you the output when the input is \(a\).
At \(a+h\), the output is \(f(a+h)\), so the output change is \(f(a+h)-f(a)\).
The derivative \(f'(a)\) is the limit of the output change divided by the input change as \(h\) approaches zero.
This gives a short justification for the interpretation: \(f(a)\) is one output, while \(f'(a)\) comes from how outputs change near \(a\). They are produced by different operations and have different meanings.
Worked Examples in Context
Worked Example: Population at a Particular Time
A model for a town’s population \(P(t)\), measured in people, \(t\) years after a starting point, is \(P(t)=1000+80t-2t^2\). Find the population and its instantaneous rate of change at \(t=10\).
Step 1: Find the function value. Substitute \(t=10\) into the population model:
The model gives a population of \(1600\) people at \(t=10\).
Step 2: Find the derivative. Differentiate the function:
Step 3: Evaluate the derivative at \(t=10\).
Thus, the population’s instantaneous rate of change at \(t=10\) is \(40\) people per year. The two answers describe different things: \(1600\) is the population at that time, and \(40\) is its rate of change then.
Check: Differentiating \(1000+80t-2t^2\) gives \(0+80-4t\), which is \(80-4t\). Substituting \(10\) gives \(40\), as calculated.
Worked Example: Temperature and Its Rate of Change
Suppose a temperature model is \(T(t)=20+6t-t^2\), where \(T(t)\) is measured in degrees Celsius and \(t\) is time in hours. Find the temperature and its instantaneous rate of change at \(t=2\).
Step 1: Evaluate the temperature model.
The modeled temperature at \(t=2\) is \(28\) degrees Celsius.
Step 2: Differentiate.
Step 3: Evaluate the derivative.
The instantaneous rate of change of the temperature at \(t=2\) is \(2\) degrees Celsius per hour. It is not the temperature: the temperature is \(28\) degrees Celsius.
Check: Differentiating \(20+6t-t^2\) term by term gives \(0+6-2t\). At \(t=2\), this derivative is \(6-4=2\), confirming the rate calculation.
Worked Example: Revenue and Marginal Change
A revenue model is \(R(q)=120q-2q^2\), where \(q\) is the number of items sold and \(R(q)\) is revenue in dollars. Find the modeled revenue and the derivative value at \(q=20\).
Step 1: Evaluate the revenue function.
The modeled revenue at \(q=20\) is \(\$1600\).
Step 2: Differentiate the revenue function.
Step 3: Evaluate the derivative at \(q=20\).
The derivative value is \(40\) dollars per item at \(q=20\). It describes a rate of change of revenue with respect to the number of items, whereas \(R(20)=1600\) is the revenue itself.
Check: Differentiating \(120q-2q^2\) gives \(120-4q\). Substituting \(q=20\) gives \(120-80=40\), confirming the derivative value.
A Common Mix-Up
A frequent error is to calculate the derivative correctly and then report it as the quantity’s value. For the population model above, someone might write:
The derivative calculation is correct, but the conclusion is not. \(P'(10)=40\) is the rate of change at \(t=10\); the population is found by evaluating the original function:
To correct this kind of error, first identify the question being asked. If it asks for the amount or state of the quantity at an input, evaluate the function. If it asks how quickly the quantity is changing at that input, evaluate the derivative.
It is also possible for the numerical answers to happen to be the same. That coincidence does not make them the same kind of quantity. For example, if \(f(t)=t^2\), then \(f(1)=1\) and \(f'(1)=2\), so they differ; with other functions or inputs their numerical values could match. The interpretation still depends on whether the calculation is \(f(a)\) or \(f'(a)\), not merely on the number obtained.
Why the Distinction Matters
In applications, a value and its rate often answer separate practical questions. A population value can describe how many people are present, while its derivative describes how quickly that population is changing at a particular time. A temperature value tells the temperature at an instant, while its derivative describes the temperature’s instantaneous rate of change. In a revenue model, revenue and the derivative of revenue likewise describe an amount and a rate, respectively.
When reading a contextual problem, keep the original function and its derivative separate. A useful routine is to name what each represents before doing any arithmetic:
- Identify the quantity modeled by the function.
- Interpret the function value as that quantity at the specified input.
- Interpret the derivative value as the quantity’s instantaneous rate of change at that input.
- Check that your final sentence answers the question asked.
This habit prevents a correct calculation from being paired with the wrong conclusion. It also prepares you to interpret derivative notation accurately in more involved applications.
Check Your Understanding
For each question, identify whether it asks for a function value or a derivative value before calculating or explaining.
- A model \(C(t)\) gives the number of cars in a parking lot at time \(t\). What does \(C(5)\) represent?
- In the same model, what does \(C'(5)\) represent?
- A temperature model gives \(T(3)=18\) and \(T'(3)=-2\). Which number is the temperature at \(t=3\), and which is its instantaneous rate of change?
- For \(f(x)=3x^2+1\), find \(f(2)\) and \(f'(2)\). State what kind of value each answer is.
- Explain why a derivative value cannot generally be reported as the value of the modeled quantity.
