Why Derivative Units Matter
A derivative measures how an output changes as an input changes. Its units therefore describe a rate: output units per input unit. For example, if position is measured in meters and time in seconds, a derivative of position with respect to time is measured in meters per second.
Keeping units visible helps explain what a derivative value means, detect errors in a calculation, and distinguish rates that have different meanings. A numerical value alone is not enough: \(5\) could represent \(5\) meters per second, \(5\) dollars per item, or something else, depending on the quantities involved.
The Units of a Derivative
Suppose \(f(x)\) is an output quantity and \(x\) is an input quantity. The derivative at \(x=a\), when it exists, is the limit of the difference quotient. The numerator is a change in the output, and the denominator is a change in the input.
Here \(h\) has the same units as the input \(x\). The difference \(f(a+h)-f(a)\) has the same units as the output \(f(x)\). Consequently, the quotient has output units divided by input units. The derivative value has those same units if the limit exists.
This rule applies both to a derivative at a particular input and to the derivative function \(f'\). If position is measured in meters and time in seconds, every value of the position derivative has units of meters per second. The numerical value can depend on the input; the units do not.
Why the Rule Works
The units follow directly from the definition of the derivative. To see why, assign units to each part of the difference quotient before taking the limit.
If \(f(x)\) is measured in \(U\), then both \(f(a+h)\) and \(f(a)\) are measured in \(U\). Their difference is also measured in \(U\).
Since \(h\) is a change in the input, it is measured in \(V\).
The difference quotient has units \(U/V\), because its numerator has units \(U\) and its denominator has units \(V\).
Taking a finite limit changes the numerical value, not the units attached to the quotient. Thus the derivative has units \(U/V\).
This is a units argument, not a test for whether a derivative exists. A formula may have consistent units even when the derivative fails to exist at some input. The limit must still exist for the derivative value to be defined.
Worked Example: Position Changing with Time
Worked Example: Position Changing with Time
A moving object’s position \(s(t)\) is measured in meters and time \(t\) in seconds. Suppose its position is described by
Find and interpret the derivative at \(t=2\mathrm{s}\).
Step 1: Determine the units. The output is position in meters and the input is time in seconds, so \(s'(t)\) must have units \(\mathrm{m}/\mathrm{s}\).
Step 2: Differentiate. Differentiating each term with respect to \(t\) gives
The first term has units \((\mathrm{m}/\mathrm{s}^2)\mathrm{s}=\mathrm{m}/\mathrm{s}\), so both terms have the required units.
Step 3: Evaluate at \(2\mathrm{s}\).
At that time, the instantaneous rate of change of position is \(11\) meters per second.
Check: At \(t=2\mathrm{s}\), the difference quotient for an increment \(h\) seconds simplifies to \(11\mathrm{m}/\mathrm{s}+(2\mathrm{m}/\mathrm{s}^2)h\). As \(h\) approaches zero, the second term approaches zero, leaving \(11\mathrm{m}/\mathrm{s}\), as calculated.
Worked Example: Cost per Item
Worked Example: Cost per Item
Let \(C(n)\) be the total cost, in dollars, of producing \(n\) items. Suppose
Find the derivative at \(n=6\) items and state its units.
Step 1: Determine the units. The output is dollars and the input is items, so \(C'(n)\) is measured in dollars per item.
Step 2: Differentiate.
Since \(n\) is measured in items, the first term also has units dollars per item.
Step 3: Evaluate at \(n=6\) items.
The derivative value is \(14\) dollars per item. It describes the instantaneous rate at which the modeled cost changes with the number of items near \(n=6\). Its units are not simply dollars: the rate compares a change in dollars with a change in items.
Check: For an increment \(h\) items, \(C(6+h)=86\mathrm{dollars}+(14\mathrm{dollars}/\mathrm{item})h+(0.5\mathrm{dollars}/\mathrm{item}^2)h^2\), while \(C(6)=86\mathrm{dollars}\). Dividing their difference by \(h\) gives \(14\mathrm{dollars}/\mathrm{item}+(0.5\mathrm{dollars}/\mathrm{item}^2)h\). The limit as \(h\to0\) is \(14\) dollars per item.
Worked Example: Converting Rate Units
Worked Example: Converting Rate Units
A derivative value is \(72\) kilometers per hour. Express the same rate in meters per second.
Step 1: Convert the output unit. Since \(1\mathrm{km}=1000\mathrm{m}\),
Step 2: Convert the input unit. One hour is \(3600\) seconds, so a rate per hour becomes a rate per \(3600\) seconds:
Check: Converting \(20\mathrm{m}/\mathrm{s}\) back gives \(20(3600)\mathrm{m}/\mathrm{h}=72000\mathrm{m}/\mathrm{h}=72\mathrm{km}/\mathrm{h}\). The rate is \(20\) meters per second.
Units Must Match the Variables
A derivative’s units depend on how both quantities are measured. For example, changing the time unit from seconds to minutes changes the numerical value of a rate, even though the underlying physical relationship is the same. If a position rate is \(2\mathrm{m}/\mathrm{s}\), then in meters per minute it is
The numerical value changes from \(2\) to \(120\) because the unit of input time changed from seconds to minutes. Always name the input and output units before interpreting a derivative. If a model uses time in minutes, a rate calculated from it is per minute, not per second.
Units also help check the terms of a model. Quantities added together must have matching units. In the position example, each term in \(s(t)\) has units of meters; in its derivative, each term has units of meters per second. This consistency is a useful check, although it cannot by itself establish that the model or calculation is correct.
A Common Units Error
A frequent mistake is to reverse the units and report input per output. For a cost function with output in dollars and input in items, the derivative is dollars per item, not items per dollar. This follows from the difference quotient: the change in cost is in dollars, and the input change is in items.
Correction: The quotient is change in cost divided by change in item count, so its units are dollars per item.
Another error is to give a derivative the same units as the original output. If position is measured in meters, the derivative with respect to time is not measured in meters; dividing the position change by a time change gives meters per unit of time.
Summary
To find the units of a derivative value, identify the units of the output and input, then divide output units by input units. The result describes how much the output changes per unit change in the input. Keep those units attached through differentiation, substitution, interpretation, and any conversion.
Check Your Understanding
Answer each question by identifying the relevant output and input units.
- A tank’s volume is measured in liters and time in minutes. What are the units of the derivative of volume with respect to time?
- A revenue function outputs dollars and takes quantity sold in items as its input. What are the units of its derivative?
- A length is measured in centimeters and the input is measured in seconds. What are the units of the derivative?
- A rate is \(3\mathrm{km}/\mathrm{h}\). Convert it to meters per hour.
- Explain why consistent units in a formula do not, by themselves, prove that the derivative exists.
