Why Derivatives Have Units
A derivative describes how an output changes as an input changes. If a model gives a distance as a function of time, its derivative describes change in distance per change in time. It therefore has units of distance divided by time.
The same idea applies to any polynomial model. The polynomial’s output and input determine the units of its derivative, while the units of its coefficients ensure that the polynomial itself is meaningful. Keeping track of all three helps catch errors that arithmetic alone may not reveal.
Units in a Polynomial Model
Suppose \(x\) is the input, with units \(U\), and \(f(x)\) is the output, with units \(V\). A polynomial model can be written as
Every term being added must have the same units as the output. Thus \(a_0\) has units \(V\). Since \(a_kx^k\) must also have units \(V\), the coefficient \(a_k\) has units \(V/U^k\). This does not mean every coefficient has the same units: the coefficient’s units depend on the power of the input it multiplies.
For example, if \(x\) is measured in seconds and \(f(x)\) in metres, then \(a_1\) has units metres per second, \(a_2\) has units metres per second squared, and \(a_0\) has units metres. The derivative’s units are metres per second.
Why the Derivative Units Are Output per Input
The difference quotient used to define a derivative compares a change in output with a change in input. If \(h\) is an input increment, then \(f(x+h)-f(x)\) is an output change. Dividing by \(h\) gives an output change per input change.
When the difference quotient approaches a limit as \(h\) approaches zero, the resulting derivative retains those units. In other words, the derivative is the instantaneous rate of change, and its units follow from the same output-change-per-input-change ratio. This argument applies wherever the derivative exists.
For a polynomial, the usual differentiation rules make the same units visible term by term. The derivative of \(a_kx^k\) is \(ka_kx^{k-1}\). The number \(k\) has no units, so this term has units
Thus every nonzero term in the derivative has the same units, \(V/U\). The derivative of the constant term is zero; in context, the derivative as a whole is still a rate measured in output-units per input-unit.
Worked Example: Distance Changing with Time
Worked Example: Distance Changing with Time
A distance model is \(s(t)=(1.5\text{ m})+(2\text{ m/s})t+(0.5\text{ m/s}^2)t^2\), where \(t\) is measured in seconds. Find the derivative and its value at \(t=4\text{ s}\).
First check the units of the model’s terms. The constant term has units m. The linear term has units \((\text{m/s})(\text{s})=\text{m}\). The quadratic term has units \((\text{m/s}^2)(\text{s}^2)=\text{m}\). Each term therefore represents a distance.
Differentiate term by term:
Both derivative terms have units m/s: the second term has units \((\text{m/s}^2)(\text{s})=\text{m/s}\). At \(t=4\text{ s}\),
As a check, differentiating the original three terms gives \(0\), \(2\text{ m/s}\), and \((1\text{ m/s}^2)t\), respectively. Their sum is the derivative above, and each nonzero term has units m/s.
Worked Example: A Polynomial with Three Powers
Worked Example: A Polynomial with Three Powers
Let \(x\) be measured in metres, and let an output \(M(x)\) be measured in kilograms:
Each term has units kg: the factors of metres from \(x\), \(x^2\), or \(x^3\) cancel the corresponding powers in the coefficient. Differentiate:
Each term in the derivative has units kg/m. For example, the last term has units \((\text{kg/m}^3)(\text{m}^2)=\text{kg/m}\).
At \(x=10\text{ m}\), the rate is
Check the derivative by differentiating each original term: the first gives \(2\text{ kg/m}\); the second gives \((2)(0.1\text{ kg/m}^2)x=(0.2\text{ kg/m}^2)x\); and the third gives \((3)(-0.002\text{ kg/m}^3)x^2=-(0.006\text{ kg/m}^3)x^2\). These are exactly the three terms used in the evaluation.
Worked Example: Converting the Units of a Rate
Worked Example: Converting the Units of a Rate
A volume \(V(t)\), in litres, is modelled as a polynomial in time \(t\), measured in hours:
The constant, linear, and quadratic terms each have units L. Differentiating gives
At \(t=5\text{ h}\),
This is a rate of \(0.2\) litres per hour. To express it in millilitres per minute, use \(1\text{ L}=1000\text{ mL}\) and \(1\text{ h}=60\text{ min}\):
The units cancel to mL/min, as required. Differentiating again from the displayed derivative gives \(V'(5\text{ h})=0.4-0.2=0.2\text{ L/h}\), confirming the value before conversion.
A Common Units Mistake
A frequent error is to treat coefficients as unitless numbers, even when the input and output represent measured quantities. For the distance model in the first example, it would be wrong to write
and then evaluate this as \(6\text{ m/s}\) at \(t=4\). The number \(2\) in the derivative came from a coefficient with units m/s. The \(t\) term came from a coefficient with units m/s² multiplied by a time in seconds. Without those units, adding \(2\) and \(t\) is not meaningful: they do not even represent the same kind of quantity.
The correction is to retain the coefficient units through differentiation:
At \(t=4\text{ s}\), the second term is \(4\text{ m/s}\), so both terms can be added and the result is \(6\text{ m/s}\). A reliable check is to ask whether every term in the original polynomial has output units, and whether every term in its derivative has output-units per input-unit.
What the Units Tell You
Derivative units communicate what a rate describes. A derivative of distance with respect to time is a distance-per-time rate; a derivative of mass with respect to length is a mass-per-length rate. The units do not determine the numerical value, but they tell you what that value measures.
They also provide a practical error check. If a proposed derivative contains terms with different units, those terms cannot be added as written. If the final rate has units of output rather than output per input, a factor of input may have been lost. Unit checks do not replace differentiation, but they can reveal mistakes in a calculation or in the way a model has been written.
Check Your Understanding
For each question, keep the units attached to coefficients and evaluated quantities.
- If \(x\) is measured in seconds and \(f(x)\) in metres, what units must the coefficient of \(x^3\) have in a polynomial model?
- A polynomial has output units kilograms and input units metres. What units should each term in its derivative have?
- For \(q(t)=3\text{ L}+(2\text{ L/min})t+(0.5\text{ L/min}^2)t^2\), write the units of the coefficient of each power of \(t\).
- Differentiate the model in question 3 and state the units of every derivative term.
- Why is it invalid to add a quantity measured in metres per second to a quantity measured in metres per second squared?
