Why Units Matter in a Linear Approximation
A linear approximation predicts a function value near a chosen input. Its numerical calculation may be simple, but its units carry essential information: they tell us what the derivative measures and whether the predicted change makes sense.
Suppose \(f\) takes an input \(x\) near a base point \(a\). The linearization at \(a\) is
Here \(x-a\) is the change in the input. The product \(f'(a)(x-a)\) estimates the change in the output, and adding it to \(f(a)\) gives the approximate output. For this addition to be meaningful, both terms must have the units of \(f\).
Where the Derivative’s Units Come From
The derivative describes output change per input change. For a function measured in units \(V\), with input measured in units \(U\), its difference quotient has units \(V/U\): the numerator is an output difference, and the denominator is an input difference. Taking the limit to obtain the derivative does not change those units.
Thus the units follow directly from the definition of the derivative. They are not an optional label added after calculation. They describe what the numerical slope means. For example, a derivative measured in square metres per metre represents an area change per metre of input change.
The linearization formula also shows why the units must fit. If \(x\) and \(a\) are both inputs expressed in \(U\), then \(x-a\) has units \(U\). Multiplying by \(f'(a)\), whose units are \(V/U\), gives an estimated output change in \(V\). That can be added to \(f(a)\), which is also in \(V\).
Worked Example: Estimating an Area Change
Worked Example: Estimating an Area Change
Let \(A(r)=r^2\) be the area of a square in terms of its side length \(r\). Suppose \(r\) is measured in metres and is close to \(2\text{ m}\). Estimate the area when the side length is \(2.1\text{ m}\), using the linearization at \(a=2\text{ m}\).
The output is an area, measured in square metres. Differentiate the formula:
Since \(A\) has units \(\text{m}^2\) and \(r\) has units \(\text{m}\), the derivative has units \(\text{m}^2/\text{m}=\text{m}\). At the base point,
The input change is
Its estimated effect on area is
The units are correct: a derivative in metres times an input change in metres gives square metres. Therefore
Check the calculation by substituting \(r=2.1\text{ m}\) into the original function. The exact area is \((2.1\text{ m})^2=4.41\text{ m}^2\), close to the estimate \(4.4\text{ m}^2\). The estimate and exact value both have area units.
Changing Units Changes the Numerical Slope
A derivative’s numerical value depends on the units used to express the input and output. Changing the unit changes the numerical coordinates, so the slope number can change too. The physical relationship does not change, and the units attached to the slope explain how to interpret its new numerical value.
For example, a length of \(2\text{ m}\) is also \(200\text{ cm}\). A change of \(0.1\text{ m}\) is also \(10\text{ cm}\). The derivative’s numerical value cannot be carried unchanged between these descriptions: its units must change with the input and output units.
Worked Example: Rewriting the Same Approximation in Centimetres
Use \(A(r)=r^2\) again, but now measure both side length and area in centimetres and square centimetres. At the same physical base side length, \(r=200\text{ cm}\), the area and derivative are
The derivative’s units are \(\text{cm}^2/\text{cm}=\text{cm}\). The target side length, \(2.1\text{ m}\), is \(210\text{ cm}\), so the input change from the base point is
The estimated area change is
Consequently, the linear approximation in these units is
To check that this is the same estimate as in metres, convert the result: \(44{,}000\text{ cm}^2=4.4\text{ m}^2\), because \(1\text{ m}^2=10{,}000\text{ cm}^2\). This matches the previous approximation. Substitution into the original function gives the exact area \((210\text{ cm})^2=44{,}100\text{ cm}^2\), so the estimate is consistent in centimetres as well.
The derivative’s numerical value changed from \(4\) when expressed in metres to \(400\) when expressed in centimetres. That is not a contradiction: in the first description the slope is \(4\text{ m}^2/\text{m}\), and in the second it is \(400\text{ cm}^2/\text{cm}\). Both describe the same physical rate of change.
A Units Check for Every Linearization
Before using a linear approximation, identify the units of the function output and input. Then check each part of the formula. This does not establish how accurate the approximation is; it checks that the quantities being combined are compatible.
Record the units \(U\) for \(x\) and the units \(V\) for \(f(x)\).
Confirm that \(f'(a)\) has units \(V/U\).
Make sure \(x\) and \(a\) are expressed in the same input unit, so \(x-a\) has units \(U\).
The product \(f'(a)(x-a)\) must have units \(V\), matching \(f(a)\), before the two are added.
If an input change is reported in a different unit from the base point, convert it before subtracting or multiplying. The same rule applies to the output: quantities in different output units must be converted to a common unit before addition.
Worked Example: Estimating a Length from an Area
Worked Example: Estimating a Length from an Area
Suppose \(s(B)=\sqrt{B}\) gives the side length of a square from its area \(B\). Measure \(B\) in \(\text{cm}^2\) and \(s\) in \(\text{cm}\). Estimate the side length when the area is \(103\text{ cm}^2\), using the nearby base area \(B=100\text{ cm}^2\).
Differentiate:
The derivative has units \(\text{cm}/\text{cm}^2=1/\text{cm}\). At the base point,
The area change is \(103\text{ cm}^2-100\text{ cm}^2=3\text{ cm}^2\). Its estimated effect on side length is
Therefore the estimated side length is
Check the units: \(1/\text{cm}\) times \(\text{cm}^2\) gives \(\text{cm}\), so the estimated change can be added to the base side length. Substitution into the original function gives the exact value \(\sqrt{103}\text{ cm}\), approximately \(10.1489\text{ cm}\). Rounded to two decimal places, this is \(10.15\text{ cm}\), agreeing with the linear estimate at that precision.
Common Mistake: Mixing Units in the Product
A units check catches arithmetic that may look plausible when units are omitted. In the square-area example, the derivative at \(2\text{ m}\) is \(4\text{ m}^2/\text{m}\). If the input change is given as \(10\text{ cm}\), multiplying the numbers \(4\) and \(10\) directly and reporting \(40\text{ m}^2\) is wrong. The derivative is per metre, but the change was entered in centimetres.
The corrected product has the required area units. Alternatively, convert the derivative to \(400\text{ cm}^2/\text{cm}\) and multiply by \(10\text{ cm}\), obtaining \(4{,}000\text{ cm}^2=0.4\text{ m}^2\). Both routes agree when the units are handled consistently.
Check Your Understanding
For each question, use units as part of the reasoning, not just as a label on the final answer.
- If \(f\) is measured in litres and \(x\) in minutes, what units does \(f'(a)\) have?
- A derivative has units \(\text{m}^2/\text{m}\), and the input changes by \(0.2\text{ m}\). What units should the estimated output change have?
- Why can the numerical value of a derivative change when the input is converted from metres to centimetres?
- A base input is given in seconds, but the target input is given in minutes. What must be done before forming \(x-a\)?
- In a linear approximation, why must \(f'(a)(x-a)\) and \(f(a)\) have matching units?
