From Input Uncertainty to Output Uncertainty
A measurement is rarely exact. If an input \(x\) is reported as a nominal value \(x_0\) with an uncertainty of \(\delta x\), then the actual input may differ from \(x_0\). When the quantity of interest is an output \(y=f(x)\), the question is how much that input uncertainty may affect the output.
For a small input change, the derivative describes the local change in the function: an input change \(\Delta x\) produces an output change that is approximately \(f'(x_0)\Delta x\). The derivative therefore converts an input change into a predicted output change. Since uncertainty is reported as a nonnegative magnitude, the corresponding first-order estimate uses the absolute value of the derivative.
What the Estimate Means
The symbol \(\delta x\) denotes an uncertainty magnitude, not a signed change. For example, if a measured input is \(x_0=4.0\) seconds with uncertainty \(0.04\) seconds, the actual input is understood to be near \(4.0\) seconds, and the size of the input uncertainty is \(0.04\) seconds. The estimate does not say whether the input is above or below its nominal value.
The value \(f'(x_0)\) is the local rate of change of the output with respect to the input. If it is large in magnitude, a small input uncertainty can produce a larger output uncertainty. If it is small, the output is locally less sensitive to that input. The absolute value ensures that the estimated uncertainty is nonnegative whether the function is increasing or decreasing.
For the estimate to be meaningful, the function must be differentiable at the nominal input, and the uncertainty must be small enough for a local linear approximation to be useful. If the derivative changes rapidly across the range of possible inputs, or the uncertainty is not small, the first-order estimate may be inaccurate.
Why the Derivative Gives This Estimate
By differentiability at \(x_0\), a small change \(\Delta x\) gives an output change of the form
where the remainder \(r(\Delta x)\) becomes small relative to \(|\Delta x|\) as \(\Delta x\) approaches zero. In other words, the linear term \(f'(x_0)\Delta x\) captures the leading change near \(x_0\).
Taking the magnitude of that leading term gives
If the input uncertainty has magnitude \(\delta x\), replace \(|\Delta x|\) by \(\delta x\) in this first-order expression. This gives \(\delta y\approx |f'(x_0)|\delta x\). The approximation sign matters: the omitted remainder means the actual output change need not equal this estimate exactly.
Write down the measured value \(x_0\) at which the output is to be estimated.
Use \(\delta x\) as a nonnegative magnitude, in the input’s units.
Compute \(f'(x_0)\), not the derivative at some other input.
Calculate \(|f'(x_0)|\delta x\); the result has the output’s units.
Worked Examples
Worked Example: A Linear Calibration
A calibration model gives an output distance \(d\), in metres, from an input time \(t\), in seconds: \(d=f(t)=2.5t+1.0\). The measured time is \(t_0=4.00\text{ s}\), with uncertainty \(\delta t=0.04\text{ s}\). Estimate the output uncertainty.
Differentiate the model:
The rate has units of metres per second, since it is output distance divided by input time. At the nominal input, \(f'(4.00)=2.5\text{ m/s}\). Therefore,
The estimated output uncertainty is \(0.10\) metres. As a check, the model is linear, so this change is exact: changing the time by \(0.04\text{ s}\) changes the distance by \(2.5(0.04)=0.10\text{ m}\). The units also check: \((\text{m/s})(\text{s})=\text{m}\).
Worked Example: A Nonlinear Calibration Near the Nominal Input
Suppose a pressure model, measured in kilopascals, depends on a temperature input \(u\), measured in degrees Celsius: \(p=f(u)=10+0.6(u-20)-0.01(u-20)^2\). The input is \(u_0=24.0^\circ\text{C}\), with uncertainty \(\delta u=0.2^\circ\text{C}\). Estimate the output uncertainty.
Differentiate with respect to \(u\):
Evaluate this derivative at the nominal input:
Thus,
The estimated output uncertainty is \(0.104\text{ kPa}\). To check the first-order estimate, first calculate the nominal output:
At the two inputs \(24.2^\circ\text{C}\) and \(23.8^\circ\text{C}\), the model gives
These differ from \(12.24\text{ kPa}\) by \(0.1036\text{ kPa}\) and \(0.1044\text{ kPa}\), respectively. Both are close to the first-order estimate \(0.104\text{ kPa}\), but neither has to equal it exactly. The units check: \((\text{kPa}/^\circ\text{C})(^\circ\text{C})=\text{kPa}\).
Worked Example: Using a Known Local Rate
A differentiable sensor model reports an output \(S\) in millivolts as a function of an input \(q\) in grams. At the nominal input \(q_0=12.0\text{ g}\), its local rate is \(f'(12.0)=-3.2\text{ mV/g}\). If the input uncertainty is \(\delta q=0.15\text{ g}\), estimate the output uncertainty.
The derivative is already evaluated at the nominal input. Take its magnitude and multiply by the input uncertainty:
The estimated output uncertainty is \(0.48\text{ mV}\). The negative rate indicates that the model locally decreases as the input increases; it does not make the uncertainty negative. As a check, \(3.2\times0.15=0.48\), and \((\text{mV/g})(\text{g})=\text{mV}\), the output unit.
Units and Interpretation
Units provide a useful check on the calculation. If the input is measured in units \(U_x\) and the output in units \(U_y\), then the derivative has units \(U_y/U_x\). Multiplying by an input uncertainty in \(U_x\) leaves units \(U_y\):
For example, a derivative measured in kilopascals per degree Celsius must be multiplied by an uncertainty in degrees Celsius to produce an estimated uncertainty in kilopascals. A result with the wrong units usually signals that the input and output have been confused, or that the derivative has been evaluated or interpreted incorrectly.
The formula estimates an absolute output uncertainty: it gives a change in the output’s own units. It does not, by itself, give a relative or percentage uncertainty. Those quantities require comparing the absolute uncertainty with an appropriate output value.
Common Pitfall: Keeping the Sign of the Derivative
A common error is to multiply by \(f'(x_0)\) without taking its absolute value and then report a negative uncertainty. For instance, if \(f'(x_0)=-3.2\text{ mV/g}\) and \(\delta x=0.15\text{ g}\), the signed product is
This is not a valid uncertainty magnitude. It is the signed first-order output change for a positive input change of \(0.15\text{ g}\). For an uncertainty magnitude, the correction is
The sign of the derivative describes direction of change; the absolute value gives the size of the predicted change.
When to Use Care
The estimate is local. It is most useful when the uncertainty in the input is small relative to the scale on which the function changes. If the derivative varies substantially across the range around \(x_0\), a single derivative value may not represent changes throughout that range well.
If \(f'(x_0)=0\), the first-order estimate is zero. This means the linear term predicts no output change at that point; it does not prove that the actual output is unchanged when the input varies. The function may still change through higher-order effects. Likewise, a very small estimated uncertainty should not be interpreted as a guarantee of exactness.
In practice, record the nominal input, the input uncertainty, the derivative at that nominal input, the calculated output uncertainty, and the units. This makes clear both what was estimated and which local rate was used.
Check Your Understanding
For each question, use the derivative at the stated nominal input and keep track of units.
- A function has \(f'(x_0)=4.5\text{ cm/s}\), and the input uncertainty is \(0.03\text{ s}\). What is the first-order output uncertainty, including units?
- Why does the estimate use \(|f'(x_0)|\) rather than \(f'(x_0)\) alone?
- A model has a negative derivative at its nominal input. What does the negative sign describe, and what is the sign of the estimated uncertainty magnitude?
- State the differentiability and local-size conditions that make the estimate useful.
- If \(f'(x_0)=0\), what does the first-order estimate say, and what does it not establish?
