Two Questions About an Uncertain Input
Suppose an input \(x\) is measured near a nominal value \(a\), but its actual value may differ from \(a\). There are two useful questions: Which direction is the output expected to change? And how large could that change be? The first question is about a signed change; the second is about its magnitude.
The derivative helps answer both questions, but in different ways. Its sign, combined with the sign of the input change, gives a first-order prediction of the output's direction. Its absolute value, or an upper bound for its absolute value across a range of inputs, helps estimate output uncertainty.
Signed Change and Its Linear Estimate
Let \(f\) be differentiable at the nominal input \(a\). Write the actual input as \(a+h\), where \(h\) is the signed input change. Define the signed output change by \(\Delta y=f(a+h)-f(a)\). A positive \(\Delta y\) means the output increased; a negative \(\Delta y\) means it decreased.
Thus, if \(f'(a)>0\), a positive input change has a positive first-order output estimate, while a negative input change has a negative estimate. If \(f'(a)<0\), the predicted directions are reversed. A negative derivative does not mean that the output change is always negative: the direction of the input change matters too.
The linear estimate follows directly from the definition of the derivative. For nonzero \(h\), write the difference quotient as
where \(r(h)\) is the difference between the quotient and \(f'(a)\). Differentiability at \(a\) means \(r(h)\to 0\) as \(h\to 0\). Multiplying by \(h\) gives
The first term is the linear estimate; the second is the error in that estimate. Since \(r(h)\to 0\), the error is small relative to \(h\) for sufficiently small changes. This is a local statement: it does not say that \(f'(a)h\) equals the actual change for every finite \(h\).
Worked Example: Reading the Sign of the Change
Worked Example: Reading the Sign of the Change
Let \(f(x)=x^2-4x\), with nominal input \(a=3\). Estimate the output change when the input increases by \(0.1\), and when it decreases by \(0.1\). Then compare each estimate with the actual change.
First calculate the nominal output and derivative:
For an input increase of \(0.1\), \(h=0.1\). The first-order estimate is
The positive estimate predicts an increase. The actual new input is \(3.1\), and
The actual change is positive, as predicted, and differs from the estimate by \(0.21-0.2=0.01\).
For a decrease of \(0.1\), \(h=-0.1\). The estimate is
The negative estimate predicts a decrease. The actual input is \(2.9\), so
Again, the sign agrees, and the estimate differs from the actual change by \(-0.19-(-0.2)=0.01\). As a direct check, expanding at \(3+h\) gives \(f(3+h)-f(3)=2h+h^2\). Substituting \(h=0.1\) gives \(0.21\), and substituting \(h=-0.1\) gives \(-0.19\), matching both computed changes.
From a Signed Estimate to Worst-Case Uncertainty
An uncertainty statement often gives a range of possible inputs rather than a single signed change. Suppose \(|x-a|\leq\delta\), where \(\delta\geq 0\). At the nominal input, the linear estimate suggests an output-change magnitude of approximately \(|f'(a)|\delta\). The absolute value is essential: an uncertainty size is nonnegative, regardless of whether the corresponding signed output change is positive or negative.
However, \(|f'(a)|\delta\) is a local estimate, not automatically a guaranteed upper bound for every actual output change in the interval. The derivative may become larger in magnitude away from \(a\). A bound on the derivative throughout the whole interval provides a rigorous worst-case estimate.
To justify the result, apply the Mean Value Theorem to the interval with endpoints \(a\) and \(x\), when \(x\ne a\). It gives a point \(c\) between them such that
Taking absolute values and using the derivative bound yields
If \(x=a\), both sides of the claimed inequality are zero, so it holds in that case as well. The interval conditions matter: continuity on the closed interval and differentiability in its interior are what allow the Mean Value Theorem to apply.
This bound controls the magnitude, not the direction, of the change. If the derivative also has one strict sign throughout the interval, the Mean Value Theorem shows that the actual output change has the direction predicted by that sign and by \(x-a\). If the derivative changes sign, a sign prediction from the derivative at \(a\) alone need not hold over the whole interval.
Worked Example: A Decreasing Output and a Worst-Case Bound
Worked Example: A Decreasing Output and a Worst-Case Bound
Let \(f(x)=10-x^2\), and suppose the nominal input is \(a=2\) with uncertainty \(|x-2|\leq 0.05\). Find the signed first-order estimates for input changes of \(+0.05\) and \(-0.05\), then find a guaranteed bound for every output change in the uncertainty interval.
Differentiate and evaluate at the nominal input:
For \(h=0.05\), the signed first-order estimate is \((-4)(0.05)=-0.20\), predicting a decrease. The actual input is \(2.05\), and the actual change is
For \(h=-0.05\), the estimate is \((-4)(-0.05)=0.20\), predicting an increase. The actual input is \(1.95\), and
To obtain a guaranteed bound, the full input interval is \([1.95,2.05]\). On this interval, \(f'(x)=-2x\), so \(|f'(x)|=2x\leq 2(2.05)=4.1\). We may take \(M=4.1\), and the theorem gives
The actual changes have magnitudes \(0.2025\) and \(0.1975\), both no greater than \(0.205\). This checks the bound at the two endpoints; the derivative-bound argument guarantees it for every input in the interval, not just those endpoints.
When the Derivative at the Nominal Input Is Zero
If \(f'(a)=0\), the first-order estimate is zero for every \(h\). This means the linear part predicts no change; it does not prove that the actual change is zero. Higher-order effects may still produce a nonzero change, and a derivative bound across the uncertainty interval can still give a useful worst-case estimate.
Worked Example: Zero First-Order Estimate, Nonzero Uncertainty
Let \(f(x)=x^3-3x\), with nominal input \(a=1\) and uncertainty \(|x-1|\leq 0.1\). Find the first-order estimate and a guaranteed output-change bound.
The nominal output and derivative are
The first-order estimate is \(f'(1)h=0\). To bound actual changes, the possible inputs lie in \([0.9,1.1]\). On this interval, \(x^2\) ranges from \(0.81\) to \(1.21\), so \(3x^2-3\) ranges from \(3(0.81)-3=-0.57\) to \(3(1.21)-3=0.63\). Therefore \(|f'(x)|\leq 0.63\), and
Check the endpoint changes. At \(x=1.1\), \(f(1.1)=1.331-3.3=-1.969\), so the change is \(-1.969-(-2)=0.031\). At \(x=0.9\), \(f(0.9)=0.729-2.7=-1.971\), so the change is \(-1.971-(-2)=0.029\). Both are nonzero, and both have magnitude less than \(0.063\), as required by the bound. The zero linear estimate should therefore not be mistaken for a zero worst-case uncertainty.
A Common Mistake: Treating the Local Estimate as a Guarantee
A frequent error is to use \(|f'(a)|\delta\) as a guaranteed bound without checking the derivative elsewhere in the input interval. For example, take \(f(x)=x^2\), \(a=1\), and \(\delta=0.1\). Since \(f'(x)=2x\), the derivative at the nominal input is \(f'(1)=2\). A tempting calculation is
This is the first-order magnitude estimate, but it is not a valid upper bound for all changes: at \(x=1.1\), the actual change is \(1.1^2-1^2=1.21-1=0.21\), which exceeds \(0.2\).
The correction is to bound the derivative over the full interval \([0.9,1.1]\). There, \(|f'(x)|=2x\leq 2.2\), giving
This is a guaranteed bound, and \(0.21\leq 0.22\). The nominal derivative remains useful for a local estimate; the interval derivative bound supplies the guarantee.
Using Both Ideas Carefully
For a signed input change, begin with the product \(f'(a)h\). Its sign gives the first-order direction, and its magnitude estimates the local output change. For a range of possible inputs, use an absolute derivative bound over that entire range to bound the magnitude of every output change.
- If the derivative at the nominal input is positive, the local estimate has the same sign as the input change.
- If the derivative at the nominal input is negative, the local estimate has the opposite sign from the input change.
- If the derivative is zero, the first-order estimate is zero, but actual changes may still be nonzero.
- If \(|f'|\leq M\) throughout the uncertainty interval, then the exact output-change magnitude is at most \(M\) times the input-change magnitude.
The distinction is useful whenever an estimate must communicate both direction and reliability. A signed estimate says which way the output is expected to move locally. A worst-case bound says how far it can move under stated assumptions, even when the estimate is not exact.
Check Your Understanding
For each question, distinguish a signed first-order estimate from a guaranteed magnitude bound.
- If \(f'(a)<0\) and \(h>0\), what is the sign of the first-order output estimate?
- Why does the estimate \(|f'(a)|\delta\) not automatically bound every actual output change for \(|x-a|\leq\delta\)?
- State the derivative-bound conditions that imply \(|f(x)-f(a)|\leq M|x-a|\).
- If \(f'(a)=0\), what does the linear estimate say, and what does it not establish?
- For \(f(x)=10-x^2\) at \(a=2\), why does an input decrease have a positive first-order output estimate?
