Tutorials › Calculus I › Relative Sensitivity Near Zero And At Large Inputs

Linearization and Differentials · Tutorial 628 of 1495

Relative Sensitivity Near Zero And At Large Inputs

Learn to calculate and interpret absolute and relative sensitivity, including their limits and important exceptions near zero inputs or outputs.

Intermediate 10 min read

What You'll Learn

  • Calculate absolute sensitivity from the magnitude of the derivative.
  • Calculate relative sensitivity when both input and output are nonzero.
  • Derive how relative input changes produce first-order relative output changes.
  • Compare sensitivity near zero and at large inputs using powers and affine functions.
  • Recognize why relative sensitivity is undefined at zero input or zero output.

Two Ways to Measure a Function’s Response

A small change in an input can produce a small or large change in the output. The derivative describes this response in output units per input unit. But when the question is how large an output change is relative to the output itself, an absolute change may not be enough. A change of one unit is substantial for an output near two, but negligible for an output near a million.

These two perspectives lead to absolute and relative sensitivity. They are local measures: they describe the first-order response to a sufficiently small change at a specified input. Neither gives a universal description of how a function behaves at every input. In particular, sensitivity near zero or at large inputs depends on the function being studied.

Definitions and Interpretation

Let \(f\) be differentiable at \(x\). For a small input change \(h\), the derivative gives the first-order estimate \(f(x+h)-f(x)\approx f'(x)h\). The size of the output change per unit input change is therefore measured by \(|f'(x)|\).

Definition: The absolute sensitivity of \(f\) at \(x\) is \(|f'(x)|\). It measures the first-order magnitude of the output change per unit change in the input.

For relative sensitivity, we compare relative changes: the output change divided by the original output, and the input change divided by the original input. These ratios require both original values to be nonzero.

Definition: If \(x\ne 0\) and \(f(x)\ne 0\), the signed relative sensitivity of \(f\) at \(x\) is \(x f'(x)/f(x)\). Its magnitude, \(\left|x f'(x)/f(x)\right|\), is the relative sensitivity. It compares the first-order relative output change with the relative input change.

Absolute sensitivity has units of output per unit input. Relative sensitivity is dimensionless: it compares two relative changes, each of which is a ratio of like units. The signed quantity retains directional information. When it is positive, the relative input and relative output changes have the same sign to first order; when it is negative, those relative changes have opposite signs. This is a statement about relative changes, not necessarily about the signs of the absolute changes.

The formula for relative sensitivity is not defined under this convention if \(x=0\) or \(f(x)=0\). Even when the algebraic expression \(x f'(x)/f(x)\) has an extension to \(x=0\), it cannot there represent a relative input change: dividing an input change by the original input would require division by zero.

Why the Relative Formula Works

Differentiability at \(x\) means that the change in the function can be written as its linear part plus a remainder that is small compared with \(h\):

$$ f(x+h)-f(x)=f'(x)h+o(h)\qquad\text{as }h\to 0. $$

For absolute sensitivity, divide the magnitude of this change by \(|h|\). The remainder divided by \(|h|\) tends to zero, so the ratio tends to \(|f'(x)|\). For relative sensitivity, suppose \(x\ne0\) and \(f(x)\ne0\). Divide the relative output change by the relative input change. The resulting ratio tends to the signed formula:

$$ \frac{\dfrac{f(x+h)-f(x)}{f(x)}}{\dfrac{h}{x}} = \frac{x}{f(x)}\frac{f(x+h)-f(x)}{h} \longrightarrow \frac{x f'(x)}{f(x)}. $$

The limit follows because differentiability gives \(\bigl(f(x+h)-f(x)\bigr)/h\to f'(x)\), and \(x/f(x)\) is a fixed finite factor under the stated conditions. Taking magnitudes gives relative sensitivity. This derivation also shows why the conditions \(x\ne0\) and \(f(x)\ne0\) matter.

Worked Example: A Power Near Zero and at Large Inputs

Worked Example: A Power Near Zero and at Large Inputs

Let \(f(x)=x^2\). Find its absolute and relative sensitivities at \(x=-2\), then compare the first-order prediction with an input change \(h=0.01\).

Differentiate: \(f'(x)=2x\), so \(f'(-2)=-4\). The absolute sensitivity is \(|-4|=4\). Since \(-2\ne0\) and \(f(-2)=4\ne0\), the relative sensitivity is defined. Its signed value and magnitude are

$$ \frac{x f'(x)}{f(x)}\bigg|_{x=-2} = \frac{(-2)(-4)}{4}=2, \qquad \left|\frac{x f'(x)}{f(x)}\right|=2. $$

The relative input change is \(h/x=0.01/(-2)=-0.005\). The first-order relative output change is therefore \(2(-0.005)=-0.01\), or a decrease of \(1\%\). The first-order absolute output change is \(f'(-2)h=(-4)(0.01)=-0.04\).

Check against the exact change: \(x+h=-1.99\), and \(f(-1.99)=(-1.99)^2=3.9601\). Thus the exact output change is \(3.9601-4=-0.0399\), close to \(-0.04\). Its relative change is \(-0.0399/4=-0.009975\), close to the predicted \(-0.01\). The derivative check is \(f'(x)=2x\), so \(f'(-2)=-4\), as used.

More generally, for \(f(x)=x^2\), the absolute sensitivity is \(2|x|\), which approaches zero as \(x\) approaches zero and grows without bound as \(|x|\) grows. For every \(x\ne0\), the relative sensitivity is \(\left|x(2x)/x^2\right|=2\). Thus the two measures can behave very differently.

Worked Example: A Nonzero Output Near a Zero Input

Worked Example: A Nonzero Output Near a Zero Input

Let \(g(x)=3+x\). Find both sensitivities at \(x=0.1\), and check the relative-change prediction for \(h=0.01\).

The derivative is \(g'(x)=1\), so the absolute sensitivity is \(1\) at every input. At \(x=0.1\), the output is \(g(0.1)=3.1\); both the input and output are nonzero. The relative sensitivity is

$$ \left|\frac{xg'(x)}{g(x)}\right| = \left|\frac{0.1(1)}{3.1}\right| = \frac{1}{31}. $$

The relative input change is \(0.01/0.1=0.1\). The signed relative sensitivity is \(1/31\), so the predicted relative output change is \((1/31)(0.1)=1/310\). Since \(g\) is affine, its exact output change is \(g(0.11)-g(0.1)=3.11-3.1=0.01\). The exact relative output change is \(0.01/3.1=1/310\), matching the prediction exactly. The derivative check is \(g'(x)=1\), and the exact change also equals \(g'(0.1)h=1(0.01)\).

At \(x=0\), the algebraic expression \(xg'(x)/g(x)\) has value \(0(1)/3=0\). However, relative sensitivity as a comparison of relative input and output changes is undefined there: a relative input change would require division by the zero input. The algebraic value is an extension of the expression, not a relative-change interpretation at zero.

Large Inputs Do Not Have One Universal Pattern

A large input does not by itself determine whether either sensitivity is large. The function matters. For a power \(f(x)=x^n\), where \(n\) is a positive integer and \(x\ne0\), differentiation gives \(f'(x)=nx^{n-1}\). Consequently, \(|f'(x)|=n|x|^{n-1}\), while the relative sensitivity is \(\left|x(nx^{n-1})/x^n\right|=n\). The absolute sensitivity grows with \(|x|\) when \(n>1\), but the relative sensitivity stays constant.

Worked Example: A Cubic at a Large Input

Let \(f(x)=x^3\), and examine \(x=10\) with \(h=0.1\). The derivative is \(f'(x)=3x^2\), so \(f'(10)=3(10)^2=300\). The absolute sensitivity is \(300\). Since \(f(10)=1000\), the relative sensitivity is

$$ \left|\frac{10(300)}{1000}\right|=3. $$

The relative input change is \(0.1/10=0.01\), so the first-order relative output change is \(3(0.01)=0.03\). The first-order absolute change is \(300(0.1)=30\).

For a check, \(10.1^3=1030.301\), so the exact output change is \(1030.301-1000=30.301\). The exact relative output change is \(30.301/1000=0.030301\), close to \(0.03\). The derivative check is \(3(10)^2=300\), confirming the value used for both predictions. The linear prediction is close because the input step is small; it is not an exact change for a cubic.

Other functions behave differently. For example, if \(f(x)=\sqrt{x}\) with \(x>0\), then \(f'(x)=1/(2\sqrt{x})\), so absolute sensitivity decreases as \(x\) grows. Its relative sensitivity is \(\left|x f'(x)/f(x)\right|=1/2\), which remains constant. By contrast, for \(f(x)=e^x\), the absolute sensitivity is \(e^x\), while relative sensitivity is \(|x|\). There is no rule that large inputs must make both measures increase, decrease, or stay fixed.

Input Near Zero and Output Near Zero Are Different Cases

It is important not to treat “near zero” as one situation. When the input approaches zero while the output remains nonzero, relative sensitivity may become small. For \(g(x)=3+x\), for example, \(|xg'(x)/g(x)|=|x/(3+x)|\), which tends to zero as \(x\to0\). The expression is defined for nonzero \(x\) near zero, but at \(x=0\) the relative-input interpretation is still unavailable.

In contrast, relative sensitivity can become large when the output approaches zero. For \(g(x)=3+x\), consider \(x=-2.9\). Then \(g(-2.9)=0.1\), \(g'(-2.9)=1\), and the absolute sensitivity is \(1\), while the relative sensitivity is \(|(-2.9)(1)/0.1|=29\). If the input changes by \(h=0.001\), the exact output changes from \(0.1\) to \(0.101\), an absolute change of \(0.001\). The relative input change is \(0.001/(-2.9)=-1/2900\), while the relative output change is \(0.001/0.1=1/100\). Their ratio is \((1/100)/(-1/2900)=-29\), in agreement with the signed sensitivity. Near an output of zero, even a small absolute output change can be large relative to the original output.

Key takeaway: Absolute sensitivity is undefined only when the derivative does not exist. Relative sensitivity additionally requires a nonzero input and a nonzero output. Its behavior near zero depends on which quantity is approaching zero and on the function.

A Common Mistake: Reading Relative Sensitivity as Absolute Direction

The sign of \(x f'(x)/f(x)\) compares the signs of relative changes, not directly the signs of the absolute changes \(\Delta x\) and \(\Delta f\). For instance, take \(f(x)=x^2\) at \(x=-1\). Then \(f'(-1)=-2\) and \(f(-1)=1\), so the signed relative sensitivity is \((-1)(-2)/1=2>0\). If \(h>0\) is small, however, the absolute input change is positive and the first-order absolute output change is \(f'(-1)h=-2h<0\). The absolute changes have opposite signs. The relative input change \(h/(-1)\) is negative, and the relative output change is approximately \((-2h)/1\), also negative; those relative changes have the same sign, as the positive sensitivity indicates.

A related mistake is to use relative sensitivity when the output is zero or when the input is zero. In either case, stop and check the definition before interpreting the formula. Absolute sensitivity may still be useful, but a relative comparison with a zero reference value is not defined.

Using the Measures Carefully

When estimating the effect of a small input uncertainty, use absolute sensitivity to assess the approximate output change in output units. Use relative sensitivity to compare percentage-scale changes, provided the input and output are nonzero. The relative measure is especially informative when the scale of the output varies substantially across inputs.

Both measures are local. They use the derivative at one base input, so they describe first-order behavior for small changes around that point. A large sensitivity signals a potentially large first-order response under the corresponding measure; it does not, by itself, guarantee that a finite change is accurately estimated by the linear approximation. The size of the change and how the derivative varies nearby also matter.

Check Your Understanding

Use the definitions and first-order change formula to answer these questions.

  1. For \(f(x)=x^4\) at \(x=2\), calculate the absolute and relative sensitivities.
  2. Why is relative sensitivity, interpreted as a ratio of relative changes, undefined at an input of zero even if the algebraic expression has a finite extension there?
  3. For \(g(x)=3+x\), what happens to relative sensitivity as \(x\) approaches \(-3\), and why?
  4. If the signed relative sensitivity is negative, what does that say about the relative input and output changes to first order?
  5. Can a function have decreasing absolute sensitivity but constant relative sensitivity at large inputs? Give a function discussed above that has this behavior.

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