Comparing Local Rates
A derivative value describes how a function changes near one particular input. Comparing derivative values at two inputs lets us compare those local rates: the function may be increasing at one input and decreasing at another, or it may change more rapidly in one location than in the other.
Two features must be kept separate. The sign of the derivative indicates the direction of local change as the input increases. The absolute value of the derivative indicates the size of the local rate. A larger signed number is not necessarily a faster rate of change.
This comparison assumes that the derivative values use the same input and output units. If they do, both derivatives have the same units, and their magnitudes can be compared directly. If the units differ, convert them to a common system before interpreting a numerical comparison.
Why the Comparison Works
The derivative at an input is defined by a limit of difference quotients. If \(f\) is differentiable at \(a\), then
Here \(h\) is a small input change, and \(f(a+h)-f(a)\) is the corresponding output change. The quotient describes output change per unit of input change. Its limit is the local rate at \(a\).
More precisely, for every \(\varepsilon>0\), differentiability means that there is a \(\delta>0\) such that
Thus, for sufficiently small input changes, the difference quotient is as close as desired to \(f'(a)\). If \(h\) is positive, its sign tells us whether the nearby output change is positive or negative; its magnitude tells us the size of that output change per unit of input. Comparing derivatives at two inputs compares these local rates, not necessarily the changes over a fixed, finite interval.
A Reliable Comparison Method
For a pair of derivative values, use the following sequence. This keeps direction, magnitude, and units from being confused.
Confirm that both values measure the same output quantity per the same input quantity. Convert if necessary.
A positive derivative indicates local increase as the input increases; a negative derivative indicates local decrease. A zero derivative indicates no first-order change at that input.
Compare absolute values to decide where the function changes more rapidly in magnitude.
Report the direction at each input and which rate has greater magnitude, if either does.
Worked Examples
Worked Example: Opposite Directions and Different Rates
Let \(f(x)=x^2-4x\). Compare the local rates at \(x=1\) and \(x=4\).
Differentiate the function:
At \(x=1\), substitute into the derivative:
At \(x=4\):
The negative value \(f'(1)=-2\) means that the function is locally decreasing as \(x\) increases near \(1\). The positive value \(f'(4)=4\) means it is locally increasing near \(4\). Their magnitudes are
Therefore, the function changes more rapidly in magnitude near \(x=4\), and its local direction there is increasing rather than decreasing. Checking the substitutions in \(f'(x)=2x-4\) gives \(-2\) at \(1\) and \(4\) at \(4\), as calculated.
Worked Example: Comparing Rates With Units
Suppose \(x\) is the numerical value of a length measured in centimeters, and \(M(x)\) is a mass measured in grams. Consider the model
Compare the local rate of mass change at lengths of \(10\) cm and \(30\) cm. Since the output is in grams and the input is in centimeters, the derivative is measured in grams per centimeter.
Differentiate:
At \(x=10\), the calculation is
At \(x=30\):
Both rates are positive, so the mass is locally increasing with length at both inputs. Their magnitudes are \(3.4\ \text{g/cm}\) and \(4.2\ \text{g/cm}\), respectively. Because \(4.2>3.4\), mass changes more rapidly per centimeter near \(30\) cm than near \(10\) cm.
The units are identical, so the numerical comparison is valid. As a check, substituting each input back into \(M'(x)=0.04x+3\) gives \(3.4\) and \(4.2\), respectively. These are local rates; they should not be read as exact mass changes for a finite increase of one centimeter.
Worked Example: Reading a Derivative Table
A differentiable quantity \(Q\) is measured in milligrams and depends on an input \(u\) measured in seconds. A derivative table gives the following values:
| Input | Derivative value |
|---|---|
| \(u=2\) s | \(Q'(2)=1.2\) mg/s |
| \(u=5\) s | \(Q'(5)=-0.5\) mg/s |
| \(u=9\) s | \(Q'(9)=-2.0\) mg/s |
At \(u=2\), the derivative is positive, so \(Q\) is locally increasing. At \(u=5\) and \(u=9\), the derivatives are negative, so \(Q\) is locally decreasing at both inputs.
To compare how rapidly \(Q\) changes, compare absolute values:
Since \(2.0>1.2>0.5\), the greatest rate magnitude is at \(u=9\) s, and the smallest is at \(u=5\) s. At \(u=9\), the quantity is decreasing; at \(u=2\), it is increasing. All three values have units of milligrams per second, so the magnitudes can be compared directly. The table values themselves provide the derivative readings; the comparison follows by taking their absolute values and checking their signs.
Signed Order Is Not the Same as Faster Change
A common error is to treat the larger signed derivative as the faster rate. Suppose at two inputs \(a\) and \(b\) the derivative values are \(-5\) units per input and \(3\) units per input. It is true that
But concluding that the function changes faster at \(b\) because \(3\) is the larger signed number is incorrect. The magnitudes are
So the rate magnitude is greater at \(a\), where the function is decreasing. At \(b\), it is increasing. The correction is to use signed values for direction and absolute values for rate magnitude. The signed values can still be ordered algebraically, but that ordering answers a different question.
What a Comparison Does Not Tell You
A larger derivative magnitude at one input does not guarantee that the function's total output change is larger over a chosen interval. Derivatives describe local behavior. Over an interval, the function may have different derivative values at different points, and those changes can accumulate in different ways.
Also, a derivative of zero at one input does not mean that the function is constant nearby. It means only that the instantaneous rate at that input is zero. For example, \(f(x)=x^2\) has \(f'(0)=0\), but \(f'(x)=2x\) is nonzero for \(x\ne 0\). A comparison involving a zero derivative should therefore be stated locally rather than as a claim of no change over an interval.
Finally, unit conversion matters. A rate written in one set of units can have a different numerical value in another. For instance, \(1\) gram per centimeter equals \(100\) milligrams per millimeter: one gram is \(1000\) milligrams, and one centimeter is \(10\) millimeters. The physical rate is unchanged, but the number changes. Compare rates only after expressing them in compatible units.
Check Your Understanding
For each question, distinguish the direction of change from the magnitude of the rate.
- If \(f'(a)=-2\) and \(f'(b)=5\) in the same units, where is the function increasing, and where does it change more rapidly in magnitude?
- For \(g(x)=x^2-6x\), find and compare \(g'(1)\) and \(g'(5)\). State the direction at each input and compare the magnitudes.
- Why is it not enough to compare signed derivative values when deciding where a function changes faster?
- If one derivative is measured in grams per centimeter and another in milligrams per millimeter, what must you do before comparing their numerical magnitudes?
- Does a larger derivative magnitude at one input determine the total change in the function over an interval? Explain briefly.
