Why Check a Derivative’s Scale?
A symbolic calculation can be correct in form yet still contain an arithmetic or coefficient error. A scale check asks whether the size of the result makes sense for the model: if the input changes a little, does the derivative predict an output change of a reasonable size? This check does not replace differentiation. It uses the model and the meaning of a derivative to help assess the result.
For a polynomial model, the derivative at an input is the local rate at which the output changes as the input changes. Its units are output units divided by input units. For example, if a model gives distance in metres from time in seconds, its derivative has units of metres per second. A value of \(12\) then means a local rate of about \(12\) metres of distance per second of time, not a distance of \(12\) metres.
From a Derivative to an Expected Output Change
The derivative is defined by the limit of nearby difference quotients. In particular, when \(f'(x)\) exists, the quotient \(\frac{f(x+\Delta x)-f(x)}{\Delta x}\) approaches \(f'(x)\) as nonzero \(\Delta x\) approaches zero. Rearranging the nearby quotient motivates the scale estimate below.
This is an approximation for a sufficiently small change, not an assertion that the output change always equals the derivative times the input change. A finite change can include additional effects, especially when the input step is not small or the derivative changes rapidly nearby. The units are a useful first check: multiplying output-units-per-input-unit by input units gives output units, as required for a change in the model’s output.
For a polynomial, the earlier power rule gives a convenient way to examine the scale term by term. If \(f(x)=c_nx^n+\cdots+c_1x+c_0\), then the contribution of \(c_kx^k\) to the derivative is \(kc_kx^{k-1}\). The derivative at a particular input is the sum of those contributions. Their sizes can help explain whether the total derivative should be large, small, or subject to cancellation.
Why the Term Contributions Change with the Input
Consider one nonconstant term \(c x^n\). Its size in the original polynomial at a nonzero input \(x\) is \(c x^n\), while its contribution to the derivative is \(n c x^{n-1}\). Dividing the latter by the former gives a simple comparison of their relative sizes.
In magnitude, the derivative contribution is \(n/|x|\) times the magnitude of the original term. This relation is a comparison at the same input, not a direct comparison of quantities with identical units: the derivative contribution has units of output per input, whereas the original term has output units. The factor \(1/x\) accounts for that difference. At \(x=0\), the ratio is undefined, so inspect the derivative terms directly instead.
This calculation also explains why input size matters. A high power can have a modest contribution at a small input and a much larger one at a larger input, even when its coefficient is small. Conversely, different derivative contributions can partly cancel. Therefore, checking only the largest coefficient, or only the largest term in the original polynomial, can be misleading.
Worked Examples
Worked Example: Predicting a Small Distance Change
Suppose a distance model, measured in metres, is \(d(t)=0.5t^3-2t^2+3t+4\), where \(t\) is measured in seconds. Check the scale of the derivative at \(t=2\) seconds, then compare its prediction with the model’s actual change over \(0.1\) seconds.
Differentiate each term using the power rule:
At \(t=2\), the separate contributions are \(1.5(2^2)=6\), \(-4(2)=-8\), and \(3\). Thus \(d'(2)=6-8+3=1\) metre per second. The contributions partly cancel, so a result much larger than their net value would warrant a closer check. For \(\Delta t=0.1\) seconds, the derivative predicts an output change of approximately \(1(0.1)=0.1\) metres.
Now calculate the actual change. First, \(d(2)=0.5(8)-2(4)+3(2)+4=4-8+6+4=6\) metres. Next, \(2.1^2=4.41\) and \(2.1^3=9.261\), so \(d(2.1)=0.5(9.261)-2(4.41)+3(2.1)+4=4.6305-8.82+6.3+4=6.1105\) metres. The actual change is \(6.1105-6=0.1105\) metres. Its secant slope over this step is \(0.1105/0.1=1.105\) metres per second, reasonably close to the derivative \(1\) metre per second. The derivative is a local rate; the finite-step slope need not be exactly equal to it.
Worked Example: Checking a Derivative Against a Nearby Change
Let \(V(t)=0.02t^3+4t^2-3t+8\) be an output measured in litres, with \(t\) in minutes. Find and assess the derivative at \(t=10\) minutes using a \(0.1\)-minute input step.
The power rule gives \(V'(t)=0.06t^2+8t-3\). At \(t=10\), the three contributions are \(0.06(100)=6\), \(8(10)=80\), and \(-3\), so \(V'(10)=6+80-3=83\) litres per minute. For a change of \(0.1\) minute, this predicts an output change of approximately \(83(0.1)=8.3\) litres.
Compute the model’s actual output change term by term. The cubic term changes by \(0.02(10.1^3-10^3)=0.02(1030.301-1000)=0.60602\) litres. The quadratic term changes by \(4(10.1^2-10^2)=4(102.01-100)=8.04\) litres. The linear term changes by \(-3(10.1-10)=-0.3\) litres, and the constant term does not change. Therefore, \(\Delta V=0.60602+8.04-0.3=8.34602\) litres. The secant slope is \(8.34602/0.1=83.4602\) litres per minute. This is close in scale to \(83\) litres per minute, and its predicted output change, \(8.3\) litres, is close to the actual \(8.34602\) litres.
Worked Example: Seeing How the Input Changes the Scale
Consider the dimensionless polynomial \(p(x)=0.01x^3+2x^2-5x+4\). Compare its derivative at \(x=2\) and \(x=20\). The small cubic coefficient does not by itself determine how important the cubic term’s derivative contribution will be.
Differentiating gives \(p'(x)=0.03x^2+4x-5\). At \(x=2\), the contributions are \(0.03(4)=0.12\), \(4(2)=8\), and \(-5\). Hence \(p'(2)=0.12+8-5=3.12\). At \(x=20\), they are \(0.03(400)=12\), \(4(20)=80\), and \(-5\), giving \(p'(20)=12+80-5=87\). In particular, the cubic term’s derivative contribution grows from \(0.12\) to \(12\) as the input increases.
Check each scale with a \(0.01\) input step. At \(x=2\), the cubic term changes by \(0.01(2.01^3-2^3)=0.01(8.120601-8)=0.00120601\). The quadratic term changes by \(2(2.01^2-2^2)=2(4.0401-4)=0.0802\), and the linear term changes by \(-5(0.01)=-0.05\). Thus \(\Delta p=0.00120601+0.0802-0.05=0.03140601\); the secant slope is \(3.140601\), close to \(p'(2)=3.12\).
At \(x=20\), the cubic term changes by \(0.01(20.01^3-20^3)=0.01(8012.006001-8000)=0.12006001\). The quadratic term changes by \(2(20.01^2-20^2)=2(400.4001-400)=0.8002\), and the linear term changes by \(-0.05\). Therefore, \(\Delta p=0.12006001+0.8002-0.05=0.87026001\). Dividing by \(0.01\) gives a secant slope of \(87.026001\), close to \(p'(20)=87\). Both nearby checks support the calculated derivative scales.
A Common Mistake: Ignoring the Exponent’s Effect
A coefficient can be small while its derivative contribution is not. For example, in \(p(x)=0.01x^3+2x^2-5x+4\) at \(x=20\), the cubic term in the derivative contributes \(3(0.01)(20^2)=12\). A tempting but incorrect check is to look only at the coefficient \(0.01\) and conclude that this term should have almost no effect. That ignores both the exponent and the input size.
There is also a concrete calculation error to watch for: forgetting the exponent-decrement step. Incorrectly writing the derivative of \(0.01x^3\) as \(0.01x^3\) would make its supposed contribution at \(x=20\) equal to \(0.01(8000)=80\). The correction is \(\frac{d}{dx}(0.01x^3)=3(0.01)x^2=0.03x^2\), which contributes \(0.03(400)=12\). This correction is consistent with the nearby change check: over \(0.01\), the cubic term changes by \(0.12006001\), a secant contribution of \(12.006001\), close to \(12\), not \(80\).
Check Your Understanding
Use units and nearby changes as well as symbolic calculations to assess each answer.
- A model outputs kilometres from an input measured in hours. What units should its derivative have, and what output change does a derivative of \(4\) predict for an input increase of \(0.2\) hours?
- For a polynomial term \(c x^4\), write its contribution to the derivative and state how its magnitude compares with the original term’s magnitude at nonzero \(x\).
- Explain why a nearby secant slope should be close to a derivative for a sufficiently small input step, but need not equal it for a finite step.
- In a polynomial derivative, why can a small coefficient still produce a large contribution at some inputs?
- What should you check if a derivative predicts an output change very different from the model’s actual change over a small input step?
