From Nearby Points to a Derivative Estimate
A derivative describes the slope of a curve at one input. A table or graph, however, usually gives only a limited number of points. The practical question is how to use those nearby points to estimate the slope at the input of interest.
The basic method is to calculate secant slopes using points closer and closer to the input. A secant slope measures the average change in the function over an interval. If the function has a derivative at the input, these slopes approach that derivative as the interval shrinks toward the input.
Finite data do not usually determine an exact derivative by themselves. They provide estimates. Comparing estimates from both sides, and from more than one interval size, helps reveal whether the values are settling toward a plausible result.
Secant Slopes Near the Input
For two points \((x_1,f(x_1))\) and \((x_2,f(x_2))\), where \(x_1\ne x_2\), the secant slope is the change in output divided by the change in input:
To estimate the derivative at \(x=a\), one useful choice is the secant through \((a,f(a))\) and \((a+h,f(a+h))\). Its slope is
Positive \(h\) uses a point to the right of \(a\); negative \(h\) uses a point to the left. As the point approaches \(a\), the interval gets smaller. The derivative is the limit of these slopes when that limit exists.
Worked Example: Estimating from a Plotted Curve
Worked Example: Estimating from a Plotted Curve
A graph shows a smooth curve near \(x=2\), with approximate read-offs \(f(2)=4\), \(f(1.8)\approx 3.24\), \(f(2.2)\approx 4.84\), \(f(1.9)\approx 3.61\), and \(f(2.1)\approx 4.41\). Estimate the derivative at \(x=2\).
First use the wider interval to the left, from \(1.8\) to \(2\):
On the wider interval to the right, from \(2\) to \(2.2\):
Now use points closer to \(2\). The left secant slope is
The right secant slope is
The estimates move closer together as the intervals narrow: the wider slopes are \(3.8\) and \(4.2\), while the closer slopes are \(3.9\) and \(4.1\). A reasonable estimate from these read-offs is \(f'(2)\approx 4\). As a check, multiplying each calculated slope by its interval width reproduces the corresponding change in output: \(3.9(0.1)=0.39\) on the left and \(4.1(0.1)=0.41\) on the right. Since the values came from a graph, their precision is limited; the estimate should not be reported as an exact result.
Comparing Left and Right Estimates
When data are available on both sides of \(a\), it is helpful to calculate left and right secant slopes separately. If both sets of slopes approach the same number as their intervals shrink, that number is a well-supported estimate of the derivative. If the slopes differ noticeably, check whether the intervals are still too wide, whether the data are rounded, or whether the slopes are not approaching a common value.
A centered secant uses one point on each side of \(a\). With \(h>0\), its slope is
This single calculation can be a useful summary, but it combines information from both sides. If left and right behavior differs, the centered slope may conceal that difference. When possible, calculate the one-sided slopes as well.
Worked Example: Using a Table for a Cubic Function
Worked Example: Using a Table for a Cubic Function
Suppose \(f(x)=x^3\), and use the table values near \(x=2\) to estimate \(f'(2)\).
| \(x\) | \(f(x)=x^3\) |
|---|---|
| 1.9 | 6.859 |
| 1.99 | 7.880599 |
| 2 | 8 |
| 2.01 | 8.120601 |
| 2.1 | 9.261 |
For the left point \(x=1.9\), the slope from \(x=1.9\) to \(x=2\) is
Using the closer left point \(x=1.99\) gives
On the right, the slope from \(2\) to \(2.1\) is
The closer right point \(x=2.01\) gives
The two closer slopes, \(11.9401\) and \(12.0601\), suggest an estimate near \(12\). To check that this pattern is consistent with the derivative definition, write the difference quotient for this function:
As \(h\) approaches \(0\), the terms \(6h\) and \(h^2\) approach \(0\), so the quotient approaches \(12\). Thus \(f'(2)=12\), agreeing with the estimates. For a direct arithmetic check of the closest table slopes, \(11.9401(0.01)=0.119401\) and \(12.0601(0.01)=0.120601\), exactly the output changes used above.
Why the Estimates Approach the Derivative
The connection between shrinking secants and the derivative follows directly from the derivative definition. The one-sided slopes are not a separate approximation rule: they are the difference quotients whose limit defines the derivative.
For a point \(a+h\), the secant through \((a,f(a))\) and \((a+h,f(a+h))\) has slope \(\dfrac{f(a+h)-f(a)}{h}\), with \(h\ne 0\).
Making \(|h|\) smaller brings the second point closer to the point where the derivative is wanted.
If the difference quotients approach one finite number from both sides, their limit is \(f'(a)\) by definition.
There is also a useful justification for centered secants. Suppose \(f'(a)\) exists. Rewrite the centered slope as the average of a right-hand and a left-hand difference quotient:
As \(h\) approaches \(0\), the first quotient approaches \(f'(a)\). The second does too: it is the difference quotient from the left, written with a positive denominator \(h\). Therefore their average approaches the average of \(f'(a)\) and \(f'(a)\), which is \(f'(a)\). This shows why centered slopes can estimate the derivative when it exists. It does not guarantee that a centered estimate will reveal what is happening on each side separately.
Worked Example: Centered Slopes from a Table
Worked Example: Centered Slopes from a Table
A table gives approximate values of a function near \(x=0\). Estimate the derivative at \(0\) using centered secants.
| \(x\) | Approximate \(f(x)\) |
|---|---|
| \(-0.1\) | 0.9048 |
| \(0.1\) | 1.1052 |
| \(-0.01\) | 0.99005 |
| \(0.01\) | 1.01005 |
For the interval from \(-0.1\) to \(0.1\), the input change is \(0.1-(-0.1)=0.2\). The centered slope is
For the narrower interval from \(-0.01\) to \(0.01\), the input change is \(0.01-(-0.01)=0.02\). Its slope is
These centered estimates are \(1.002\) and \(1\), so the data suggest \(f'(0)\approx 1\). Check the computations by multiplying each slope by its interval width: \(1.002(0.2)=0.2004\) and \(1(0.02)=0.02000\). Each product matches the table’s output change. Because the entries are approximate, the table supports an estimate, not a claim that the exact derivative must equal \(1\).
Common Mistakes and Limits of the Method
A secant slope requires dividing the change in output by the change in input. Forgetting the denominator produces a change in function values, not a slope.
It is also a mistake to treat one secant slope as the derivative without considering the interval. A wider interval gives an average change over more of the curve, and that average may differ from the slope right at the input. Comparing smaller intervals helps assess the estimate.
Smaller intervals are not automatically more reliable when the data are rounded or noisy. If two recorded outputs differ only slightly, rounding can have a large effect after the output change is divided by a very small input change. Use the precision actually supplied, avoid reporting unsupported digits, and compare more than one interval when possible.
Most importantly, a finite set of data cannot establish by itself that the derivative exists or that the estimates will keep approaching the same number. It can motivate and support an estimate. The derivative, when it exists, is defined by the limiting behavior of secant slopes, not by any single pair of recorded points.
Check Your Understanding
Use the derivative-estimation ideas from this tutorial to answer the following questions.
- For \(f(3)=7\) and \(f(3.1)=7.42\), calculate the secant slope and state which side of \(x=3\) the nearby point is on.
- What would it suggest if left and right secant slopes get closer to the same value as their intervals shrink?
- For data at \(a-h\) and \(a+h\), write the centered secant slope and identify its input-interval width.
- Why can a very small interval make a slope estimate less dependable when table values are rounded?
- Can a finite list of nearby secant slopes prove that the derivative exists? Explain briefly.
