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Basic Differentiation Rules · Tutorial 432 of 1495

Estimating Polynomial Derivatives With Secant Slopes

Use nearby secant slopes to estimate a polynomial’s derivative, and understand why shrinking the input interval makes the estimates approach the tangent slope.

Beginner 8 min read

What You'll Learn

  • Calculate a secant slope between a point and a nearby input
  • Use shrinking intervals to form a sequence of derivative estimates
  • Compare numerical estimates with a polynomial’s derivative at the point
  • Explain algebraically why secant slopes approach the derivative
  • Recognize why the denominator must be the change in input

From a Secant Slope to a Derivative Estimate

A derivative at a point gives the slope of the tangent line there. A secant line, by contrast, passes through two points on the graph. Its slope is an average rate of change over the interval between those points. If one point is fixed and the other moves closer to it, the secant slope can estimate the tangent slope.

For a polynomial, earlier differentiation rules let us calculate the derivative directly. Secant slopes provide a second way to assess the expected value: compute slopes over progressively smaller intervals and observe whether they approach the derivative.

Definition: For a function \(f\), a fixed input \(a\), and a nonzero increment \(h\), the secant slope from \(a\) to \(a+h\) is \[ \frac{f(a+h)-f(a)}{h}. \] The increment \(h\) is the change in input, since \((a+h)-a=h\). The formula applies when both \(a\) and \(a+h\) are in the domain of \(f\).

When \(h\) is positive, the second input lies to the right of \(a\); when \(h\) is negative, it lies to the left. In either case, making \(|h|\) smaller brings the second point closer to the fixed point. A single secant slope is an average rate over a nonzero interval, not the instantaneous rate at \(a\).

How to Build an Estimate

For a polynomial \(f\), first identify the point \(a\) and choose a nonzero increment \(h\). Evaluate the polynomial at both inputs, subtract the function values in the same order as the inputs, and divide by \(h\). Repeat with smaller increments. If the resulting slopes settle toward a value, that value is a numerical estimate of the derivative at \(a\).

1
Fix the point.
Choose the input \(a\) at which the derivative is being estimated.
2
Choose nearby inputs.
Use nonzero increments \(h\) whose absolute values decrease, such as \(0.5\), \(0.1\), and \(0.01\).
3
Calculate each secant slope.
For each increment, evaluate \([f(a+h)-f(a)]/h\).
4
Assess the trend.
Compare the slopes with one another and, when available, with the derivative found using polynomial differentiation rules.

Smaller increments often give closer estimates, but a decimal table alone is not a proof that the values converge. For the polynomial examples below, algebra shows exactly how the secant slopes differ from the derivative. This also explains why shrinking the interval is informative.

Worked Examples

Worked Example: Estimating the Derivative of a Quadratic

Estimate the derivative of \(f(x)=x^2+2x\) at \(a=1\) using forward increments \(h=1\), \(h=0.5\), and \(h=0.1\). Then compare with the derivative from the polynomial rules.

First, evaluate the fixed function value:

$$f(1)=1^2+2(1)=3.$$

For \(h=1\), the nearby input is \(1+1=2\), and

$$f(2)=2^2+2(2)=8,\qquad \frac{f(2)-f(1)}{1}=\frac{8-3}{1}=5.$$

For \(h=0.5\), the nearby input is \(1+0.5=1.5\), and

$$f(1.5)=(1.5)^2+2(1.5)=2.25+3=5.25,\qquad \frac{f(1.5)-f(1)}{0.5}=\frac{5.25-3}{0.5}=\frac{2.25}{0.5}=4.5.$$

For \(h=0.1\), the nearby input is \(1+0.1=1.1\), and

$$f(1.1)=(1.1)^2+2(1.1)=1.21+2.2=3.41,\qquad \frac{f(1.1)-f(1)}{0.1}=\frac{3.41-3}{0.1}=\frac{0.41}{0.1}=4.1.$$

The estimates are \(5\), \(4.5\), and \(4.1\). By the power, constant-multiple, and sum rules, \(f'(x)=2x+2\), so \(f'(1)=2(1)+2=4\). The estimates move toward \(4\) as the increment shrinks. The comparison checks the numerical trend against the derivative value.

Worked Example: A Cubic with a Larger Initial Error

Estimate the derivative of \(g(x)=x^3-2x\) at \(a=2\) using \(h=0.5\), \(h=0.1\), and \(h=0.01\).

The fixed function value is

$$g(2)=2^3-2(2)=8-4=4.$$

For \(h=0.5\), the nearby input is \(2.5\):

$$g(2.5)=(2.5)^3-2(2.5)=15.625-5=10.625,$$
$$\frac{g(2.5)-g(2)}{0.5}=\frac{10.625-4}{0.5}=\frac{6.625}{0.5}=13.25.$$

For \(h=0.1\), the nearby input is \(2.1\):

$$g(2.1)=(2.1)^3-2(2.1)=9.261-4.2=5.061,$$
$$\frac{g(2.1)-g(2)}{0.1}=\frac{5.061-4}{0.1}=\frac{1.061}{0.1}=10.61.$$

For \(h=0.01\), the nearby input is \(2.01\):

$$g(2.01)=(2.01)^3-2(2.01)=8.120601-4.02=4.100601,$$
$$\frac{g(2.01)-g(2)}{0.01}=\frac{4.100601-4}{0.01}=\frac{0.100601}{0.01}=10.0601.$$

The polynomial rules give \(g'(x)=3x^2-2\), and hence \(g'(2)=3(2)^2-2=12-2=10\). The three secant estimates approach \(10\); the closest of these is \(10.0601\). As a check on the estimates, expansion at \(2+h\) gives \(g(2+h)=4+10h+6h^2+h^3\). Subtracting \(g(2)=4\) and dividing by \(h\) gives \(10+6h+h^2\), which yields \(13.25\), \(10.61\), and \(10.0601\) for the stated increments.

Worked Example: Estimating at a Negative Input

Estimate the derivative of \(p(x)=2x^3-x^2+3\) at \(a=-1\), using forward increments \(h=0.2\) and \(h=0.1\).

At the fixed input,

$$p(-1)=2(-1)^3-(-1)^2+3=-2-1+3=0.$$

For \(h=0.2\), the nearby input is \(-1+0.2=-0.8\):

$$p(-0.8)=2(-0.8)^3-(-0.8)^2+3=2(-0.512)-0.64+3=-1.024-0.64+3=1.336.$$
$$\frac{p(-0.8)-p(-1)}{0.2}=\frac{1.336-0}{0.2}=6.68.$$

For \(h=0.1\), the nearby input is \(-1+0.1=-0.9\):

$$p(-0.9)=2(-0.9)^3-(-0.9)^2+3=2(-0.729)-0.81+3=-1.458-0.81+3=0.732.$$
$$\frac{p(-0.9)-p(-1)}{0.1}=\frac{0.732-0}{0.1}=7.32.$$

Using the polynomial rules, \(p'(x)=6x^2-2x\), so \(p'(-1)=6(-1)^2-2(-1)=6+2=8\). The estimates \(6.68\) and \(7.32\) move toward \(8\) as \(h\) decreases. Their direction is also reasonable here: both forward secants are below the derivative value.

Why the Secant Slopes Approach the Derivative

The approach can be justified directly for any polynomial. Write \(p(x)=\sum_{k=0}^{n}c_kx^k\), where \(n\) is a nonnegative integer and the \(c_k\) are constants. The binomial theorem expands each power at \(a+h\). Subtracting \(p(a)\) removes the terms that do not contain \(h\); division by \(h\) then leaves the derivative value plus terms that contain positive powers of \(h\).

$$ \begin{aligned} \frac{p(a+h)-p(a)}{h} &=\sum_{k=1}^{n}c_k\sum_{j=1}^{k}\binom{k}{j}a^{k-j}h^{j-1}\\ &=\sum_{k=1}^{n}k c_k a^{k-1} +\sum_{k=1}^{n}c_k\sum_{j=2}^{k}\binom{k}{j}a^{k-j}h^{j-1}. \end{aligned} $$

The first sum is \(p'(a)\), by the power and sum rules. Every term in the second sum contains \(h^{j-1}\) with \(j-1\geq 1\). It is a finite sum, so it tends to zero as \(h\) tends to zero. Therefore the secant slopes approach \(p'(a)\). This explains the estimates: for a polynomial, their difference from the derivative is made up of terms that vanish with the increment.

Takeaway: For a polynomial, the secant slope over an increment \(h\) is the derivative at the fixed point plus terms that become small as \(h\) becomes small. Numerical estimates can suggest the derivative’s value; the polynomial rules give the value to compare against.

A Common Denominator Error

The denominator is the change in input, not the nearby input itself. For example, take \(f(x)=x^2\), \(a=2\), and \(h=0.5\). The nearby input is \(2.5\), so the change in input is \(2.5-2=0.5\). The correct calculation is

$$\frac{f(2.5)-f(2)}{2.5-2}=\frac{6.25-4}{0.5}=\frac{2.25}{0.5}=4.5.$$

A wrong calculation would divide by \(2.5\), the endpoint:

$$\frac{f(2.5)-f(2)}{2.5}=\frac{2.25}{2.5}=0.9.$$

This is not the secant slope between inputs \(2\) and \(2.5\), because its denominator is not their difference. As a check, the derivative is \(f'(x)=2x\), so \(f'(2)=4\); the correct secant slope \(4.5\) is near \(4\), while \(0.9\) is not a plausible nearby estimate.

Using Estimates Carefully

When using secant slopes to assess an expected derivative, keep the inputs and calculations visible. State the fixed point, list each increment, and show the corresponding slope. A table can make a trend easier to see, but it should not hide the computations or imply more precision than the function values support.

Also distinguish “closer to the point” from “closer to the derivative value.” For the polynomial examples, shrinking the increment makes the secant slopes approach the derivative, as the algebraic justification shows. But estimates can approach from above or below, and an early estimate may be noticeably inaccurate. A sequence of numerical results is evidence of a trend, not a substitute for the definition or a proof.

Check Your Understanding

Use the definition of a secant slope and the polynomial rules where useful.

  1. For \(f(x)=x^2+1\) at \(a=2\), calculate the forward secant slope for \(h=0.5\).
  2. For \(g(x)=x^3\) at \(a=1\), calculate the secant slopes for \(h=0.1\) and \(h=0.01\). What derivative value do they suggest?
  3. In \([f(a+h)-f(a)]/h\), what quantity does the denominator represent?
  4. For a polynomial, why do the terms beyond \(p'(a)\) in the secant-slope expression vanish as \(h\) shrinks?
  5. Is one computed secant slope enough to establish the exact derivative at a point? Explain briefly.

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