Learning module Derivatives · Lesson 1 of 5
From average rate to instantaneous rate
Move from measurable average rates to the finite limit defining a derivative.
Allow 60–90 minutes, plus problem work.
Table of contents
Guiding question. A moving object has position \(s(t)=t^2\) metres after \(t\) seconds. What is its speed at exactly \(t=1\)?
Between two distinct times \(a\) and \(b\), we can calculate its average rate of change:
\[\frac{s(b)-s(a)}{b-a}.\]The units are metres per second. On a position–time graph, this is the slope of the secant line joining \((a,s(a))\) and \((b,s(b))\). The numerator measures change in position; the denominator measures elapsed time. Their quotient is a rate, not a position.
For an interval with one endpoint at \(1\) and the other at \(1+h\), where \(h\ne0\),
\[\frac{s(1+h)-s(1)}{h} =\frac{(1+h)^2-1}{h}=2+h.\]| Increment \(h\) | Second time \(1+h\) | Average rate \(2+h\) |
|---|---|---|
| \(1\) | \(2\) | \(3\) |
| \(0.1\) | \(1.1\) | \(2.1\) |
| \(0.01\) | \(1.01\) | \(2.01\) |
| \(-0.1\) | \(0.9\) | \(1.9\) |
| \(-0.01\) | \(0.99\) | \(1.99\) |
Predict. Will approaching \(1\) from earlier times give the same rate as approaching from later times? What would it mean if the two answers differed?
Try. Move the second point from the right, then from the left. Compare \(h=0.5\) and \(h=-0.5\). Finally set \(h=0\). At zero the two points coincide, so there is no secant slope to calculate. The limiting slope can still exist.
This motivates the definition. Let \(a\) be an interior point of the domain of \(f\). Its derivative at \(a\) is the finite real number
\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h},\]provided this two-sided limit exists. Equivalently, replace \(a+h\) by \(x\) and let \(x\to a\). We call \(f\) differentiable at \(a\) when this finite limit exists.
For our example, \(s'(1)=2\) metres per second. The tangent line has equation \(y=1+2(t-1)\). We found its slope by taking a limit; substituting \(h=0\) into the original quotient would only produce an undefined expression.
The derivative function \(f'\) assigns \(f'(a)\) to every point where it exists. The symbols \(f'(a)\) and \(\frac{df}{dx}(a)\) denote the same number. At an endpoint, only a one-sided derivative may be available; throughout the module, an unqualified derivative at a point means a finite two-sided derivative at an interior point.
Problems: rates and limits
1.1 — An average that hides a change. For \(f(x)=x^2\) and \(a\in\mathbb R\), calculate the average rates on \([a-h,a]\), \([a,a+h]\), and \([a-h,a+h]\) for \(h>0\). Which one already equals \(f'(a)\) for every \(h\)? Explain why that does not make the function linear.
Hint
Factor a difference of squares. Keep track of the length of each interval.
Solution
The three rates are \(2a-h\), \(2a+h\), and \(2a\). The symmetric interval’s rate equals the derivative because the quadratic errors on opposite sides cancel. Rates on arbitrary intervals still vary with their endpoints, so the function is not linear.
1.2 — Two endpoints moving at once. Let \(f(x)=x^3\). Find the secant slope between \(x=a-h\) and \(x=a+2h\) for \(h\ne0\), and its limit as \(h\to0\). Why is the denominator \(3h\) rather than \(h\)?
Hint
Use \((u^3-v^3)/(u-v)=u^2+uv+v^2\).
Solution
The horizontal displacement is \((a+2h)-(a-h)=3h\). The slope is \(3a^2+3ah+3h^2\), which tends to \(3a^2\). This is a secant limit with both endpoints moving, not the defining quotient with one endpoint fixed. Here the explicit calculation shows they have the same limit.
1.3 — A tempting replacement definition. Someone proposes defining the derivative using only
\[\lim_{h\to0}\frac{f(a+h)-f(a-h)}{2h}.\]Test this proposal on \(f(x)=\lvert x\rvert\) at \(a=0\). Can this symmetric quotient certify differentiability?
Hint
First compute the symmetric quotient. Then compute the original quotient separately for positive and negative \(h\).
Solution
The symmetric quotient is always zero. The defining quotient is \(\lvert h\rvert/h\), equal to \(1\) on the right and \(-1\) on the left. The derivative therefore does not exist. A symmetric difference can hide incompatible one-sided behavior.
1.4 — How close is close enough? For \(f(x)=x^3\) at \(a=1\), find a positive \(\delta\) such that every \(0<\lvert h\rvert<\delta\) gives a secant slope within \(0.01\) of the instantaneous rate. Justify your choice for both signs of \(h\).
Hint
The quotient is \(3+3h+h^2\). Bound the absolute error instead of checking a few decimal values.
Solution
For \(\lvert h\rvert<1\), the error is at most \(3\lvert h\rvert+h^2\le4\lvert h\rvert\). Take \(\delta=0.0025\). Then every nonzero increment of smaller magnitude gives error strictly less than \(0.01\). A smaller positive \(\delta\) also works.
Checkpoint. Explain the distinction between a secant slope, a derivative at a point, and the derivative function. Explain why a list of slopes approaching a number suggests a limit without proving it.