Chapter 3 — Front Matter
Purpose and Overview
Chapter 3 introduces one of the two central ideas of calculus: the derivative. The derivative is a precise mathematical tool for measuring instantaneous rate of change. It answers questions such as: How fast is this object moving right now? How steeply does this curve rise at this point? How sensitive is this output to a small change in input? What is the slope of a curve at exactly one location?
These are questions that algebra alone cannot answer. Algebra gives the slope of a straight line and the average rate of change between two points. But when the change in input shrinks to zero — when the two points on a curve collapse into a single point — ordinary division fails. The derivative resolves this problem through the concept of a limit.
This chapter builds the derivative carefully. It starts with the idea of average rate of change, which students have seen before. It then asks what happens when the interval over which change is measured becomes arbitrarily small. That limiting process produces the derivative — the instantaneous rate of change at a single point. The chapter then develops derivative notation, shows how to estimate derivatives from graphs and tables, introduces the derivative as a function in its own right, and explains when derivatives fail to exist.
Students should leave this chapter understanding that the derivative is not just a formula or a calculation. It is a measurement. It measures change at a point, and it carries meaning: units, direction, context, and interpretation.
Opening Question
A car leaves a traffic light and travels according to the position function \(s(t)\), where \(s\) is measured in meters and \(t\) in seconds. During the first ten seconds, the car accelerates steadily. You can calculate how far the car traveled in total over those ten seconds. You can calculate its average speed. But can you say precisely how fast the car is moving at exactly \(t = 4\) seconds? Not with algebra alone. The speedometer in the car gives a reading at every instant — that instantaneous reading is precisely what the derivative describes. By the end of this chapter, you will be able to define, calculate, interpret, and reason about instantaneous rates of change using the derivative.
14Sections
10Objectives
16Key Terms
12+Worked Examples
Learning Objectives
- 1Distinguish average rate of change from instantaneous rate of change and explain why the distinction matters.
- 2Describe the geometric relationship between secant lines, tangent lines, and the derivative.
- 3State the formal limit definition of the derivative and use it to calculate derivatives from first principles.
- 4Interpret derivative notation including \(f'(x)\), \(\frac{dy}{dx}\), and \(\frac{d}{dx}[f(x)]\).
- 5Estimate derivatives numerically from tables and visually from graphs.
- 6Attach correct units to a derivative and explain what those units measure.
- 7Understand the derivative as a function and connect the graphs of \(f\) and \(f'\).
- 8Identify points where a function is not differentiable and explain why differentiability fails.
- 9Interpret derivatives in the context of motion: position, velocity, and acceleration.
- 10Recognize common mistakes in derivative reasoning and avoid them.
Why This Chapter Matters
The derivative is one of the most powerful and widely used ideas in mathematics. Every branch of science and engineering depends on it. Physicists use derivatives to describe velocity, acceleration, force, and fields. Engineers model stress, heat, current, and structural behavior. Economists analyze marginal cost, marginal revenue, and optimal decision-making. Computer scientists use derivatives to train machine learning models, optimize algorithms, and run simulations. Every time a system changes, derivatives measure how fast and in what direction. Learning to compute derivatives is important. Understanding what they mean is essential.
Sections 3.1 – 3.2
From Average to Instantaneous: Secant and Tangent Lines
Average Rate of Change · Secant Lines · Tangent Lines · The Limiting Process
Section 3.1
Average Versus Instantaneous Rate of Change
Rate of change describes how one quantity changes relative to another. In algebra, the rate of change of a linear function \(f(x) = mx + b\) is the slope \(m\) — constant everywhere. For nonlinear functions, the rate of change varies from point to point. A curve like \(f(x) = x^2\) rises slowly near \(x = 0\) and steeply for large values of \(x\). There is no single slope that describes the whole curve. This is where calculus begins.
Definition
Average Rate of Change
The average rate of change of a function \(f\) over an interval from \(x = a\) to \(x = b\) is:
\[\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}\]
This is the slope of the secant line connecting the two points \((a, f(a))\) and \((b, f(b))\).
Example 3.1.1
Average Speed of a Car
- ProblemA car's position is \(s(t) = 5t^2\), where \(s\) is in meters and \(t\) in seconds. Find the average rate of change of position from \(t = 1\) to \(t = 4\).
- ThinkAverage rate of change of position with respect to time is average velocity. Use the slope formula over \([1, 4]\).
- Calculate
s(1) = 5(1)² = 5 meters
s(4) = 5(4)² = 80 meters
Average rate of change = (80 − 5) / (4 − 1) = 75 / 3 = 25 m/s
- CheckThe car travels 75 meters in 3 seconds. An average of 25 m/s is reasonable for a car starting from low speed and accelerating.
- InterpretOver \(t = 1\) to \(t = 4\), the car's average velocity is 25 meters per second. This does not mean the car was always traveling at 25 m/s — it was accelerating throughout the interval.
Answer: Average rate of change = 25 m/s (average velocity over the interval).
The Problem with Average Rate of Change
Average rate of change cannot tell you the rate of change at a single instant. What speed would the speedometer display at exactly \(t = 2.73\) seconds? To answer this, think about smaller and smaller intervals. As \([a, b]\) shrinks and both endpoints approach the same point, the average rate of change approaches the instantaneous rate of change at that point. This limiting process is the foundation of the derivative.
Definition
Instantaneous Rate of Change
The instantaneous rate of change of \(f\) at \(x = a\) is:
\[\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]
This limit — when it exists — is the derivative of \(f\) at \(a\).
Section 3.2
Secant Lines and Tangent Lines
When you compute the average rate of change of \(f\) between \(x = a\) and \(x = b\), you are computing the slope of the secant line through the two points \((a, f(a))\) and \((b, f(b))\) on the graph. A secant line is a straight line that intersects a curve at two points. It captures the average behavior of the curve between those points.
Graph Instruction — Secant Line
Draw the graph of \(f(x) = x^2\). Mark two points: \(A = (1, 1)\) and \(B = (3, 9)\). Draw the secant line through \(A\) and \(B\). Label its slope: \(\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4\).
Now imagine keeping point \(A = (1, f(1))\) fixed and letting \(B\) slide along the curve toward \(A\). As \(B\) gets closer to \(A\), the secant line pivots. Its slope changes. In the limit, as \(B\) reaches \(A\), the secant line becomes the tangent line at \(A\).
Graph Instruction — Secant Lines Approaching the Tangent
Using the graph of \(f(x) = x^2\), draw several secant lines through \(A = (1, 1)\) and progressively closer second points: \(B_1 = (3, 9)\), then \(B_2 = (2, 4)\), then \(B_3 = (1.5, 2.25)\), then \(B_4 = (1.1, 1.21)\). Observe how the secant lines rotate toward a single limiting line — the tangent line at \(A\). Label this tangent line's slope as the instantaneous rate of change at \(x = 1\).
Definition
Tangent Line
A tangent line touches a curve at exactly one point (locally) and has the same direction as the curve at that point. Its slope equals the instantaneous rate of change — the derivative — of the function at the point of tangency:
\[m_{\text{tan}} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]
Key Geometric Idea
A curve looks like a straight line if you zoom in closely enough at any smooth point. The tangent line is the limit of what you see when you zoom in infinitely. This is the geometric heart of differential calculus.
Sections 3.3 – 3.4
The Formal Derivative and Its Notation
Limit Definition · Difference Quotient · Computing from First Principles · Notation
Section 3.3
The Derivative as a Limit
Formal Definition
The Derivative
The derivative of a function \(f\) at the point \(x = a\), written \(f'(a)\), is defined as:
\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]
provided this limit exists. The expression \(\frac{f(a+h) - f(a)}{h}\) is called the difference quotient.
Why the Limit Is Not "Dividing by Zero"
When we write \(\lim_{h \to 0}\), we ask: what value does the expression approach as \(h\) gets arbitrarily small? We simplify the difference quotient algebraically while \(h \neq 0\), then evaluate what it approaches. This is a fundamentally different question than setting \(h = 0\).
Example 3.3.1
Derivative of \(f(x) = x^2\) at \(x = 3\)
- ProblemUse the limit definition to find \(f'(3)\) for \(f(x) = x^2\).
- RepresentThe difference quotient is \(\dfrac{f(3+h) - f(3)}{h}\).
- Calculate
f(3+h) = (3+h)² = 9 + 6h + h²
f(3) = 9
[f(3+h) − f(3)] / h = (9 + 6h + h² − 9) / h
= (6h + h²) / h
= h(6 + h) / h
= 6 + h
f'(3) = lim_{h→0} (6 + h) = 6
- CheckWe divided by \(h\) while \(h \neq 0\). After simplification, \(6 + h\) is well-defined at \(h = 0\), so the limit is 6. ✓
- InterpretThe instantaneous rate of change of \(f(x) = x^2\) at \(x = 3\) is 6. The slope of the tangent line to the parabola at \((3, 9)\) is 6.
Answer: \(f'(3) = 6\)
Example 3.3.2
Derivative of \(f(x) = x^2\) at a General Point
- ProblemUse the limit definition to find \(f'(x)\) for \(f(x) = x^2\).
- ThinkKeep the input variable as \(x\) rather than a specific number. This produces a formula for the derivative at any point.
- Calculate
f(x+h) = (x+h)² = x² + 2xh + h²
[f(x+h) − f(x)] / h = (x² + 2xh + h² − x²) / h
= (2xh + h²) / h
= 2x + h
f'(x) = lim_{h→0} (2x + h) = 2x
- Check\(f'(x) = 2x\) gives \(f'(3) = 6\) ✓ and \(f'(0) = 0\) (horizontal tangent at the vertex) ✓.
- InterpretFor every point on \(y = x^2\), the slope of the tangent line is \(2x\). The derivative is not a single number — it is a function of \(x\).
Answer: \(f'(x) = 2x\)
Example 3.3.3
Derivative of a Linear Function
- ProblemFind \(f'(x)\) for \(f(x) = 3x - 5\).
- Calculate
[f(x+h) − f(x)] / h = [3(x+h) − 5 − (3x − 5)] / h
= [3x + 3h − 5 − 3x + 5] / h
= 3h / h = 3
f'(x) = lim_{h→0} 3 = 3
- InterpretThe derivative of a linear function is its slope constant. Derivatives generalize the idea of slope.
Answer: \(f'(x) = 3\)
Example 3.3.4
Derivative of \(f(x) = \sqrt{x}\)
- ProblemFind \(f'(x)\) for \(f(x) = \sqrt{x}\) using the limit definition.
- ThinkThe difference quotient involves \(\sqrt{x+h} - \sqrt{x}\). To evaluate the limit, rationalize the numerator by multiplying by the conjugate.
- Calculate
[f(x+h) − f(x)] / h = (√(x+h) − √x) / h
Multiply by (√(x+h) + √x) / (√(x+h) + √x):
= [(x+h) − x] / [h(√(x+h) + √x)]
= h / [h(√(x+h) + √x)]
= 1 / (√(x+h) + √x)
f'(x) = lim_{h→0} 1/(√(x+h)+√x) = 1/(2√x)
- CheckThis requires \(x > 0\). At \(x = 0\), the formula produces a zero denominator — the derivative does not exist there. The graph of \(y = \sqrt{x}\) has a vertical tangent at the origin. ✓
- InterpretFor \(x > 0\), the slope of the tangent line to \(y = \sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\). As \(x\) increases, the slope decreases — the curve becomes flatter.
Answer: \(f'(x) = \dfrac{1}{2\sqrt{x}}\), for \(x > 0\)
Section 3.4
Derivative Notation
The derivative has several notations. Each is useful in different contexts. Every calculus student should be comfortable reading and writing all of them.
| Name | Form | Read As | Best Used For |
|---|
| Lagrange (prime) | \(f'(x)\) or \(y'\) | "f prime of x" | Compact; derivative as a function |
| Leibniz | \(\dfrac{dy}{dx}\) or \(\dfrac{d}{dx}[f(x)]\) | "dee y dee x" | Units, related rates, integration |
| Newton (dot) | \(\dot{y}\) | "y dot" | Physics; time derivatives |
| Operator | \(Df(x)\) | "D of f" | Advanced mathematics |
To denote the derivative evaluated at a specific point \(x = a\):
\[f'(a) \qquad \text{or} \qquad \left.\frac{dy}{dx}\right|_{x=a}\]
Important Warning — Leibniz Notation
\(\frac{dy}{dx}\) is not a fraction. It is a limit of a ratio — a single symbol representing a limit. It often behaves like a fraction (and you will use it that way with the chain rule and implicit differentiation), but those manipulations require careful justification. Do not split \(\frac{dy}{dx}\) into separate pieces without a mathematical reason.
Sections 3.5 – 3.9
Reading, Estimating, and Visualizing Derivatives
Graphs · Tables · Units · Derivative as a Function · Graphs of f and f′
Section 3.5
Estimating Derivatives from Graphs
If you have a graph of a function, you can estimate the derivative at a point by estimating the slope of the tangent line there. The key: the derivative at a point equals the slope of the tangent line at that point. If the tangent line rises steeply, the derivative is large and positive. If it falls, the derivative is negative. If it is horizontal, the derivative is zero.
Example 3.5.1
Interpreting Derivative Sign from a Graph
- SituationA smooth curve \(y = f(x)\) rises from left to right, reaches a peak at \(x = 2\), then falls — shaped like a hill.
- Analyze
For x < 2: curve is rising → tangent has positive slope → f'(x) > 0
At x = 2: curve has a peak → tangent is horizontal → f'(2) = 0
For x > 2: curve is falling → tangent has negative slope → f'(x) < 0
- InterpretThe sign of the derivative tells you whether the function is increasing, decreasing, or momentarily flat — a foundational idea developed fully in Chapter 7.
Example 3.5.2
Estimating a Derivative from a Graph
- SituationA graph of \(f(x) = x^2\) is drawn on a coordinate grid. A tangent line at \((1, 1)\) appears to pass through \((0, -1)\) and \((2, 3)\).
- Calculate\(\text{Slope} = \dfrac{3 - (-1)}{2 - 0} = \dfrac{4}{2} = 2\)
- CheckThe algebraic result: \(f'(x) = 2x\), so \(f'(1) = 2\). ✓
Answer: \(f'(1) \approx 2\)
Section 3.6
Estimating Derivatives from Tables
When you have numerical data rather than a formula or graph, estimate the derivative using the symmetric difference quotient:
\[f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}\]
Why Symmetric?
Using a symmetric interval centered at the point gives a more accurate estimate than using only a one-sided difference, because it averages the behavior on both sides of the point.
Example 3.6.1
Estimating a Derivative from a Table
- ProblemUse the table to estimate \(f'(1.0)\).
- Table
| \(x\) | 0.90 | 0.95 | 1.00 | 1.05 | 1.10 |
|---|
| \(f(x)\) | 0.810 | 0.903 | 1.000 | 1.103 | 1.210 |
- CalculateUse \(h = 0.05\): \[f'(1.0) \approx \frac{f(1.05) - f(0.95)}{2(0.05)} = \frac{1.103 - 0.903}{0.10} = \frac{0.200}{0.10} = 2.00\]
- CheckIf \(f(x) = x^2\), then \(f'(1) = 2(1) = 2\). The data is consistent. ✓
Answer: \(f'(1.0) \approx 2.00\)
Section 3.7
Derivatives and Units
Units of a Derivative
The units of \(f'(x)\) are always units of f(x) per units of x. This is not bookkeeping — it reveals what the derivative is actually measuring.
| Context | Function | Derivative | Units of \(f'\) | Meaning |
|---|
| Motion | \(s(t)\) in km, \(t\) in hr | \(s'(t)\) | km/h | Velocity |
| Economics | \(C(x)\) in dollars, \(x\) = items | \(C'(x)\) | dollars/item | Marginal cost |
| Biology | \(P(t)\) in thousands of cells, \(t\) in hr | \(P'(t)\) | thousands of cells/hr | Growth rate |
Dimensional Checking
When you compute a derivative and the units seem wrong, that is a signal to recheck the setup. A result of "7 meters per meter" should prompt you to ask whether you differentiated the right function with respect to the right variable. Units always carry physical meaning.
Section 3.8
The Derivative as a Function
In Section 3.3, the derivative was computed at a specific point \(x = a\), producing a single number \(f'(a)\). But the process can be repeated for any input value. When \(x\) is left as a variable, the result is a new function — the derivative function \(f'(x)\).
Definition
The Derivative Function
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
This assigns to each input \(x\) the slope of the tangent line to \(f\) at that point. Its domain may be smaller than the domain of \(f\) — every point where the limit fails to exist is excluded.
Illustration
The Derivative Function of \(f(x) = x^2\)
- ResultFrom Example 3.3.2: \(f'(x) = 2x\).
- Meaning
f'(−3) = −6 → tangent line at (−3, 9) has slope −6
f'( 0) = 0 → tangent line at (0, 0) is horizontal (vertex)
f'( 1) = 2 → tangent line at (1, 1) has slope 2
f'( 5) = 10 → tangent line at (5, 25) has slope 10
- InterpretThe derivative function \(f'(x) = 2x\) is linear. It describes how the slope of the parabola changes as \(x\) varies.
Section 3.9
Graphs of \(f\) and \(f'\)
Reading \(f'\) from the Graph of \(f\)
Where \(f\) is increasing → \(f'(x) > 0\) (above the axis)
Where \(f\) is decreasing → \(f'(x) < 0\) (below the axis)
Where \(f\) has a local max or min → \(f'(x) = 0\) (crosses the axis)
Where \(f\) increases rapidly → \(f'\) is large and positive
Where \(f\) decreases rapidly → \(f'\) is large and negative
Graph Instruction — Paired Graphs of \(f\) and \(f'\)
Draw two coordinate axes, one above the other, sharing the same horizontal scale. Top graph: a smooth curve \(y = f(x)\) that rises until \(x = 1\) (local max), falls until \(x = 3\) (local min), then rises again. Bottom graph: the corresponding \(y = f'(x)\) — positive (above axis) for \(x < 1\), zero at \(x = 1\), negative for \(1 < x < 3\), zero at \(x = 3\), positive for \(x > 3\). This pairing — \(f\) on top, \(f'\) below — recurs throughout the course.
Example 3.9.1
Matching \(f\) and \(f'\)
- Situation\(f\) is constant at \(y = 2\) for \(x < 0\); increases steadily from \(x = 0\) to \(x = 4\); then decreases for \(x > 4\).
- Analyze
x < 0: f is flat → f'(x) = 0
0 < x < 4: f is rising → f'(x) > 0
x = 4: f has a peak → f'(4) = 0
x > 4: f is falling → f'(x) < 0
- Answer\(f'\) is zero for \(x < 0\), a positive constant on \((0, 4)\) if \(f\) rises linearly, zero at \(x = 4\), and negative for \(x > 4\).
Sections 3.10 – 3.11
When Derivatives Fail to Exist
Differentiability · Corners · Cusps · Vertical Tangents · Discontinuities
Section 3.10
Differentiability
Definition
Differentiability
A function \(f\) is differentiable at \(x = a\) if \(f'(a)\) exists — that is, if \(\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\) exists and is a finite number.
Key Theorem
If \(f\) is differentiable at \(x = a\), then \(f\) is continuous at \(x = a\).
The contrapositive: if \(f\) is not continuous at \(a\), then \(f\) is not differentiable at \(a\).
However: continuity does not guarantee differentiability. A function can be continuous at a point and still fail to have a well-defined tangent line there.
Section 3.11
Corners, Cusps, Vertical Tangents, and Discontinuities
There are four main geometric situations in which the derivative fails to exist at a point. Recognizing them visually and algebraically is essential.
| Situation | Graph Feature | Classic Example | Derivative |
|---|
| Smooth curve | Well-defined tangent line | \(f(x) = x^2\) | Exists |
| Corner | Sharp turn; two different one-sided slopes | \(f(x) = |x|\) at \(x = 0\) | Does not exist |
| Cusp | Pointed tip; slopes blow up to \(\pm \infty\) | \(f(x) = x^{2/3}\) at \(x = 0\) | Does not exist |
| Vertical tangent | Slope approaches \(\pm \infty\) | \(f(x) = x^{1/3}\) at \(x = 0\) | Does not exist |
| Discontinuity | Jump, hole, or asymptote | Any jump discontinuity | Does not exist |
Example 3.11.1
The Corner at \(f(x) = |x|\), \(x = 0\)
- ProblemShow algebraically that \(f(x) = |x|\) is not differentiable at \(x = 0\).
- Left limit\[\lim_{h \to 0^-} \frac{|0+h| - 0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\]
- Right limit\[\lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\]
- ConclusionThe left and right one-sided limits are \(-1\) and \(+1\) respectively. They are not equal, so the two-sided limit does not exist. The derivative of \(|x|\) at \(x = 0\) does not exist.
Answer: \(f(x) = |x|\) is not differentiable at \(x = 0\) because the one-sided derivatives differ.
Graph Instruction — Corner
Draw \(y = |x|\). Mark the corner at the origin. Label the slope as \(-1\) to the left and \(+1\) to the right. Draw the two one-sided "tangent" rays pointing in different directions. Note that no single line could be called the tangent at \(x = 0\).
Student Warning — "Looks Smooth" Is Not Enough
A function may appear smooth on a rough sketch and still fail to be differentiable. Always check one-sided limits near sharp turns, pointed tips, or steep vertical sections. Visual inspection is a starting point, not a proof.
Sections 3.12 – 3.14
Motion, Common Mistakes, and the Road Ahead
Position · Velocity · Acceleration · Common Errors · Preparing for Differentiation Rules
Section 3.12
Motion: Position, Velocity, and Acceleration
The Derivative Chain for Motion
Let \(s(t)\) = position of an object at time \(t\).
Velocity: \(v(t) = s'(t) = \dfrac{ds}{dt}\) — instantaneous rate of change of position.
Acceleration: \(a(t) = v'(t) = s''(t) = \dfrac{d^2s}{dt^2}\) — instantaneous rate of change of velocity.
Speed: \(|v(t)|\) — always non-negative; the magnitude of velocity.
Speeding Up vs. Slowing Down
An object speeds up when velocity and acceleration have the same sign. It slows down when they have opposite signs. A negative velocity alone does not mean slowing down — it means moving in the negative direction.
Example 3.12.1
Position, Velocity, and Direction of Motion
- ProblemA particle moves along a straight line with \(s(t) = t^3 - 6t^2 + 9t\) (meters, seconds), \(t \geq 0\). (a) Find velocity at \(t = 2\). (b) When is the particle at rest? (c) When does it move in the positive direction?
- Differentiate\(v(t) = s'(t) = 3t^2 - 12t + 9\) (using rules from Chapter 4, which can be verified with the limit definition)
- Part a\(v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3\) m/s
- Part b
3t² − 12t + 9 = 0
t² − 4t + 3 = 0
(t − 1)(t − 3) = 0
t = 1 and t = 3
- Part cSign of \(v(t) = 3(t-1)(t-3)\): positive for \(0 \leq t < 1\) (both factors negative → product positive); negative for \(1 < t < 3\); positive for \(t > 3\).
- InterpretThe particle starts moving in the positive direction, reverses at \(t = 1\), moves backward until \(t = 3\), then moves forward again. Speed at \(t = 2\) is \(|-3| = 3\) m/s.
Answer: \(v(2) = -3\) m/s; at rest at \(t = 1\) and \(t = 3\); positive direction for \(t \in [0,1) \cup (3, \infty)\).
Example 3.12.2
Acceleration and the Second Derivative
- ProblemFor \(s(t) = t^3 - 6t^2 + 9t\), find the acceleration at \(t = 2\). Is the particle speeding up or slowing down at that moment?
- Calculate
v(t) = 3t² − 12t + 9
a(t) = v'(t) = 6t − 12
a(2) = 6(2) − 12 = 0 m/s²
- InterpretAt \(t = 2\), acceleration is zero — the velocity is momentarily constant. Near \(t = 1.5\): \(v(1.5) = -2.25\) m/s and \(a(1.5) = -3\) m/s² — both negative, so the particle is speeding up (magnitude increasing). Near \(t = 2.5\): \(v < 0\) but \(a > 0\) — opposite signs, so the particle is slowing down.
Answer: \(a(2) = 0\) m/s². The particle is momentarily at constant velocity at \(t = 2\).
Section 3.13
Common Derivative Concept Mistakes
Mistake 1 — Confusing \(f(a)\) with \(f'(a)\)
"The function value at x = 3 is 9, so the derivative at x = 3 is 9."
Correction: \(f(a)\) is the height on the graph. \(f'(a)\) is the slope of the tangent line. They are different quantities with different units. A tall, nearly flat curve can have a small derivative.
Mistake 2 — Assuming Continuity Implies Differentiability
"The function is continuous everywhere, so the derivative exists everywhere."
Correction: Continuity is necessary but not sufficient for differentiability. \(f(x) = |x|\) is continuous everywhere but not differentiable at \(x = 0\). Always check for corners, cusps, and vertical tangents.
Mistake 3 — Treating \(\frac{dy}{dx}\) as an Ordinary Fraction
"I can just move dx to the other side like any variable."
Correction: \(\frac{dy}{dx}\) is a limit of a ratio, not a ratio itself. Manipulating it as a fraction requires careful justification. Rules for when this is valid are taught in the chain rule and implicit differentiation chapters.
Mistake 4 — Confusing Secant Slope with Derivative
"I drew a line through two points on the curve and got slope 4, so the derivative is 4."
Correction: The secant line gives average rate of change, not instantaneous rate of change. The derivative requires a limit as the two points merge into one.
Mistake 5 — Thinking \(f'(a) = 0\) Means \(f(a) = 0\)
"If f'(a) = 0, then f(a) = 0."
Correction: A zero derivative means the tangent line is horizontal — the function's rate of change is zero at that point. The function can have any nonzero value. For example, \(f(x) = x^2 + 5\) has \(f'(0) = 0\) but \(f(0) = 5\).
Mistake 6 — Confusing Velocity and Speed
"Velocity is −3 m/s, so the car is slowing down."
Correction: Negative velocity means motion in the negative direction, not deceleration. Speed is \(|v(t)|\). Whether an object is speeding up or slowing down depends on whether velocity and acceleration have the same or opposite signs.
Mistake 7 — Applying the Derivative at a Discontinuity
"The difference quotient gave a number at the jump, so the derivative exists there."
Correction: The derivative does not exist at any point of discontinuity. The limit of the difference quotient cannot be finite if the function jumps to a nonzero value as \(h \to 0\).
Section 3.14
Preparing for Differentiation Rules
The limit definition of the derivative is the correct foundation. It defines what the derivative is. But computing derivatives directly from the definition — expanding algebraic expressions, rationalizing, taking limits — is slow for complicated functions. In Chapter 4, you will learn differentiation rules: formulas that allow you to compute derivatives quickly and reliably. These rules are not shortcuts invented for convenience — they are theorems proven from the limit definition. You are learning the definition first so that when you use the rules later, you understand why they work.
Readiness Checklist — Before Moving to Chapter 4
You should be able to: (1) write the limit definition from memory; (2) compute the derivative of a simple polynomial or radical using the definition; (3) identify the four situations where a derivative fails to exist; (4) estimate derivatives from graphs and tables; (5) attach units to a derivative and explain what it measures; (6) describe the relationship between the graphs of \(f\) and \(f'\); (7) solve a basic motion problem using derivatives. If any of these feel uncertain, revisit the relevant section before continuing.
Chapter 3 — Back Matter
Chapter Summary
- Sec 3.1The average rate of change of \(f\) over \([a,b]\) is \(\frac{f(b)-f(a)}{b-a}\) — the slope of the secant line. As the interval shrinks to zero, this approaches the instantaneous rate of change.
- Sec 3.2A secant line passes through two points on a curve; a tangent line touches the curve at one point. The tangent line arises as the limiting position of secant lines when one point approaches the other.
- Sec 3.3The derivative \(f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\). The difference quotient measures average rate of change; the limit makes it instantaneous. Derivatives can be computed for polynomials, radicals, and reciprocals using algebraic simplification.
- Sec 3.4Derivative notation: \(f'(x)\), \(\frac{dy}{dx}\), \(\frac{d}{dx}[f(x)]\), \(\dot{y}\). Leibniz notation is not a fraction; it is a limit of a ratio. Evaluate at a point using \(f'(a)\) or \(\left.\frac{dy}{dx}\right|_{x=a}\).
- Sec 3.5From a graph: where \(f\) rises, \(f' > 0\); where \(f\) falls, \(f' < 0\); where \(f\) has a horizontal tangent, \(f' = 0\). Estimate the derivative by estimating the slope of the tangent line using two readable points.
- Sec 3.6Estimate derivatives from tables using the symmetric difference quotient \(f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}\). Using a centered interval gives a more accurate approximation than a one-sided difference.
- Sec 3.7Units of \(f'(x)\) = units of \(f\) per unit of \(x\). Velocity, marginal cost, and growth rate are all derivatives with meaningful units. Checking units confirms physical meaning and catches setup errors.
- Sec 3.8The derivative function \(f'(x)\) assigns to each \(x\) the slope of the tangent at that point. Its domain may be smaller than the domain of \(f\).
- Sec 3.9The graph of \(f'\) shows the slope of \(f\) at every point. Where \(f\) is increasing, \(f'\) is positive; where decreasing, negative; where horizontal, zero.
- Sec 3.10Differentiability at a point means the derivative limit exists and is finite. Differentiability implies continuity; continuity does not imply differentiability.
- Sec 3.11Derivatives fail to exist at corners, cusps, vertical tangents, and discontinuities. The classic corner is \(|x|\) at \(x = 0\), where one-sided derivatives are \(-1\) and \(+1\).
- Sec 3.12\(v(t) = s'(t)\) is velocity; \(a(t) = v'(t) = s''(t)\) is acceleration. Speeding up when \(v\) and \(a\) have the same sign; slowing down when opposite. Speed is \(|v(t)|\).
- Sec 3.13Seven common errors: confusing \(f(a)\) with \(f'(a)\); assuming continuity implies differentiability; treating \(\frac{dy}{dx}\) as a fraction; confusing secant slope with derivative; assuming \(f'(a) = 0\) means \(f(a) = 0\); confusing velocity and speed; computing a derivative at a discontinuity.
- Sec 3.14The limit definition is the foundation. Chapter 4 builds differentiation rules as theorems from this definition, allowing faster and more reliable computation.
Key Terms
average rate of change
The slope of the secant line between two points: \(\frac{f(b)-f(a)}{b-a}\). Measures total change per unit of input over an interval.
difference quotient
The expression \(\frac{f(a+h)-f(a)}{h}\), representing average rate of change over an interval of width \(h\).
instantaneous rate of change
The limit of the average rate of change as the interval shrinks to zero; the derivative at a point.
secant line
A line through two points on a curve. Its slope is the average rate of change over the interval between the points.
tangent line
The limiting position of secant lines as one point approaches another; touches the curve at one point with slope equal to the derivative.
derivative
The limit \(f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\). The instantaneous rate of change; the slope of the tangent line.
derivative function
The function \(f'(x)\) that assigns to each input the slope of the tangent to \(f\) at that point.
differentiability
A function is differentiable at a point if its derivative exists there. Differentiability implies continuity; the converse is false.
corner
A point where a function is continuous but the graph changes direction sharply, causing the two one-sided derivatives to differ. Example: \(|x|\) at \(x=0\).
cusp
A pointed tip on a graph where the slopes from each side blow up to \(\pm\infty\). Derivative does not exist. Example: \(x^{2/3}\) at \(x=0\).
vertical tangent
A point where the tangent line is vertical — slope is undefined. Example: \(x^{1/3}\) at \(x=0\).
position function \(s(t)\)
A function giving the location of an object at time \(t\).
velocity \(v(t) = s'(t)\)
The derivative of position; instantaneous rate of change of position. Can be positive, negative, or zero.
speed \(|v(t)|\)
The absolute value of velocity; always non-negative. Measures how fast an object moves regardless of direction.
acceleration \(a(t) = v'(t)\)
The derivative of velocity; the second derivative of position. Measures how quickly velocity is changing.
second derivative \(f''(x)\)
The derivative of the derivative function. Describes how the slope of \(f\) is changing. Developed fully in Chapter 7.
Concept Review Questions
- Explain in your own words why the average rate of change over an interval is not the same as the instantaneous rate of change at a point. When are they equal?
- A secant line and a tangent line both pass through a point on a curve. How do they differ? Which one requires a limit to define?
- Why is it necessary to take a limit when defining the derivative? Why can't you compute the slope of the line from \((a, f(a))\) to \((a, f(a))\)?
- Explain why every differentiable function must be continuous, but not every continuous function must be differentiable. Give an example.
- A student claims: "If \(f'(3) = 0\), then \(f(3) = 0\)." Identify the error.
- Describe the four types of points where a derivative can fail to exist. Sketch one example of each.
- How does the graph of \(f'(x)\) relate to the graph of \(f(x)\)? What feature of \(f\) corresponds to \(f'\) crossing the \(x\)-axis?
- A velocity function is negative on an interval. What does this tell you about motion? Does it determine whether the particle is speeding up or slowing down?
Derivative Definition Practice
- \(f(x) = 3x + 7\)
- \(f(x) = x^2 - 4x\)
- \(f(x) = 5\)
- \(f(x) = x^3\)
- \(f(x) = \frac{1}{x}\), \(x \neq 0\)
- \(f(x) = \sqrt{2x}\), \(x > 0\)
- \(f(x) = x^2\), find \(f'(5)\)
- \(f(x) = 2x^2-3x\), find \(f'(1)\)
- \(f(x) = \sqrt{x+1}\), find \(f'(3)\)
- \(f(x) = \frac{1}{x+2}\), find \(f'(0)\)
- \(f(x) = 4x^2 - x + 2\)
- \(f(x) = x^3 - 2x\)
- \(f(x) = \dfrac{2}{x}\)
Graph-Based Derivative Practice
- A smooth curve \(y = f(x)\) is described below. State whether \(f'(x)\) is positive, negative, or zero at each location.
(a) At \(x = 2\), the function reaches a local maximum.
(b) At \(x = 5\), the function is increasing steeply.
(c) At \(x = 7\), the function is decreasing.
(d) At \(x = 10\), the function is constant.
- A function \(f\) satisfies: \(f(0) = 3\) with tangent slope \(-2\); and \(f(4) = -1\) with tangent slope \(5\).
(a) Write the equation of the tangent line at \(x = 0\).
(b) Write the equation of the tangent line at \(x = 4\).
(c) Is \(f\) increasing or decreasing at each point?
- A graph of \(f(x)\) shows: increasing for \(x < 3\); local max at \(x = 3\); decreasing for \(3 < x < 7\); local min at \(x = 7\); increasing for \(x > 7\). Sketch what the graph of \(f'(x)\) should look like.
- Use the table to estimate \(f'(2.0)\) using a symmetric difference quotient.
| \(x\) | 1.8 | 1.9 | 2.0 | 2.1 | 2.2 |
|---|
| \(f(x)\) | 2.94 | 3.48 | 4.00 | 4.52 | 5.06 |
Motion Practice
- A particle's position is \(s(t) = t^2 - 4t + 3\) (feet, seconds).
(a) Find the velocity function \(v(t)\).
(b) Find the velocity at \(t = 1\) and \(t = 4\).
(c) When is the particle at rest?
(d) When is the particle moving in the positive direction?
(e) Find the speed at \(t = 3\).
- A ball is dropped from 100 meters: \(s(t) = 100 - 4.9t^2\).
(a) Find the velocity as a function of time.
(b) What is the velocity at \(t = 3\) seconds? What does the sign mean?
(c) When does the ball hit the ground? (Set \(s(t) = 0\).)
(d) What is the velocity at the moment of impact?
- Position: \(s(t) = t^3 - 9t^2 + 15t + 2\) (cm, seconds), \(t \geq 0\).
(a) Find the velocity and acceleration functions.
(b) Find all times when the particle is at rest.
(c) Determine direction of motion on each interval.
(d) Is the particle speeding up or slowing down at \(t = 1\)?
Applications
- A company's revenue is \(R(x) = 20x - 0.05x^2\) dollars. (a) Find the marginal revenue function \(R'(x)\). (b) Find and interpret \(R'(100)\) with units. (c) Is revenue increasing or decreasing when 200 units are sold?
- A bacterial population is \(P(t) = 500 + 40t - t^2\) (thousands of cells, hours). (a) Find \(P'(t)\). (b) Find and interpret \(P'(10)\). (c) At what time does the population stop growing?
- A car's position is \(s(t) = 3t^2 + 2t\) meters, \(t\) in seconds. (a) Find the velocity at \(t = 5\). (b) Find the average velocity from \(t = 2\) to \(t = 5\). (c) Explain why the two answers differ.
Error Analysis
Each problem shows incorrect work. Identify the error and write the correct solution.
Problem 1
f'(3) for f(x)=x²: [f(3+h)−f(3)]/h = (9+h²−9)/h = h²/h = h → f'(3) = 0
What algebraic error was made in expanding \((3+h)^2\)? What is the correct answer?
Problem 2
"Since f(5) = 12, the slope of the tangent to f at x = 5 is 12."
What conceptual error is this? What additional information would you need to find the slope?
Problem 3
"Velocity at t = 3 is −8 m/s. Since velocity is negative, the particle is decelerating."
Identify the error. What information would determine whether the particle is speeding up or slowing down?
Problem 4
"f(x) = x^(2/3) is continuous at x = 0 and looks smooth, so it must be differentiable at x = 0."
Is this correct? Examine the one-sided limits of the difference quotient. What kind of point is \(x = 0\) on this graph?
Problem 5
For f(x)=√x: [f(x+h)−f(x)]/h = (√(x+h)−√x)/h. Student stops here and writes "limit = 0/0, undefined."
What technique should be applied next? Complete the computation correctly.
Chapter 3 Checkpoint
Chapter 3 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 3
Part A — Concept Questions
- State the limit definition of the derivative. In your own words, explain what each part of the formula means.
- A function \(g\) is continuous on all real numbers. Is \(g\) necessarily differentiable everywhere? Justify your answer with an example.
- Explain why the derivative of a linear function \(f(x) = mx + b\) should equal \(m\) everywhere. Verify this using the limit definition.
- Describe the geometric difference between a secant line and a tangent line. How does the tangent line arise from the secant through a limiting process?
Part B — Skill Problems
- Use the limit definition to find \(f'(x)\) for \(f(x) = 2x^2 + 3x - 1\).
- Find the derivative of \(f(x) = \frac{1}{x-1}\) at \(x = 3\) using the limit definition.
- A function \(h(x)\) has values: \(h(2.9) = 6.9\), \(h(3.0) = 7.3\), \(h(3.1) = 7.7\). Estimate \(h'(3.0)\) using a symmetric difference quotient with \(k = 0.1\).
- The tangent line to \(f\) at \((2, 5)\) has slope \(-3\). Write the equation of this tangent line.
- Identify each point below where the derivative does not exist and state the reason: "The graph of \(f\) includes a jump at \(x = 1\), a corner at \(x = 3\), a cusp at \(x = 5\), and a smooth section from \(x = 6\) to \(x = 10\)."
Part C — Applications
- A particle's position is \(s(t) = 2t^3 - 9t^2 + 12t\) meters, \(t\) in seconds. (a) Find the velocity and acceleration functions. (b) Find all times when the particle is at rest. (c) Is the particle speeding up or slowing down at \(t = 1\)?
- The cost to produce \(x\) units is \(C(x) = 0.01x^2 + 5x + 200\) dollars. (a) Find the marginal cost function \(C'(x)\). (b) Interpret \(C'(50)\) in context with units. (c) As production increases, is marginal cost increasing or decreasing?
Part D — Graph Interpretation
- The graph of \(f'(x)\) is positive for \(x < 2\), equals zero at \(x = 2\), negative for \(2 < x < 6\), equals zero at \(x = 6\), and positive for \(x > 6\). (a) On what intervals is \(f(x)\) increasing? (b) At what values of \(x\) does \(f\) have a local maximum or minimum? (c) Sketch a possible graph of \(f\).
Answers may be placed in the answer key.
Looking Ahead
Chapter 4: Differentiation Rules
Chapter 3 established what the derivative is. Chapter 4 will build differentiation rules — the constant rule, the power rule, constant multiple rule, sum and difference rules, product rule, quotient rule, and chain rule — as theorems derived from the limit definition. These rules allow you to differentiate any algebraic combination of functions efficiently and reliably. You will also compute higher-order derivatives and begin preparing for transcendental functions in Chapter 5. Every skill from Chapter 3 — recognizing derivative structure, interpreting meaning, checking units — carries directly forward.
MGU Library Connection
- Derivative Rules Formula Sheet — Appendix F of this textbook
- Common Derivatives Reference — Appendix G of this textbook
- Precalculus Readiness Review — Appendix A of this textbook
- Trigonometry and Precalculus, Chapters 1–3: Functions, Graphs, and Rates — prerequisite background
- MGU Physics Curriculum: Kinematics (velocity and acceleration as derivatives of position)
- MGU Economics Curriculum: Marginal analysis (derivatives of cost, revenue, and profit functions)
- Data Science Foundations and Machine Learning Mathematics: Gradient descent uses derivatives to minimize loss functions
- Differential Equations: Differential equations are relationships between functions and their derivatives — Chapter 13 of this textbook introduces them
- MGU Dictionary entries: derivative, difference quotient, differentiability, tangent line, secant line, instantaneous rate of change