Difficulty: Introductory to Intermediate | Prerequisites: limits (algebraic), basic algebra, equation of a line (y = mx + b)
The derivative is the central idea of differential calculus, and MAT 210 introduces it through the limit definition before moving on to shortcut rules. Before shortcuts, though, the course wants you to understand what a derivative actually measures: instantaneous rate of change. Average rate of change (the slope between two points) is the stepping stone. If you are comfortable with limits and can write the equation of a line from a point and a slope, you have everything you need here.
The average rate of change is the slope between two points on a function. The derivative shrinks that gap to zero and gives the instantaneous rate of change (the slope of the tangent line). You find it using the limit definition, then use the result to write tangent-line equations.
Average rate of change
The slope of the secant line between two points on a function: [f(b) − f(a)] / (b − a). Think of it as the "overall slope" between two input values.
Secant line
A line passing through two points on a curve. In simple terms, it connects point A to point B on the graph.
Difference quotient
[f(x + h) − f(x)] / h. The average rate of change written with the gap labelled h. This is the expression you take the limit of to get the derivative.
Derivative (f′(x))
The instantaneous rate of change of f at x, defined as lim h→0 of [f(x + h) − f(x)] / h. Think of it as the slope of the curve at a single point, not between two points.
Tangent line
A line that touches the curve at exactly one point and has the same slope as the curve at that point. In simple terms, it is the line that best approximates the curve right at that spot.
Slope-intercept form
y = mx + b, where m is the slope and b is the y-intercept. This is the form you write your tangent line equation in.
The average rate of change of f over an interval [a, b] is:
[f(b) − f(a)] / (b − a)
This is the slope of the straight line connecting the two points (a, f(a)) and (b, f(b)) on the graph.
When the function is given as a table of values, identify the rows for x = a and x = b, read off f(a) and f(b), and plug into the formula.
Worked example from the quiz:
The table gives x in days and f(x) in dollars. Find the average rate of change over [2, 6].
From the table: f(2) = 20, f(6) = 12.
Average rate of change = (12 − 20) / (6 − 2) = −8 / 4 = −2.
Answer: −2 dollars per day.
The unit of the average rate of change is always [unit of f(x)] per [unit of x]. In the quiz, f(x) is in dollars and x is in days, so the answer is dollars per day. Exams often ask you to state the unit explicitly.
The derivative of f(x) is defined as:
f′(x) = lim h→0 [f(x + h) − f(x)] / h
The process is mechanical once you have the steps down. Expand f(x + h), subtract f(x), simplify, factor out h from the numerator, cancel, then take the limit as h → 0.
Worked example from the quiz: f(x) = 3x² − 4x
Step 1: Write f(x + h).
f(x + h) = 3(x + h)² − 4(x + h) = 3(x² + 2xh + h²) − 4x − 4h = 3x² + 6xh + 3h² − 4x − 4h
Step 2: Compute f(x + h) − f(x).
(3x² + 6xh + 3h² − 4x − 4h) − (3x² − 4x) = 6xh + 3h² − 4h
Step 3: Divide by h.
(6xh + 3h² − 4h) / h = h(6x + 3h − 4) / h = 6x + 3h − 4
Step 4: Take the limit as h → 0.
lim h→0 (6x + 3h − 4) = 6x − 4
Answer: f′(x) = 6x − 4.
Forgetting to expand (x + h)² fully: it is x² + 2xh + h², not x² + h².
Sign errors when subtracting f(x): make sure you distribute the negative sign across every term of f(x).
Forgetting to factor h out of the numerator before cancelling.
Once you have f′(x), finding the tangent line at a specific point is a three-step process.
Step 1: Find the slope.
Plug the given x-value into f′(x). This gives you m, the slope of the tangent line.
Step 2: Find the point.
Plug the same x-value into f(x). This gives you the y-coordinate. The point on the curve is (x, f(x)).
Step 3: Write the line equation.
Use y = mx + b. Plug in the point and the slope to solve for b, then write the final equation.
Worked example from the quiz: f(x) = 3x² − 4x, tangent at x = 1
Slope: f′(1) = 6(1) − 4 = 2. So m = 2.
Point: f(1) = 3(1)² − 4(1) = 3 − 4 = −1. The point is (1, −1).
Equation: −1 = 2(1) + b, so b = −3.
Answer: y = 2x − 3.
Average rate of change: [f(b) − f(a)] / (b − a)
Limit definition of the derivative: f′(x) = lim h→0 [f(x + h) − f(x)] / h
Tangent line at x = a: y = f′(a)(x − a) + f(a), or equivalently, use y = mx + b with m = f′(a) and the point (a, f(a)) to solve for b.
Expansion to memorise: (x + h)² = x² + 2xh + h². Getting this wrong is the single most common source of errors in limit-definition problems.
Average rate of change is how businesses measure growth over a period: revenue this quarter minus revenue last quarter, divided by time. The derivative is the instantaneous version, the speedometer reading rather than the trip average. In finance, the derivative of a cost function tells a company exactly how much the next unit will cost to produce (marginal cost).
Students often confuse average rate of change with the derivative. Average rate of change uses two points and gives a slope over an interval. The derivative uses the limit and gives the slope at a single point.
Students sometimes skip the limit step and treat the difference quotient [f(x+h) − f(x)] / h as the derivative. It is not. You must take lim h→0.
Students forget to state units when asked for average rate of change. If f is in dollars and x is in days, the answer is in dollars per day, not just a number.
Students often expand (x + h)² as x² + h², missing the 2xh term. This error cascades through the entire derivative calculation.
⚠️ The limit definition of the derivative is almost guaranteed on Quiz 1. You will be asked to show every step, not just the final answer. Partial credit depends on the algebra being visible.
⚠️ Tangent line problems are a two-parter: find f′(x), then find the equation. Missing either half costs significant marks.
⚠️ Average rate of change problems will ask for units. Forgetting the unit costs a point on nearly every grading rubric.
⚠️ Know the difference between secant and tangent lines. The exam may ask you to explain the relationship in words.
True or false: The average rate of change and the derivative give the same value. False. The average rate of change is over an interval; the derivative is at a point.
Fill in the blank: The limit definition of the derivative is f′(x) = lim h→0 __________. [f(x + h) − f(x)] / h.
True or false: (x + h)² = x² + 2xh + h². True.
Fill in the blank: To find the tangent line at x = a, you need two things: the __________ and the __________. Slope (f′(a)) and the point (a, f(a)).
True or false: The unit of average rate of change is always just the unit of f(x). False. It is [unit of f(x)] per [unit of x].
Q: A table gives x in days and f(x) in dollars. f(2) = 20 and f(6) = 12. What is the average rate of change over [2, 6], and what are the units?
A: (12 − 20) / (6 − 2) = −8/4 = −2 dollars per day.
Q: Using the limit definition, find f′(x) for f(x) = 3x² − 4x.
A: Expand f(x+h) = 3x² + 6xh + 3h² − 4x − 4h. Subtract f(x) to get 6xh + 3h² − 4h. Divide by h to get 6x + 3h − 4. Take lim h→0 to get f′(x) = 6x − 4.
Q: Find the equation of the tangent line to f(x) = 3x² − 4x at x = 1.
A: f′(1) = 6(1) − 4 = 2 (the slope). f(1) = 3 − 4 = −1 (the y-value). Using y = mx + b: −1 = 2(1) + b, so b = −3. The tangent line is y = 2x − 3.
Q: What is the difference between a secant line and a tangent line?
A: A secant line passes through two points on the curve and its slope equals the average rate of change. A tangent line touches the curve at one point and its slope equals the derivative (instantaneous rate of change).
Q: If the average rate of change of a function over [1, 5] is 3 dollars per hour, and f(1) = 10, what is f(5)?
A: 3 = [f(5) − 10] / (5 − 1), so f(5) − 10 = 12, and f(5) = 22 dollars.
The limit definition of the derivative connects directly back to limits (the first doc in this set). Once MAT 210 moves past the limit definition, you will learn shortcut rules (power rule, product rule, chain rule), but those rules are derived from this same definition. The tangent line concept reappears in linear approximation and in optimisation problems, where you set f′(x) = 0 to find maxima and minima.
Average rate of change, instantaneous rate of change, derivative, limit definition of derivative, difference quotient, secant line, tangent line, slope, f prime, f′(x), point-slope form, slope-intercept, y = mx + b, marginal cost, MAT 210, Brief Calculus, ASU, Arizona State University, Quiz 1