Course: MAT210, Brief Calculus (Arizona State University) Section: 1.4 – The Difference Quotient and the Derivative Source: Lecture gapped notes
Difficulty: Introductory-Intermediate Prerequisites: Section 1.3 (Rates of Change), comfort with limits and basic algebra.
Tags: derivative, difference quotient, instantaneous rate of change, tangent line, secant line, limit definition of derivative, f prime, differentiability, continuity, MAT210
This section bridges the gap between approximating rates of change (what you did with tables in Section 1.3) and computing them exactly with algebra. The central idea: the slope of a secant line, which measures average rate of change, becomes the slope of a tangent line when you shrink the interval to zero. That tangent-line slope is the derivative. If you missed Section 1.3, go back and read the notes on average rate of change and secant lines first, because everything here builds directly on those ideas.
The difference quotient [f(a+h) − f(a)] / h gives the average rate of change (secant-line slope). Take its limit as h → 0 and you get the derivative f′(a), the exact instantaneous rate of change (tangent-line slope) at a single point. Replace the fixed point a with a variable x and you have the derivative function f′(x), which tells you the rate of change everywhere at once.
Average rate of change
The change in the output divided by the change in the input over an interval. Geometrically, it is the slope of the secant line through two points on the graph. In simple terms, this means: how fast something is changing on average between two points.
Secant line
A straight line that passes through two points on the graph of a function. Think of it as the line you draw between two dots on a curve to measure the overall slope between them.
Difference quotient
The expression [f(a+h) − f(a)] / h, which gives the slope of the secant line between the point (a, f(a)) and a nearby point (a+h, f(a+h)). In simple terms, this is the algebraic formula for average rate of change, written so you can let h shrink toward zero.
Instantaneous rate of change
The rate of change of a function at a single point, found by taking the limit of the difference quotient as h → 0. Think of it as: if average rate of change is your speed over a whole road trip, instantaneous rate of change is your speed at one exact moment.
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 the secant line becomes once the two points merge into one.
Derivative at a point, f′(a)
The limit of the difference quotient as h → 0 at x = a. Formally: f′(a) = lim(h→0) [f(a+h) − f(a)] / h. Read "f prime of a." This gives you the exact slope of the tangent line, and therefore the instantaneous rate of change, at the specific input a.
Derivative function, f′(x)
The function obtained by replacing the fixed value a with the variable x in the derivative definition: f′(x) = lim(h→0) [f(x+h) − f(x)] / h. In simple terms, instead of computing the derivative one point at a time, the derivative function gives you a formula that works for every x at once.
Differentiable (at a point)
A function is differentiable at x = a if f′(a) exists, meaning the limit definition produces a finite value. Think of it as: the curve is smooth enough at that point that you can draw exactly one tangent line.
Continuity (as it relates to differentiability)
A function is continuous at a point if there is no break, jump, or hole in the graph there. Continuity is a necessary condition for differentiability, but it is not sufficient on its own. In simple terms, the function must be connected at the point for the derivative to have any chance of existing, but being connected does not guarantee smoothness.
In Section 1.3, you approximated the instantaneous rate of change by plugging smaller and smaller values of Δt into the average-rate-of-change formula and reading the table.
The difference quotient formalises that same idea algebraically, so you no longer need a table.
The key substitution: replace the specific input value (e.g. 20) with a general value a, and replace Δt with h.
Definition of the difference quotient
For a function f at x = a, the difference quotient is:
\frac{f(a+h) - f(a)}{h}It represents the slope of the secant line between the points (a, f(a)) and (a+h, f(a+h)).
h is the horizontal distance between the two points. It can be positive or negative.
This expression is the starting point for finding the derivative. You simplify it algebraically, then take the limit as h → 0.
When you take the limit of the difference quotient as h → 0, the secant line becomes a tangent line, and you get the derivative:
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}This gives the instantaneous rate of change of f at x = a. The notation f′(a) is read "f prime of a."
The function must be continuous at a for this definition to apply, though continuity alone does not guarantee the derivative exists.
Worked example: f(x) = x², find f′(1)
Step 1: Write the difference quotient at a = 1.
[f(1+h) − f(1)] / h = [(1+h)² − 1] / h
Step 2: Expand and simplify.
(1 + 2h + h² − 1) / h = (2h + h²) / h = 2 + h
(This step assumes h ≠ 0, which is valid because h is approaching 0, not equal to it.)
Step 3: Take the limit as h → 0.
Since 2 + h is continuous, substitute directly: f′(1) = 2 + 0 = 2.
The slope of the tangent line to f(x) = x² at x = 1 is 2.
Worked example: R(t) = −0.02t² + 2t + 21, find R′(20)
Step 1: Compute R(20+h).
R(20+h) = −0.02(20+h)² + 2(20+h) + 21
Expand: −0.02(400 + 40h + h²) + 40 + 2h + 21
Simplify: −8 − 0.8h − 0.02h² + 40 + 2h + 21 = 53 + 1.2h − 0.02h²
Step 2: Form the difference quotient.
[R(20+h) − R(20)] / h
R(20) = −0.02(400) + 40 + 21 = 53
So: (53 + 1.2h − 0.02h² − 53) / h = (1.2h − 0.02h²) / h = 1.2 − 0.02h
Step 3: Take the limit as h → 0.
R′(20) = 1.2
Interpretation: at week 20, revenue is increasing at a rate of $1.2 million per week.
Computing f′(a) gives you the rate of change at one specific value. To get a formula that works at every x, replace the fixed number a with the variable x:
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}The result, f′(x), is itself a function. Plug in any x-value and you get the instantaneous rate of change there, no need to redo the whole difference-quotient process from scratch.
Worked example: f(x) = x² + 4, find f′(x)
Step 1: Write the difference quotient.
[f(x+h) − f(x)] / h = [(x+h)² + 4 − (x² + 4)] / h
Step 2: Expand and simplify.
(x² + 2xh + h² + 4 − x² − 4) / h = (2xh + h²) / h = 2x + h
Step 3: Take the limit as h → 0.
f′(x) = 2x
Now you can evaluate the derivative anywhere by substitution:
f′(1) = 2(1) = 2
f′(−3) = 2(−3) = −6
f′(0) = 2(0) = 0
This is far more efficient than running the difference quotient separately for each point.
If f′(a) exists, the function is called differentiable at x = a. Not every function has a derivative at every point. The graph must be both continuous and smooth (no sharp corners, no cusps) at that point.
Three cases where differentiability fails
Corner (sharp turn): The absolute value function f(x) = |x| is not differentiable at x = 0. The graph makes a sharp V, so there is no single tangent line.
Cusp: The function g(x) = (x − 2)^(2/3) is not differentiable at x = 2. The curve comes to a pointed tip and the tangent line would be vertical (undefined slope).
Discontinuity (jump, hole, or break): If the function is not continuous at a point, it cannot be differentiable there. A gap in the graph means there is no curve to be tangent to.
The continuity-differentiability relationship
Continuity at a point is a necessary condition for differentiability.
Continuity is not sufficient: a function can be continuous at a point but still fail to be differentiable there (corners, cusps).
If a function is differentiable at a point, it is guaranteed to be continuous there. (Differentiability implies continuity, but the reverse does not hold.)
In business modelling, nondifferentiable points can represent real phenomena: sudden pricing shifts, hard cutoffs in tax brackets, or threshold-based decision rules.
Difference quotient (slope of secant line):
\frac{f(a+h) - f(a)}{h}Derivative at a point (slope of tangent line):
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}Derivative function (rate of change at every x):
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}All three are built from the same skeleton. The difference quotient is the fraction; the derivative at a point adds the limit at a fixed a; the derivative function replaces a with x so the formula works everywhere.
The derivative is the mathematical tool behind any question of the form "how fast is this changing right now?" In business, R′(t) tells you how quickly revenue is growing at a specific week, which is more useful for decision-making than knowing the average growth over an entire quarter. In physics, the derivative of a position function gives velocity at an instant. In economics, marginal cost and marginal revenue are derivatives of the total cost and total revenue functions.
Students often think "h = 0" when simplifying the difference quotient. It does not. You assume h ≠ 0 while simplifying (which is what lets you cancel h from numerator and denominator), and only then take the limit as h approaches 0.
Students often assume that if a function is continuous at a point, it must be differentiable there. Continuity is necessary but not sufficient. The absolute value function at x = 0 is the classic counterexample: continuous, but not differentiable.
Students sometimes confuse the difference quotient (no limit) with the derivative (with the limit). The difference quotient alone gives the slope of a secant line, not the tangent line. You must take the limit to get the derivative.
Students sometimes forget to expand f(a+h) fully. The most common algebra error is writing f(a+h) = f(a) + f(h), which is wrong for almost every function. You must substitute (a+h) into the entire function expression.
⚠️ You will almost certainly be asked to find a derivative using the limit definition (the difference quotient method), not shortcut rules, on at least one exam problem. Know the three-step process cold: write the difference quotient, simplify algebraically, take the limit.
⚠️ Interpreting the derivative in context is a favourite exam question. If R′(20) = 1.2 and R is revenue in millions of dollars per week, you must be able to say: "At week 20, revenue is increasing at a rate of $1.2 million per week."
⚠️ Expect a question asking you to determine, from a graph, where a function is not differentiable and why (corner, cusp, or discontinuity).
⚠️ The relationship between continuity and differentiability is a common true/false or conceptual question. Know both directions: differentiable implies continuous (true), continuous implies differentiable (false).
True or false: The difference quotient gives the slope of the tangent line.
False. It gives the slope of the secant line. You need the limit to get the tangent.
Fill in the blank: f′(a) = lim(h→0) ____
[f(a+h) − f(a)] / h
True or false: If a function is continuous at x = 3, it must be differentiable at x = 3.
False. Continuity is necessary but not sufficient for differentiability.
True or false: If a function is differentiable at x = 5, it must be continuous at x = 5.
True. Differentiability implies continuity.
Fill in the blank: The derivative function f′(x) is obtained by replacing the fixed value ____ with the variable ____ in the derivative definition.
a with x
Q: Use the limit definition of the derivative to find f′(2) for f(x) = 3x² − x.
A: Difference quotient at a = 2: [f(2+h) − f(2)] / h = [3(2+h)² − (2+h) − (3·4 − 2)] / h = [3(4 + 4h + h²) − 2 − h − 10] / h = [12 + 12h + 3h² − 2 − h − 10] / h = (11h + 3h²) / h = 11 + 3h. Taking the limit as h → 0: f′(2) = 11.
Q: Find the derivative function f′(x) for f(x) = 5x + 1 using the limit definition.
A: [f(x+h) − f(x)] / h = [5(x+h) + 1 − (5x + 1)] / h = [5x + 5h + 1 − 5x − 1] / h = 5h / h = 5. Taking the limit: f′(x) = 5. (The derivative of a linear function is its slope, as expected.)
Q: A function is continuous at x = 4 but has a sharp corner there. Is it differentiable at x = 4? Explain.
A: No. A sharp corner means there is no unique tangent line at that point, so the derivative does not exist. Continuity is necessary for differentiability, but a corner makes the left-hand and right-hand slopes different, so the limit does not exist.
Q: If R(t) represents revenue in millions of dollars and R′(15) = −0.5, interpret this result.
A: At t = 15 (week 15), revenue is decreasing at a rate of $0.5 million per week. The negative sign means revenue is falling, not rising.
Q: Explain why you assume h ≠ 0 when simplifying the difference quotient, even though you eventually let h → 0.
A: If you set h = 0 directly in the difference quotient, you get the indeterminate form 0/0. By assuming h ≠ 0, you can cancel h from the numerator and denominator, simplifying the expression into a form that is continuous at h = 0. Only then do you substitute h = 0 to evaluate the limit.
This section connects directly to Section 1.3 (Rates of Change), which introduced the average rate of change and secant lines. The derivative formalises what you were approximating numerically there. Going forward, derivative shortcut rules (power rule, product rule, chain rule) will replace the difference-quotient process for efficiency, but they all rest on this limit definition. In business calculus, the derivative leads into marginal analysis (Sections 2.x), where you interpret f′(x) as marginal cost, marginal revenue, or marginal profit.
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