Unit 2Differentiability

Concept 4.248Built for May 2027 AP Exam

Differentiability

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Differentiability is the stronger property. If $f$ has a derivative at $a$ , then $f$ is continuous at $a$ , and the reason is structural rather than case-by-case: the difference quotient can only converge if its numerator shrinks at least as fast as its denominator. The implication runs one way only. A continuous function may still fail to have a derivative at a point, and a corner, a cusp, or a vertical tangent is the standing witness. On the BC papers this is rarely a computation. It appears as a statement to classify, or as a justification the grader is told to look for: when $r$ is twice differentiable, $r'$ is differentiable, so $r'$ is continuous, which is exactly what a theorem applied to $r'$ needs. That chain earned a point in 2022, so the sentence is worth practising until it writes itself.

On the AP Exam

Scoring on these items is about logic, not algebra. The grader looks for one thing: whether the direction of the arrow in your sentence matches the direction the hypothesis allows. Because nothing needs computing, points are lost inside a single word, usually by writing "continuous, therefore differentiable" in a justification that was otherwise correct.

What you’ll see: A statement to classify as always, sometimes, or never true; a piecewise function with a join to test; or a theorem that needs continuity as a precondition.
What you’ll do: Name which of the two properties you were given, name which one you were asked to supply, then check that the arrow points the right way.
Memory hook: Differentiable is the expensive one. It buys continuity at the same point, and it buys nothing in the other direction.
Use it on the exam: For MCQs: Read the statement as an implication and test it against a single counterexample, usually $f(x) = \lvert x \rvert$ at the origin. One counterexample kills a universal claim. For FRQs: No released part asks you to prove the implication. What the rule buys you is a sentence: a function given as twice differentiable has a derivative that is differentiable, hence continuous, which is the hypothesis a value-guaranteeing theorem needs.

Concept Review

The Arrow Runs One Way

Figure 1 — The solid arrow is the theorem and the crossed arrow is the error: differentiability forces continuity, and no amount of continuity produces a derivative. The three chips are continuous functions that fail at a single input.

Two labelled boxes joining differentiability to continuity by a solid arrow labelled the theorem, a crossed-out dashed arrow running back labelled the error, and three chips naming functions that are continuous but not differentiable at the origin

Four Statements, Two Truths

Statement	Symbolic form	Verdict
Differentiable at $a$ implies continuous at $a$	$D \Rightarrow C$	True
Continuous at $a$ implies differentiable at $a$	$C \Rightarrow D$	False
Not continuous at $a$ implies not differentiable at $a$	$\lnot C \Rightarrow \lnot D$	True — the contrapositive of row 1
Not differentiable at $a$ implies not continuous at $a$	$\lnot D \Rightarrow \lnot C$	False — the inverse, not the contrapositive

Rows 1 and 3 say the same thing. Rows 2 and 4 say the same wrong thing. An implication and its contrapositive are logically identical; an implication and its inverse are not, and the inverse is the mistake almost every student makes at least once.

Why the Implication Holds

Write the difference quotient and clear its denominator:

$\frac{f(a+h)-f(a)}{h} = Q(h) \qquad\text{so}\qquad f(a+h)-f(a) = h \cdot Q(h)$

If $f$ is differentiable at $a$ , then $Q(h) \to f'(a)$ as $h \to 0$ , while $h \to 0$ on its own. The product of something converging to a finite number and something converging to zero converges to zero, so

$\lim_{h \to 0}\bigl[f(a+h)-f(a)\bigr] = 0 \qquad\text{which is}\qquad \lim_{h\to 0} f(a+h) = f(a)$

That last line is continuity at $a$ . Nothing in the argument survives if $Q(h)$ fails to converge, which is exactly why the converse collapses: the algebra needs a limit that may not exist.

Three Witnesses Against the Converse

Function	Point	Continuous?	Differentiable?	What fails
$f(x) = \lvert x \rvert$	$x = 0$	Yes	No	One-sided quotients settle at $-1$ and $1$
$f(x) = x^{2/3}$	$x = 0$	Yes	No	One-sided quotients diverge with opposite signs
$f(x) = \sqrt[3]{x}$	$x = 0$	Yes	No	One-sided quotients both diverge to $+\infty$

Each function is unbroken at the point in question, and each fails the derivative for a different reason. That variety is the point: continuity controls whether the graph has a hole or a jump, and it says nothing about whether the graph has a corner.

Where the Continuity Is Actually Spent

The theorem is rarely the answer; it is the licence for the next step. Three places the licence gets used:

Mean Value Theorem. Requires continuity on $[a,b]$ and differentiability on $(a,b)$ . When a problem grants differentiability on the closed interval, the continuity hypothesis is already met, and saying so costs one clause.
Extreme Value Theorem. Requires continuity on a closed interval. A function described as differentiable on $[-3, 4]$ therefore attains both a maximum and a minimum there.
Intermediate Value Theorem on $f'$ . The BC version of the chain: $f$ twice differentiable means $f'$ is differentiable, so $f'$ is continuous, so $f'$ takes every value between any two of its values. The 2022 scoring guideline pays a point for exactly this sentence.

Common Misconception Check

Myth 1: "The graph has no break, so the derivative exists." Truth: No break gives continuity. A corner is unbroken and still has no derivative, because the two one-sided quotients disagree.
Myth 2: "If $f$ is not differentiable at $a$ , then $f$ is not continuous at $a$ ." Truth: This is the inverse, not the contrapositive. Non-differentiability may come from a corner, which is continuous.
Myth 3: "Continuity at every point implies differentiability at every point." Truth: Weierstrass built continuous functions that are differentiable nowhere. On the exam, one corner on a continuous curve is enough to defeat the claim.

Topic-Specific Study Method

The Direction Check

Borrowed idea: Name the direction. Differentiability is the harder property, so it hands continuity over for free, and continuity hands nothing back. Spend the gift at once, before the theorem that needs continuity appears. A corner is unbroken and still has no derivative, because the two one-sided quotients settle at different values while the function value never moves, which is why continuity answers a question about breaks and differentiability answers a question about slopes. Keep that asymmetry in front of you and the classification items stop being guesses.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A true/false statement	"If $f$ is continuous at $a$ , then $f'(a)$ exists"	The converse, stated as fact	Reject it with the corner counterexample
A piecewise join	Two formulas meeting at $x = 2$	Continuity and differentiability are two separate tests	Run both tests, in that order
A theorem setup	"The function is differentiable on $[0, 5]$ "	Continuity is a gift you should claim	Write the continuity clause before quoting the theorem
A graph with a sharp turn	An unbroken curve with a point	A break rules out differentiability, a corner does not	Check the one-sided slopes, not just the height

① Take $f(x) = \lvert x - 2 \rvert$ and answer two separate questions: is $f$ continuous at $2$ , and is $f$ differentiable at $2$ ? Write both answers in full sentences, and notice that the first is a limit of values while the second is a limit of quotients.

② For any piecewise definition in your practice set, compute the two one-sided limits of $f$ at the join, then compute the two one-sided limits of the difference quotient. Two pairs of numbers, two verdicts, and the order matters: a failed continuity test ends the analysis immediately.

③ Whenever a solution you are reading says "since $f$ is differentiable", write in the margin what that sentence is buying: continuity, and which theorem needs it. If nothing downstream needs it, the sentence was decoration.

Training & Error Diagnosis

Core Archetype (Simplest Form): A function $f$ is differentiable at $x = 3$ . A classmate concludes that $f'(3)$ is positive because the graph passes through the point $(3, 5)$ . Identify the error.

Standard Execution:

Separate the two claims. Differentiability is a statement that a limit exists, not a statement about its sign.

Name what is guaranteed. Continuity at $3$ , and the existence of a finite $f'(3)$ .

Name what is not. The sign of $f'(3)$ is unconstrained; the derivative may be negative, zero, or positive.

Name the second error. The height $f(3) = 5$ carries no information about the slope at $3$ .

Repair the sentence. "Differentiability at $3$ guarantees that $f$ is continuous at $3$ and that $f'(3)$ is a finite number."

Exam Variation (Context / Parameter Shift): Let $f(x) = \begin{cases} 3x - 1 & x \le 2 \cr x^2 + kx + 1 & x > 2 \end{cases}$ . Find the value of $k$ that makes $f$ continuous at $x = 2$ , then determine whether $f$ is differentiable there.

Solution: Continuity comes first. The left branch gives $f(2) = 5$ , and the right-hand limit is $4 + 2k + 1 = 5 + 2k$ , so continuity requires $k = 0$ . Differentiability comes second. The left-hand slope is $3$ throughout; the right-hand derivative is $2x + k$ , which is $4$ at $x = 2$ . The one-sided slopes disagree, so $f$ is continuous at $2$ but not differentiable there. The order is not optional: differentiability presupposes continuity, so the value of $k$ has to be settled before the slopes can be compared.

Common Traps & Failure Signals:

The converse swap (Signal: the word "therefore" joining continuity to differentiability): the implication was used backwards.
The single-test stop (Signal: slopes compared but values never checked): the join was tested for differentiability without testing continuity, and the answer passed a condition that was never verified.
The unclaimed hypothesis (Signal: a theorem quoted with no continuity sentence): differentiability was granted and never converted, leaving a gap the grader is instructed to look for.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Converse Shortcut $\rightarrow$ Write the direction before using it.
Silent Assumption: You called the curve differentiable because it looked unbroken.
Unclaimed Hypothesis: You were handed differentiability and never converted it into the continuity that the next theorem needs, so the justification went missing at exactly the step the grader was reading.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; The implication is judged inside multiple-choice statements, and it surfaces on free response only as a one-point justification inside a larger question
10-Year Total Count: 1 of 9 Exams · 1 FRQ Part; All 54 free-response questions from the nine released BC sets were read part by part
FRQ Years Appeared: 2022 (Q4b)

Practice Examples

MCQ 1

Reading the Implication

The function $h$ is differentiable at $x = -1$ . Which statement must be true?

MCQ 2

A Join That Works

Let $f(x) = \begin{cases} x^2 & x \le 1 \cr 2x - 1 & x > 1 \end{cases}$ . Which statement about $f$ at $x = 1$ is true?

MCQ 3

The Contrapositive

The graph of $g$ has a jump discontinuity at $x = 1$ . Which conclusion is justified?

FRQ 1 • A Join With a Parameter

Let $f$ be the function defined by

$f(x) = \begin{cases} 3x - 1 & x \le 2 \cr x^2 + kx + 1 & x > 2 \end{cases}$

where $k$ is a constant.

(a) Find the value of $k$ for which $f$ is continuous at $x = 2$ . Show the work that leads to your answer.

(b) Using the value of $k$ from part (a), determine whether $f$ is differentiable at $x = 2$ . Justify your answer.

(c) Explain why a function that is not continuous at a point cannot be differentiable there.

(d) The function $g$ is differentiable on $[-2, 5]$ with $g(-2) = 3$ and $g(5) = 3$ . Explain why $g$ must attain a maximum value on $[-2, 5]$ , and name the theorem that supplies the guarantee.

Cheatsheet