A differentiable function is a function whose derivative exists at every point in its domain, meaning its graph is smooth with a well-defined tangent line slope everywhere. In AP Calc, differentiability at a point guarantees continuity at that point, but continuity does not guarantee differentiability.
A differentiable function has a derivative at every point in its domain. In plain terms, you can find the slope of the tangent line anywhere on the graph. For the derivative to exist at a point, the limit of the difference quotient has to exist there, which means the left-hand and right-hand limits of that difference quotient must be equal.
The big idea from Topic 2.4 is a one-way street. If a function is differentiable at a point, it must be continuous there. But a continuous function can still fail to be differentiable. The classic failures are a sharp corner like f(x) = |x| at x = 0 (the left and right slopes disagree) and a vertical tangent like f(x) = ∛x at x = 0 (the slope is undefined because the tangent line is vertical). Think of differentiability as continuity plus smoothness. The graph can't jump, and it also can't have corners, cusps, or vertical tangents.
This term lives in Topic 2.4 (Connecting Differentiability and Continuity) in Unit 2, and it directly supports learning objective 2.4.A: Explain the relationship between differentiability and continuity. The essential knowledge here is exam gold. Differentiable implies continuous, the converse is false, and if a point isn't in the domain of f, it isn't in the domain of f'. Beyond Unit 2, the word "differentiable" is quietly doing work all over the exam. Theorems like the Mean Value Theorem require it as a hypothesis, and FRQs constantly say things like "a twice-differentiable function H" to license you to take derivatives without worry. Knowing exactly what that word guarantees (and what it doesn't) is the skill being tested.
Keep studying AP Calculus Unit 2
Visual cheatsheet
view galleryContinuous (Unit 1)
Differentiability and continuity are linked by a one-way implication. Differentiable at a point means continuous there, but f(x) = |x| is continuous at x = 0 and still not differentiable. AP loves testing students who flip this implication backwards.
Derivative (Unit 2)
A function is differentiable at a point exactly when the derivative exists there, meaning the limit of the difference quotient exists. "Differentiable function" is just the global version of "the derivative exists," applied to every point in the domain.
Piecewise function (Unit 2)
Piecewise functions are the exam's favorite differentiability trap. At the seam, you have to check two things in order. First the pieces must meet (continuity), then the left-hand and right-hand derivatives must match. Matching values alone isn't enough.
Critical Point (Unit 5)
Critical points happen where the derivative is zero or doesn't exist. Corners and cusps, the spots where a function fails to be differentiable, are exactly where 'derivative doesn't exist' critical points come from. So Unit 2's failure cases become Unit 5's candidates for maxes and mins.
Multiple-choice questions hit this two ways. One style shows you a graph with a corner, like h(x) = |x − 3|, and asks you to explain why the function is continuous but not differentiable at x = 3 (correct answer: the one-sided limits of the difference quotient disagree). Another style asks you for the procedure to test differentiability at a point, which is computing the left-hand and right-hand limits of the difference quotient and checking whether they're equal. On FRQs, "differentiable" usually appears as a given hypothesis. The 2018 FRQ on tree height and the 2019 particle motion FRQ both open with "twice-differentiable function," which is the College Board's way of telling you that derivatives exist, so theorems like the Mean Value Theorem are fair game. Your job is to recognize when that hypothesis matters and to cite it when justifying your answer.
Every differentiable function is continuous, but not every continuous function is differentiable. Continuity means the graph has no breaks. Differentiability adds smoothness on top of that, ruling out corners (like |x| at x = 0) and vertical tangents (like ∛x at x = 0). When an MCQ says "f is continuous at x = c, which must be true?", differentiability at c is NOT one of the safe conclusions. The implication only runs one direction.
A differentiable function has a derivative at every point in its domain, so you can find a tangent line slope anywhere on its graph.
If a function is differentiable at a point, it is automatically continuous there, but the reverse is not true.
A function fails to be differentiable at corners and cusps, where the left-hand and right-hand limits of the difference quotient disagree, like f(x) = |x| at x = 0.
A function also fails to be differentiable where the tangent line is vertical, like f(x) = ∛x at x = 0, because a vertical line has no slope.
If a point is not in the domain of f, it cannot be in the domain of f', so discontinuities always kill differentiability.
When an FRQ says a function is 'differentiable' or 'twice-differentiable,' that's permission to apply derivative-based theorems like the Mean Value Theorem.
It's a function whose derivative exists at every point in its domain, meaning the limit of the difference quotient exists everywhere. Graphically, it's smooth with no breaks, corners, cusps, or vertical tangents. This is the focus of Topic 2.4 in Unit 2.
No. Continuity is necessary for differentiability but not sufficient. f(x) = |x| is continuous at x = 0 but not differentiable there, because the slope is −1 from the left and +1 from the right. The implication only works the other way: differentiable does mean continuous.
First confirm it's continuous at the point (if not, you're done, it's not differentiable). Then compute the left-hand and right-hand limits of the difference quotient at that point. If they exist and are equal, the function is differentiable there. This is exactly the procedure AP multiple-choice questions ask about for functions like g(x) = |x − 3| at x = 3.
Because the left-hand and right-hand limits of the difference quotient are not equal. Coming from the left the slope is −1, and from the right it's +1. The graph has a sharp corner, and there's no single tangent line slope at that point.
It's a guarantee. Saying H is twice-differentiable (like the tree-height function in the 2018 FRQ) tells you H' and H'' both exist, so you can safely apply theorems that require differentiability, like the Mean Value Theorem, and reason about concavity. Cite that hypothesis in your justification to earn the point.