No, a function is not differentiable at a hole. The definition of the derivative at a point requires the function to actually have a value at that point, so if the function is undefined there (a hole), the derivative cannot exist. This is true even if the function behaves perfectly smoothly on either side of the hole.
Why the Definition Rules It Out
The derivative of a function at a point c is defined as:
f'(c) = lim as x→c of [f(x) – f(c)] / (x – c)
Notice that f(c) appears directly in the formula. If the function has a hole at c, meaning f(c) doesn’t exist, then the expression f(x) – f(c) is meaningless. You can’t subtract a number that isn’t there. The difference quotient itself is undefined, so the limit of it can’t exist either. As Ohio University course materials put it: if x₀ is not an element of the function’s domain, then f'(x₀) is left undefined and the function is not differentiable at that point.
The Continuity Connection
There’s a deeper reason this works out. A fundamental theorem in calculus states that differentiability implies continuity. If a function is differentiable at a point, it must be continuous there. Continuity at a point c requires three things: f(c) must be defined, the limit of f(x) as x approaches c must exist, and that limit must equal f(c).
A hole fails the very first requirement. The function isn’t defined at the point, so it can’t be continuous there. And if it’s not continuous, it can’t be differentiable. This gives you a quick logical shortcut: spot a discontinuity of any kind, and you can immediately conclude the function is not differentiable at that point.
A Common Example
Consider the function f(x) = (x² – 1) / (x – 1). For every x except 1, this simplifies to x + 1. The graph looks exactly like the line y = x + 1, but with a hole at x = 1 because the original expression has a zero in the denominator there.
The function behaves beautifully near x = 1. From both sides, the values approach 2. The slope approaching from both sides is 1. Everything about the function “wants” to be smooth and differentiable at that point. But technically, f(1) doesn’t exist, so the derivative at x = 1 is undefined.
This is what makes holes frustrating for students. The function isn’t misbehaving the way it does at a vertical asymptote or a sharp corner. The problem is purely that one point is missing from the domain.
Filling the Hole Fixes the Problem
The good news is that a hole is a fixable issue. Mathematicians call it a “removable discontinuity” (or “removable singularity”) precisely because you can remove it. If you define a new function g(x) that equals f(x) everywhere the original is defined, and also define g(c) to equal the limit of f(x) as x approaches c, you’ve filled the hole.
In the example above, you’d define g(x) = x + 1 for all x, including x = 1. Now g(1) = 2, the function is continuous, and the derivative g'(1) = 1 exists with no issues. MIT course materials describe this process as defining f(x) at the missing point to be the value that makes the function well-behaved, effectively erasing the singularity.
This is worth understanding because in practice, when you’re asked to “find the derivative” of something like (x² – 1)/(x – 1), you’re often expected to simplify first and work with the continuous version. But if someone specifically asks whether the original unsimplified function is differentiable at the hole, the answer is no.
The Geometric Intuition
Geometrically, the derivative at a point represents the slope of the tangent line to the curve at that point. A tangent line touches the curve at a specific location. If the curve has a hole, there’s no point on the curve for the tangent line to touch. You can draw a line that would be tangent if the point existed, but the formal definition requires contact with an actual point on the graph.
This is the same logic that applies to any type of discontinuity. Jump discontinuities, infinite discontinuities, and removable discontinuities (holes) all prevent a tangent line from being defined at that location, which means no derivative exists there.
Holes vs. Other Non-Differentiable Points
It helps to compare holes with other situations where differentiability fails, because the reasons are different.
- Holes (removable discontinuities): The function isn’t defined at the point. The derivative formula can’t even be set up. Fixable by defining the function value.
- Jump discontinuities: The function is defined but the left and right limits disagree. Continuity fails, so differentiability fails. Not fixable by redefining one point.
- Sharp corners or cusps: The function is defined and continuous, but the slope from the left doesn’t match the slope from the right. The limit in the derivative formula doesn’t exist.
- Vertical tangent lines: The function is continuous but the slope approaches infinity. The derivative is unbounded.
Holes are actually the “mildest” type of non-differentiability. The function is doing everything right except existing at the one point that matters. That’s why they’re removable, and why in most practical calculus work, you’ll simplify the function, fill the hole, and move on.