Differentiability, and Where Derivatives Fail
A function is differentiable where its derivative exists — and the derivative fails at corners, cusps, vertical tangents, and discontinuities. Differentiable always means continuous, but continuous does not mean differentiable.
By the end you'll be able to spot the four ways a derivative fails to exist, explain why \(|x|\) has no derivative at 0, and state which way the implication between differentiability and continuity runs.
Predict: for which of these five functions will the left and right secant slopes at \(x = 0\) disagree — or run off to infinity — as h shrinks? Pick each case, drag h toward 0, and check the verdict.
Watch the continuity line too: four of the five stay continuous at 0 — only the jump breaks continuity — yet only the smooth one is differentiable there.
left slope ≈ −1.000 · right slope ≈ +1.000 — the sides disagree, so f′(0) does not exist (a corner). Continuous at 0: yes · differentiable at 0: no
Differentiability is a stronger, smoother condition than continuity: the difference quotient must settle on a single finite value from both sides, and four kinds of points refuse.
\(f\) is differentiable at \(a\) if the limit defining \(f'(a)\) exists — the same limit \(\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\) you built two lessons ago, with one demand: the left-hand and right-hand versions must agree on a single finite number. \(f\) is differentiable on an interval if that holds at every point of the interval. And here's the theorem this lesson turns on: if \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\). Why? For the difference quotient to have a finite limit, its numerator \(f(a+h) - f(a)\) must shrink to 0 as \(h \to 0\) — and that is exactly continuity. The converse is false: \(|x|\) is continuous everywhere but not differentiable at 0. "Differentiable" is the smaller, more demanding class sitting inside "continuous."
Zoom in on a differentiable point and the graph flattens into a straight line — that's what "smooth" means. The four failures each refuse to flatten. Corners (like \(|x|\) at 0): the two sides zoom into two different lines, slopes \(-1\) and \(+1\). Cusps (like \(x^{2/3}\) at 0): both sides steepen toward vertical, from opposite directions. Vertical tangents (like \(x^{1/3}\) at 0): the graph zooms into one clean line — but a vertical one, and a vertical line's slope is not a finite number. Discontinuities: any jump, hole, or asymptote from the continuity lesson breaks differentiability outright. Notice the first three are all continuous — you can draw them without lifting your pen — which is exactly why continuity isn't enough.
Is \(f(x) = |x|\) differentiable at \(x = 0\)? Test the one-sided difference quotients. From the right (\(h > 0\)): \(\frac{|0+h| - |0|}{h} = \frac{h}{h} = 1\), so the right-hand limit is \(+1\). From the left (\(h < 0\)): \(\frac{|h|}{h} = \frac{-h}{h} = -1\), so the left-hand limit is \(-1\). The two sides disagree, so \(\lim_{h \to 0} \frac{f(0+h)-f(0)}{h}\) does not exist — \(f'(0)\) does not exist: a corner. Yet \(f\) IS continuous at 0: \(\lim_{x \to 0} |x| = 0 = f(0)\). Continuous, but not differentiable.
Now \(g(x) = x^{1/3}\) at \(x = 0\). The difference quotient is \(\frac{g(0+h) - g(0)}{h} = \frac{h^{1/3}}{h} = \frac{1}{h^{2/3}}\). Since \(h^{2/3} \ge 0\) for every \(h\), as \(h \to 0\) from either side the quotient \(\to\) ____ , so \(g'(0)\) ____ .
Reveal the answer
The quotient \(\to +\infty\) from both sides, so \(g'(0)\) does not exist — the slope is not a finite number. This is the vertical tangent: the graph of \(x^{1/3}\) passes through the origin standing straight up. Note \(g\) is still continuous at 0. Pick the vertical tangent case above and drag h toward 0 to watch both secant slopes climb together.
More info — why differentiability forces continuity
Recall the three-part continuity check from the continuity lesson: \(f(a)\) is defined, \(\lim_{x \to a} f(x)\) exists, and the limit equals the value. The third part is the same as \(\lim_{h \to 0} \left[ f(a+h) - f(a) \right] = 0\). Now write that difference as a product: \(f(a+h) - f(a) = \frac{f(a+h)-f(a)}{h} \cdot h\). If \(f\) is differentiable at \(a\), the first factor \(\to f'(a)\) (a finite number) and the second factor \(\to 0\), so by the product limit law the whole thing \(\to f'(a) \cdot 0 = 0\). Continuity follows. The OpenStax section in Dive deeper gives the full proof and more examples of each failure mode.
Check your understanding
Is \(f(x) = |x - 2|\) differentiable at \(x = 2\)?
You're told only that \(f\) is differentiable at \(x = 3\). Which statement MUST be true?
True or false: if a function is continuous at a point, it must be differentiable there.
The function \(g\) is continuous at 0. Near \(x = 0\), the one-sided difference quotients of \(g\) satisfy: left-hand quotient \(\to -\infty\) and right-hand quotient \(\to +\infty\). Which failure mode is this, and is \(g\) differentiable at 0?
Recap
- \(f\) is differentiable at \(a\) when the difference-quotient limit exists — the left-hand and right-hand slopes must agree on one finite number.
- Derivatives fail four ways: corners (sides disagree, e.g. \(|x|\)), cusps (sides \(\to \mp\infty\), e.g. \(x^{2/3}\)), vertical tangents (both sides \(\to\) the same infinity, e.g. \(x^{1/3}\)), and any discontinuity.
- Differentiable \(\Rightarrow\) continuous: a finite difference-quotient limit forces \(f(a+h) - f(a) \to 0\).
- Continuous \(\not\Rightarrow\) differentiable: \(|x|\) is continuous everywhere but has no derivative at 0.
Dive deeper
- OpenStax Calculus Volume 1 — 3.2 The Derivative as a Function Establish the differentiability-implies-continuity theorem and the corner/cusp/vertical-tangent failure modes
Sources
- Differentiability, and Where Derivatives Fail