A function can be continuous at a point yet still fail to have a derivative there — differentiability is a strictly stronger requirement than continuity, never the other way around.
Three ways a derivative can fail to exist at x0:
- A corner or cusp (the graph comes to a sharp point ∨ or ∧). Example: f(x)=∣x−2∣ at x=2 — continuous there, but f′(2−)=−1=1=f′(2+), so f′(2) does not exist.
- A vertical tangent. Example: f(x)=x1/3 at x=0 — continuous, but f′(0)=limx→0x−2/3=+∞, not a finite number.
- A discontinuity. Example: f(x)=⌊x⌋ at any integer n — not even continuous there, so certainly not differentiable; or a jump function like f(x)=x (x≤0), f(x)=x+1 (x>0), where f′(0+) fails to exist because the right-hand difference quotient blows up as the jump is approached.
These three cases are exhaustive: a function fails to be differentiable at a point of its domain precisely when one of them holds. In short, discontinuity always forces non-differentiability — but continuity alone never guarantees differentiability, as cases (1) and (2) show.
The one implication that always holds:
Theorem 10.1. If f is differentiable at x0, then f is continuous at x0. …