Connecting Differentiability and Continuity: When Derivatives Do and Do Not Exist
Introduced| English |
|---|
| differentiability/ˌdɪfəˌrenʃɪəˈbɪlɪti/ |
| continuous/kənˈtɪnjuːəs/ |
| corner/ˈkɔːnə/ |
| cusp/kʌsp/ |
| vertical tangent/ˈvɜːtɪkl ˈtændʒənt/ |
Smooth implies connected — but not the reverse
- If a function has a derivative at a point, its graph is not just unbroken there — it is smooth.
- Differentiability 可微性 at a point is a stronger condition than continuity.
- Key fact: differentiable $\Rightarrow$ continuous. A tangent line can't exist at a break.
- But the reverse fails: a continuous graph can still have a sharp spot with no derivative.
A smooth, differentiable curve
y = ax² + bx
This parabola is smooth everywhere — a tangent exists at every point, unlike a corner where the slope jumps.
Why differentiable forces continuous
- A derivative is the limit of secant slopes; for that limit to be finite, the function can't jump.
- If $f$ jumped at $c$, the secant slopes near $c$ would blow up — no finite tangent slope.
- So having $f'(c)$ guarantees $f$ is continuous 连续 at $c$.
- Contrapositive (handy on exams): if $f$ is not continuous at $c$, it is not differentiable there.
Which implication is true?
Differentiable forces continuous; the reverse is false.
If $f$ is not continuous at $c$, then $f$ at $c$ is...
Contrapositive of "differentiable $\Rightarrow$ continuous": not continuous $\Rightarrow$ not differentiable.
Where the derivative fails to exist
- Continuity is not enough — three shapes break differentiability even with no gap:
- A corner 尖角: the left and right slopes differ (like $|x|$ at $0$).
- A cusp 尖点: the slopes shoot to $+\infty$ and $-\infty$ (like $x^{2/3}$ at $0$).
- A vertical tangent 竖直切线: the tangent is vertical, so its slope is undefined (like $\sqrt[3]{x}$ at $0$).
At which features does a derivative fail to exist even if the function is continuous?
Corners, cusps, and vertical tangents kill the derivative; a smooth turning point still has $f'=0$.
At a ____ tangent, the tangent line is vertical, so its slope (the derivative) is undefined.
A vertical line has undefined slope; e.g. $\sqrt[3]{x}$ at $0$.
It all comes back to the tangent
- A derivative exists only when there is one well-defined tangent line with a finite slope.
- A corner has two different tangents; a cusp/vertical tangent has an infinite slope — neither gives a single finite value.
- $|x|$ is continuous everywhere but has no derivative at $0$ — the classic counterexample.
- So: check continuity first (necessary), then check for a single finite tangent slope (the rest).
$f(x)=|x|$ is continuous at $0$ but not differentiable there.
The corner gives left slope $-1$, right slope $+1$ — no single tangent slope.
For $f(x)=|x|$, what is the right-hand slope (for $x>0$) at $0$?
For $x>0$, $f(x)=x$, so the slope is $+1$ (the left slope is $-1$, hence the corner).
Do not flip the implication. Differentiable $\Rightarrow$ continuous is true, but continuous $\Rightarrow$ differentiable is false. $f(x)=|x|$ proves it: perfectly continuous at $0$, yet the corner means $f'(0)$ does not exist (left slope $-1$, right slope $+1$).
Is $f(x)=|x|$ differentiable at $x=0$?
- Continuous at $0$? Yes — no gap. ✓
- Left slope: for $x<0$, $f(x)=-x$, slope $-1$. Right slope: for $x>0$, $f(x)=x$, slope $+1$.
- The one-sided slopes disagree ($-1\neq1$) — a corner.
- So $f'(0)$ does not exist, even though $f$ is continuous there.
Differentiability $\Rightarrow$ continuity (a tangent needs an unbroken graph), but continuity does not imply differentiability. A derivative fails to exist at a corner (slopes disagree), a cusp, or a vertical tangent (infinite slope) — anywhere there is no single, finite tangent line.