Connecting Differentiability and Continuity: When Derivatives Do and Do Not Exist · 连接可导性与连续性：导数何时存在及何时不存在
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| differentiability/ˌdɪfəˌrenʃɪəˈbɪlɪti/ | 可微性 | kě wēi xìng |
| continuous/kənˈtɪnjuːəs/ | 连续 | lián xù |
| corner/ˈkɔːnə/ | 尖角 | jiān jiǎo |
| cusp/kʌsp/ | 尖点 | jiān diǎn |
| vertical tangent/ˈvɜːtɪkl ˈtændʒənt/ | 竖直切线 | shù zhí qiè xiàn |
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.
光滑蕴含连通——但反过来不成立
- 若函数在某点有导数,它的图在那里不仅不断裂——而且光滑。
- 某点的可微性是比连续性更强的条件。
- 关键事实:可微 $\Rightarrow$ 连续。在断裂处不可能存在切线。
- 但反过来不成立:连续的图仍可能有一个无导数的尖锐处。
A smooth, differentiable curve · 一条光滑、可导的曲线
y = ax² + bx
This parabola is smooth everywhere — a tangent exists at every point, unlike a corner where the slope jumps. · 这条抛物线处处光滑——每一点都存在切线，不像角点那样斜率会发生跳跃。
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.
为何可微迫使连续
- 导数是割线斜率的极限;要让这个极限有限,函数就不能跳跃。
- 若 $f$ 在 $c$ 跳跃,$c$ 附近的割线斜率会爆炸——没有有限的切线斜率。
- 所以拥有 $f'(c)$ 就保证 $f$ 在 $c$ 处连续。
- 逆否命题(考试好用):若 $f$ 在 $c$ 处不连续,它在那里就不可微。
Which implication is true · 真? · 哪个蕴含关系是正确的？
Differentiable forces continuous; the reverse is false. · 可导必连续，反之不成立。
If $f$ is not · 不 continuous at $c$, then $f$ at $c$ is... · 若 $f$ 在 $c$ 处不连续，则 $f$ 在 $c$ 处是……
Contrapositive of "differentiable $\Rightarrow$ continuous": not continuous $\Rightarrow$ not differentiable. · “可导 $\Rightarrow$ 连续”的逆否命题：不连续 $\Rightarrow$ 不可导。
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 $0$).
导数在何处不存在
- 连续还不够——即使没有缺口,三种形状也会破坏可微性:
- 尖角:左右斜率不同(如 $|x|$ 在 $0$)。
- 尖点:斜率冲向 $+\infty$ 与 $-\infty$(如 $x^{2/3}$ 在 $0$)。
- 竖直切线:切线竖直,所以斜率无定义(如 $\sqrt[3]{x}$ 在 $0$)。
At which features does a derivative fail to exist even if the function is continuous? · 函数在哪些特征点即使连续也无法求导？
Corners, cusps, and vertical tangents kill the derivative; a smooth turning point still has $f'=0$. · 尖点、尖瓣和垂直切线会导致导数不存在；而光滑拐点仍有导数 $f'=0$。
At a ____ tangent, the tangent line is vertical, so its slope (the derivative) is undefined. · 在____切线处，切线垂直，因此其斜率（即导数）未定义。
A vertical line has undefined slope; e.g. $\sqrt[3]{x}$ at $0$. · 垂直线斜率未定义；例如 $\sqrt[3]{x}$ 在 $0$ 处。
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).
一切都回到切线
- 只有当存在一条斜率有限、定义明确的切线时,导数才存在。
- 尖角有两条不同切线;尖点/竖直切线的斜率无穷——都给不出单一的有限值。
- $|x|$ 处处连续,却在 $0$ 处无导数——经典的反例。
- 所以:先检查连续性(必要),再检查是否有单一的有限切线斜率(其余)。
$f(x)=|x|$ is continuous at $0$ but not differentiable there. · $f(x)=|x|$ 在 $0$ 处连续但不可导。
The corner gives left slope $-1$, right slope $+1$ — no single tangent slope. · 尖点处左斜率为 $-1$，右斜率为 $+1$——不存在唯一的切线斜率。
For · 支持 $f(x)=|x|$, what is the right-hand slope (for $x>0$) at $0$? · 对于 $f(x)=|x|$，右斜率（对于 $x>0$）在 $0$?
For · 支持 $x>0$, $f(x)=x$, so the slope is $+1$ (the left slope is $-1$, hence the corner). · 对于 $x>0$，$f(x)=x$，故斜率为 $+1$（左斜率为 $-1$，因此形成尖点）。
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 $+1$).
不要把蕴含关系反过来。可微 $\Rightarrow$ 连续成立,但连续 $\Rightarrow$ 可微是错的。$f(x)=|x|$ 证明了这点:在 $0$ 处完全连续,但尖角意味着 $f'(0)$ 不存在(左斜率 $-1$,右斜率 $+1$)。
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.
$f(x)=|x|$ 在 $x=0$ 处可微吗?
- 在 $0$ 处连续?是——没有缺口。✓
- 左斜率:当 $x<0$,$f(x)=-x$,斜率 $-1$。右斜率:当 $x>0$,$f(x)=x$,斜率 $+1$。
- 两个单侧斜率不一致($-1\neq1$)——一个尖角。
- 所以 $f'(0)$ 不存在,尽管 $f$ 在那里连续。
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.
可微 $\Rightarrow$ 连续(切线需要不断裂的图),但连续不蕴含可微。导数在尖角(斜率不一致)、尖点、或竖直切线(斜率无穷)处不存在——任何没有单一、有限切线之处。