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Lesson 59 — Differentiability and Smoothness

Differentiable implies continuous. Corner points, cusps, vertical tangents. C^k and C^∞ classes. Weierstrass function.

Used in: 2nd year advanced HS (calculus) · Japanese Equiv. Math III · German Klasse 12 Leistungskurs Equiv.

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Differentiability at $a$ : The function $f$ is differentiable at $a$ if and only if this limit exists as a finite real number. If it exists, its value is the derivative $f'(a)$ — the slope of the tangent line to the graph at the point $(a, f(a))$ .

Choose your door

Rigorous notation, full derivation, hypotheses

Definitions and Theorems

Differentiability at a Point

"If $f'(a)$ exists, we say that $f$ is differentiable at $a$ . If $f$ is differentiable at every number in an open interval $(a, b)$ , then $f$ is differentiable on $(a, b)$ ." — OpenStax Calculus Volume 1, §3.2

Fundamental Theorem (Differentiability Implies Continuity)

"If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ ." — Active Calculus, §1.7, Boelkins 2024 (Theorem 1.7.1)

Types of Non-Differentiability Points

The four main types of non-differentiable points. From left: corner (finite, distinct one-sided derivatives), cusp (opposite infinite one-sided derivatives), vertical tangent (derivative = +∞ from both sides), jump (function not continuous).

Type	Example at $0$	What Occurs
Corner	$\lvert x \rvert$	$f'_+(0) = 1 \neq -1 = f'_-(0)$
Cusp	$x^{2/3}$	$f'_\pm(0) = \pm\infty$
Vertical Tangent	$x^{1/3}$	$f'(0) = +\infty$
Jump Discontinuity	$\text{sgn}(x)$	$f$ is not continuous
Non-removable Oscillation	$x\sin(1/x),\ f(0)=0$	limit of quotient does not exist

$C^k$ Hierarchy

Example: $C^\infty$ but not $C^\omega$ (Cauchy)

f(x) = \begin{cases} e^{-1/x^2}, & x > 0 \\ 0, & x \leq 0 \end{cases}

This function is $C^\infty(\mathbb{R})$ and $f^{(n)}(0) = 0$ for all $n \geq 0$ , but $f \not\equiv 0$ . Thus $f \notin C^\omega$ — it definitively separates the smooth and analytic classes.

Weierstrass Function

$W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x), \quad 0 < a < 1,\ b \text{ an odd integer},\ ab > 1 + \tfrac{3}{2}\pi.$

Solved Examples

Example— 1· Verifying Differentiability of |x| at Zero (Basic)

Problem. Prove that $f(x) = |x|$ is continuous at $0$ but not differentiable at $0$ .

Strategy. Check continuity via limit and then calculate the one-sided derivatives. If they differ, the derivative does not exist.

Solution.

Continuity. $\lim_{x \to 0} |x| = 0 = f(0)$ . Continuous.
Right-hand derivative. $f'_+(0) = \lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1.$
Left-hand derivative. $f'_-(0) = \lim_{h \to 0^-} \frac{|0+h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1.$
Conclusion. $f'_+(0) = 1 \neq -1 = f'_-(0)$ . The two-sided limit does not exist. $f$ is not differentiable at $0$ .

Verification. Geometrically: the graph forms a V-shape corner at $0$ . Any line through $(0, 0)$ cuts the graph on one side but doesn't tangent from both.

Source. Active Calculus §1.7 — Boelkins, 2024 · CC-BY-NC-SA.

Example— 2· Differentiability of a Piecewise Function with Discontinuity

Problem. Determine if the function

g(x) = \begin{cases} x^2 + 1, & x \leq 2 \\ 5x - 7, & x > 2 \end{cases}

is differentiable at $x = 2$ .

Strategy. Check continuity first; if it's not continuous, it's already not differentiable — saves work.

Solution.

Continuity at $2$ .

$\lim_{x \to 2^-} (x^2 + 1) = 4 + 1 = 5, \qquad \lim_{x \to 2^+} (5x - 7) = 10 - 7 = 3.$

The one-sided limits are $5 \neq 3$ . The function is not continuous at $2$ .
Conclusion. Discontinuity implies non-differentiability. $g$ is not differentiable at $x = 2$ .

Verification. By the theorem: differentiable implies continuous. Therefore non-continuous implies non-differentiable (contrapositive). We don't need to calculate one-sided derivatives.

Source. OpenStax Calculus Volume 1 §3.2 — OpenStax, 2016 · CC-BY-NC-SA.

Example— 3· Differentiability via Definition with Squeeze Theorem

Problem. Let

h(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}

Prove that $h$ is differentiable at $0$ and find $h'(0)$ .

Strategy. Use the definition directly — differentiation rules don't apply at $x = 0$ because $1/x$ is undefined there. Estimate using the Squeeze Theorem.

Solution.

Apply the definition. $h'(0) = \lim_{h \to 0} \frac{h(0+h) - h(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h\sin(1/h).$
Estimate. $|h \sin(1/h)| \leq |h| \cdot 1 = |h| \to 0 \quad \text{as } h \to 0.$
By the Squeeze Theorem. $-|h| \leq h\sin(1/h) \leq |h|$ , and $|h| \to 0$ , therefore $h\sin(1/h) \to 0$ .
Conclusion. $h'(0) = 0$ . $h$ is differentiable at $0$ .

Verification. For $x \neq 0$ : $h'(x) = 2x\sin(1/x) - \cos(1/x)$ . The term $\cos(1/x)$ oscillates as $x \to 0$ , so $h' \notin C^0$ — $h$ is differentiable but not $C^1$ .

Source. Active Calculus §1.7 — Boelkins, 2024 · CC-BY-NC-SA.

Example— 4· Finding Parameters to Ensure C¹ (Advanced)

Problem. Find the values of $a$ and $b$ such that the function

p(x) = \begin{cases} ax^2 + b, & x \leq 1 \\ 2x^3 + 3, & x > 1 \end{cases}

is $C^1$ on $\mathbb{R}$ .

Strategy. To be $C^1$ at the junction point $x = 1$ , we require (1) continuity and (2) equality of derivatives. These two requirements form a $2 \times 2$ system in $a, b$ .

Solution.

Continuity condition at $x = 1$ .

$p(1^-) = a \cdot 1^2 + b = a + b, \qquad p(1^+) = 2 \cdot 1^3 + 3 = 5.$

Equation (I): $a + b = 5$ .
Equality of derivative condition at $x = 1$ .

For $x < 1$ : $p'(x) = 2ax$ , so $p'_-(1) = 2a$ .

For $x > 1$ : $p'(x) = 6x^2$ , so $p'_+(1) = 6$ .

Equation (II): $2a = 6 \Rightarrow a = 3$ .
Solve the system. From (II): $a = 3$ . Substituting into (I): $3 + b = 5 \Rightarrow b = 2$ .
Verification.
- $p(1) = 3 + 2 = 5 = 2 + 3$ . Continuous.
- $p'_-(1) = 2 \cdot 3 \cdot 1 = 6 = p'_+(1)$ . Continuous derivative.

Source. APEX Calculus §2.1 — Hartman et al., 2024 · CC-BY-NC.

Example— 5· Analyzing Cusp and C^k Hierarchy for x^(2/3) (Challenge)

Problem. Let $f(x) = x^{2/3}$ .

(a) Show that $f$ is not differentiable at $0$ .

(b) Determine the maximum regularity class of $f$ around $0$ .

Strategy. Use the definition of the derivative at $0$ and power rules for $x \neq 0$ .

Solution.

(a) Derivative at $0$ via definition.

$f'(0) = \lim_{h \to 0} \frac{h^{2/3} - 0}{h} = \lim_{h \to 0} h^{2/3 - 1} = \lim_{h \to 0} h^{-1/3}.$

As $h \to 0^+$ : $h^{-1/3} \to +\infty$ . As $h \to 0^-$ : $h^{-1/3} = -|h|^{-1/3} \to -\infty$ .

The limit is not a finite real number. Thus $f$ is not differentiable at $0$ — a cusp.

(b) Hierarchy. On any interval $I$ not containing $0$ , $f \in C^\infty$ . But on any interval containing $0$ , $f \in C^0$ (continuous) but $f \notin C^1$ (not differentiable at $0$ ). Maximum regularity: $C^0$ .

(c) For $x \neq 0$ .

$f'(x) = \frac{2}{3} x^{-1/3} = \frac{2}{3x^{1/3}}.$

As $x \to 0^+$ : $f'(x) \to +\infty$ . As $x \to 0^-$ : $f'(x) \to -\infty$ .

The graph has a cusp pointing upwards at $0$ : the tangent lines become increasingly vertical as you approach $0$ , diverging to $+\infty$ on the right and $-\infty$ on the left.

Verification. $f(0) = 0^{2/3} = 0$ , $\lim_{x\to 0} x^{2/3} = 0$ . Continuous at $0$ . The one-sided derivatives $\to \pm\infty$ : confirms a cusp, not a vertical tangent (which would have both one-sided derivatives going to $+\infty$ ).

Source. OpenStax Calculus Volume 1 §3.2 — OpenStax, 2016 · CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

22 9 5 2 2

Sources

Active Calculus — Boelkins, 2024 · §1.7 "Limits, Continuity, and Differentiability" · CC-BY-NC-SA. Primary source.
Calculus Volume 1 — OpenStax, 2016 · §3.2 "The Derivative as a Function" · CC-BY-NC-SA.
APEX Calculus — Hartman et al., 2024 · v5 · §2.1 "Instantaneous Rates of Change: The Derivative" · CC-BY-NC.