Differentiability & Continuity | AP Calculus Topic 2.4

1. Start with continuity and derivative definitions

Continuity at an interior input requires the function value to exist, the two-sided function limit to exist, and those values to agree:

lim_{x \to a} f (x) = f (a) .

Differentiability at an interior input requires a finite real limit of the difference quotient:

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

Continuity compares nearby outputs with a base output. Differentiability compares limiting rates. At a domain endpoint, continuity uses the available side; derivative endpoint conventions must also be identified explicitly. A point must belong to the original function's domain before either its continuity there or its derivative there can be established.

Use Topic 2.2: derivative definition and notation and continuity at a point to review these definitions. Derivative values estimated in Topic 2.3 do not themselves prove differentiability.

2. Differentiability implies continuity

Core concept: A finite derivative at a point forces continuity there, but continuity alone does not guarantee a derivative. A continuous function can fail to be differentiable at a corner, cusp or vertical tangent; a point of discontinuity also cannot have a derivative. Use definitions and one-sided difference quotients to justify each conclusion.

DIFFERENTIABILITY IMPLIES CONTINUITY

If $f$ is differentiable at $x = a$ , then $f$ is continuous at $x = a$

Symbolic Form:

f^{'} (a) exists \Rightarrow f is continuous at x = a

Contrapositive (Also True):

f is discontinuous at x = a \Rightarrow f^{'} (a) does NOT exist

Plain English: If you can take the derivative at a point, the function must be continuous there. Any break in the function means no derivative at that point!

THE CONVERSE IS FALSE!

It is NOT true that: If $f$ is continuous at $x = a$ , then $f$ is differentiable at $x = a$

Continuity does NOT guarantee differentiability! A function can be continuous but still fail to have a derivative (corners, cusps, vertical tangents).

Differentiable \Rightarrow Continuous (TRUE)

Continuous \Rightarrow Differentiable (FALSE)

Visual Summary:

All differentiable functions are continuous
Some continuous functions are NOT differentiable
At a point where a function is discontinuous, it is not differentiable there. It may still be differentiable at other inputs.

A stronger condition: Differentiability requires a single finite limiting rate, not merely connected-looking behavior. A graph that appears smooth at a chosen scale is not a proof that the derivative exists.

3. Why differentiability forces continuity: the proof

Formal Proof (AP® Level)

Given: $f^{'} (a)$ exists

To Prove: $f$ is continuous at $x = a$ , i.e., $lim_{x \to a} f (x) = f (a)$

Proof:

We need to show: $lim_{x \to a} [f (x) - f (a)] = 0$
Multiply and divide by $(x - a)$ :
$lim_{x \to a} [f (x) - f (a)] = lim_{x \to a} [\frac{f (x) - f (a)}{x - a} \cdot (x - a)]$
Use limit properties to separate:
$= (lim_{x \to a} \frac{f (x) - f (a)}{x - a}) \cdot (lim_{x \to a} (x - a))$
Recognize the derivative:
$= f^{'} (a) \cdot 0 = 0$

Therefore: $lim_{x \to a} f (x) = f (a)$ , so $f$ is continuous at $x = a$ ∎

Why the proof works: The difference-quotient limit exists as a finite real number. For $x \neq a$ , multiplying the quotient by $x - a$ recovers $f (x) - f (a)$ . The product tends to zero by the limit laws. This does not use an undefined quotient at $x = a$ .

4. Four common ways a finite derivative can fail

When a finite $f' (a)$ does not exist

The following are four common reasons a finite derivative fails to exist at an input. They are not an exhaustive list: an oscillating difference quotient can also fail to converge.

Case 1: DISCONTINUITY (Any Type)

If $f$ has any discontinuity at $x = a$ (jump, removable, infinite), then $f^{'} (a)$ does NOT exist.

Why: By the theorem, differentiability requires continuity. No continuity → no derivative!

Examples:

Jump discontinuity: $f (x) = {\begin{matrix} x & x < 1 \\ x + 1 & x \geq 1 \end{matrix}$ at $x = 1$
Removable discontinuity: $f (x) = \frac{x^{2} - 1}{x - 1}$ at $x = 1$ (if not defined there)
Infinite discontinuity: $f (x) = \frac{1}{x}$ at $x = 0$

Case 2: CORNER (Sharp Point)

At a corner, the left and right difference-quotient limits are finite but unequal. The function is continuous at the corner, but there is no single tangent slope.

Test: Check if $f_{-}^{'} (a) \neq f_{+}^{'} (a)$ (one-sided derivatives don't match)

Classic Example: $f (x) = | x |$ at $x = 0$

Left derivative: $f_{-}^{'} (0) = - 1$
Right derivative: $f_{+}^{'} (0) = + 1$
Since $- 1 \neq + 1$ , $f^{'} (0)$ does NOT exist

Other Examples:

$f (x) = | x - 2 |$ at $x = 2$
$f (x) = | x^{2} - 4 |$ at $x = \pm 2$
Any piecewise function with different slopes meeting at a point

Case 3: CUSP (Pointed Peak or Valley)

At the standard cusp illustrated here, the function is continuous while its one-sided difference quotients diverge with opposite signs. There is no finite derivative.

Characteristic: A pointed peak or valley with oppositely unbounded one-sided rates. This differs from both a finite-slope corner and a same-sign vertical-tangent example.

Classic Example: $f (x) = x^{\frac{2}{3}}$ or $f (x) = \sqrt[3]{x^{2}}$ at $x = 0$

As $x \to 0^{-}$ , slope $\to - \infty$
As $x \to 0^{+}$ , slope $\to + \infty$
The derivative does not exist (infinite from both sides)

Difference from a corner: A corner has two distinct finite one-sided rates. A cusp of this type has oppositely unbounded one-sided rates. Infinity is not a real derivative value.

Case 4: VERTICAL TANGENT

If the function is continuous at an interior input and both one-sided difference quotients diverge to the same signed infinity, the graph has a vertical tangent there. The tangent line exists geometrically, but it has no finite slope, so the ordinary derivative does not exist.

Use the defining quotient: Test $\frac{f (a + h) - f (a)}{h}$ from each side. The statement $lim_{x \to a} | f^{'} (x) | = \infty$ alone does not distinguish a cusp from a vertical tangent, nor does it establish that the original function is defined and continuous at the base point.

Classic Example: $f (x) = \sqrt[3]{x}$ or $f (x) = x^{\frac{1}{3}}$ at $x = 0$

$f^{'} (x) = \frac{1}{3} x^{\frac{- 2}{3}} = \frac{1}{3 \sqrt[3]{x^{2}}}$
As $x \to 0$ , $f^{'} (x) \to \infty$
The function is continuous but not differentiable at $x = 0$

Other Examples:

$f (x) = \sqrt{x}$ at $x = 0$ (from the right)
$f (x) = x^{\frac{1}{5}}$ at $x = 0$

Summary Table:

Case	What Happens	Classic Example
Discontinuity	Function has break	Step function, $\frac{1}{x}$ at 0
Corner	Slopes differ (finite)	$\| x \|$ at 0
Cusp	Slopes infinite, opposite	$x^{\frac{2}{3}}$ at 0
Vertical Tangent	Slope infinite (same direction)	$x^{\frac{1}{3}}$ at 0

For the cusp $f (x) = \sqrt[3]{x^{2}} = | x |^{\frac{2}{3}}$ , the base-point quotient is $\frac{| h |^{\frac{2}{3}}}{h} = \frac{1}{\sqrt[3]{h}}$ . It tends to $- \infty$ from the left and $+ \infty$ from the right. For the cube-root vertical tangent, $\frac{\sqrt[3]{h}}{h} = \frac{1}{(\sqrt[3]{h})^{2}}$ tends to $+ \infty$ from both sides.

The first three functions are continuous at zero. The jump graph has an open left-hand limit point and a filled function value at input one. Markers show actual values or exclusions; plot edges are not domain endpoints.

Undefined points and removable holes

The original rational function $\frac{x^{2} - 1}{x - 1}$ excludes one. Cancelling its common factor does not create a value or a derivative there. If a new extension explicitly assigns the value two at one, it becomes the line $f (x) = x + 1$ everywhere, and that extended function has derivative one there. The derivative belongs to the specified function, not just a simplified expression.

The square-root example has an endpoint at zero. Its right-hand quotient $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$ diverges as $h \to 0^{+}$ . Thus it has no finite right-hand derivative; there is also no available left side for an ordinary two-sided derivative.

5. Define and compare one-sided derivatives

Definition of One-Sided Derivatives

Left-Hand Derivative (from the left):

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

Right-Hand Derivative (from the right):

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

Key Fact:

f^{'} (a) exists \Leftrightarrow f_{-}^{'} (a) = f_{+}^{'} (a) (both finite)

At an interior point, the ordinary derivative exists exactly when the two one-sided difference-quotient limits exist as finite real numbers and are equal. At a domain endpoint, identify a requested one-sided derivative separately; an unavailable side cannot silently be supplied.

How to Use One-Sided Derivatives:

For piecewise functions: Verify the defined boundary value and continuity first. Then calculate the one-sided difference quotients using that same boundary value. With polynomial or other smooth branches that meet continuously, their endpoint branch slopes give those one-sided derivatives.
For corners: Calculate both one-sided derivatives; if they differ, no derivative exists
For continuity: Even if function is continuous, one-sided derivatives must match for differentiability

Here $f_{-}^{'} (a)$ and $f_{+}^{'} (a)$ denote the left and right derivative limits using the actual value $f (a)$ . They are not simply the limits of a formula for $f^{'}$ as its input approaches the point. A finite derivative can exist even when the derivative function is not continuous nearby.

6. Five worked examples with complete justification

Example 1: Testing for Differentiability (Absolute Value)

Problem: Determine if $f (x) = | x - 3 |$ is differentiable at $x = 3$

Solution:

Check continuity:
- $f (3) = 0$
- $lim_{x \to 3} | x - 3 | = 0$
- Continuous at $x = 3$
Rewrite as piecewise:
$f (x) = {\begin{matrix} - (x - 3) = 3 - x & x < 3 \\ x - 3 & x \geq 3 \end{matrix}$
Find left-hand derivative:
- For $x < 3$ : $f (x) = 3 - x$ , so $f^{'} (x) = - 1$
- $f_{-}^{'} (3) = - 1$
Find right-hand derivative:
- For $x > 3$ : $f (x) = x - 3$ , so $f^{'} (x) = 1$
- $f_{+}^{'} (3) = 1$
Compare: $f_{-}^{'} (3) = - 1 \neq 1 = f_{+}^{'} (3)$

Conclusion: NOT differentiable at $x = 3$ because the one-sided derivatives don't match. This is a corner.

Example 2: Piecewise Function Differentiability

Problem: For which $k$ is the following function differentiable at two?

f (x) = {\begin{matrix} x^{2} & x \leq 2 \\ k x & x > 2 . \end{matrix}

Preserve the base value: $f (2) = 4$ .
Continuity: The left limit is four and the right limit is $2 k$ , so continuity requires $2 k = 4$ , or $k = 2$ .
Use that continuity value: With $k = 2$ , the one-sided derivative limits are
$f_{-}^{'} (2) = lim_{h \to 0^{-}} \frac{(2 + h)^{2} - 4}{h} = lim_{h \to 0^{-}} (4 + h) = 4,$

$f_{+}^{'} (2) = lim_{h \to 0^{+}} \frac{2 (2 + h) - 4}{h} = 2 .$
They differ, so the only continuous choice gives a corner.
Explain the conflicting conditions: Matching the two branch slopes would require $k = 4$ , but then the right limit would be eight while the defined value remains four. Branch-slope matching is insufficient when continuity fails.

Answer: No value of $k$ makes this specified function differentiable at two. Continuity requires $k = 2$ , whereas matching branch slopes would require $k = 4$ ; these cannot both hold.

Example 3: Better Piecewise Function

Problem: Find $a$ and $b$ so that $f (x) = {\begin{matrix} x^{2} & x \leq 1 \\ a x + b & x > 1 \end{matrix}$ is differentiable at $x = 1$

Solution:

Continuity condition:
- $lim_{x \to 1^{-}} x^{2} = 1$
- $lim_{x \to 1^{+}} (a x + b) = a + b$
- Equation 1: $a + b = 1$
Differentiability condition:
- $f_{-}^{'} (1) = 2 x |_{x = 1} = 2$
- $f_{+}^{'} (1) = a$
- Equation 2: $a = 2$
Solve the system:
- From Equation 2: $a = 2$
- Substitute into Equation 1: $2 + b = 1 \Rightarrow b = - 1$

Answer: $a = 2$ and $b = - 1$

Result: $f (x) = {\begin{matrix} x^{2} & x \leq 1 \\ 2 x - 1 & x > 1 \end{matrix}$ is differentiable everywhere

Example 4: Identifying Non-Differentiable Points

Problem: Identify all points where $f (x) = \sqrt[3]{(x - 2)^{2}}$ is not differentiable

Solution:

Check continuity: $f (x)$ is continuous everywhere (cube root is defined for all real numbers)
Find the derivative:
$f^{'} (x) = \frac{2}{3} (x - 2)^{\frac{- 1}{3}} = \frac{2}{3 \sqrt[3]{x - 2}}$
Check where $f^{'} (x)$ doesn't exist:
- When $x - 2 = 0$ , i.e., $x = 2$
- At $x = 2$ : $f^{'} (2)$ is undefined (division by zero)
- As $x \to 2$ , $| f^{'} (x) | \to \infty$

Answer: Not differentiable at $x = 2$

Reason: This is a cusp (the function has a sharp point where slopes from both sides go to ±∞)

Example 5: Vertical Tangent

Problem: Show that $f (x) = x^{\frac{1}{3}}$ has a vertical tangent at $x = 0$

Solution:

Check continuity:
- $lim_{x \to 0} x^{\frac{1}{3}} = 0 = f (0)$
- Continuous at $x = 0$
Find the derivative:
$f^{'} (x) = \frac{1}{3} x^{\frac{- 2}{3}} = \frac{1}{3 x^{\frac{2}{3}}} = \frac{1}{3 \sqrt[3]{x^{2}}}$
Check behavior at $x = 0$ :
- $lim_{x \to 0^{+}} f^{'} (x) = lim_{x \to 0^{+}} \frac{1}{3 \sqrt[3]{x^{2}}} = + \infty$
- $lim_{x \to 0^{-}} f^{'} (x) = lim_{x \to 0^{-}} \frac{1}{3 \sqrt[3]{x^{2}}} = + \infty$
- The slopes blow up to $+ \infty$ from both sides

Conclusion: The function is continuous at $x = 0$ but not differentiable because the tangent line is vertical (infinite slope)

Check the base-point quotients behind the conclusions

In Example 3, continuity requires $a + b = 1$ . Only after imposing that relation does the right-hand quotient simplify to the branch slope $a$ :

\frac{a (1 + h) + b - 1}{h} = \frac{a + b - 1}{h} + a = a, a + b = 1 .

The left-hand quotient tends to two. Thus $a = 2$ and $b = - 1$ , preserving both the base value and the derivative.

In Example 4, the quotient at the cusp input two is

\frac{f (2 + h) - f (2)}{h} = \frac{\sqrt[3]{h^{2}}}{h} = \frac{1}{\sqrt[3]{h}}, h \neq 0 .

It diverges negatively from the left and positively from the right, proving the claimed failure. Away from two the displayed derivative formula is valid.

In Example 5, the quotient at zero is $\frac{\sqrt[3]{h}}{h} = \frac{1}{(\sqrt[3]{h})^{2}}$ . Both sides diverge positively. The tangent is the vertical line $x = 0$ , but a finite derivative value does not exist.

The left panel is Example 2 with the only continuous choice. The right panel is Example 3 after solving both conditions. A filled join value comes from the left branch; the right branch approaches that same point. Curves continue beyond the plotting windows.

7. A reliable differentiability checklist

The Complete Differentiability Test (Step-by-Step)

To determine if $f$ is differentiable at $x = a$ :

Step 1: Check if $a$ is in the domain
- If $f (a)$ is undefined → NOT differentiable
Step 2: Check continuity at $x = a$
- Verify $lim_{x \to a} f (x) = f (a)$
- If discontinuous → NOT differentiable (by theorem)
Step 3: Check for corners/cusps
- For piecewise: Check if $f_{-}^{'} (a) = f_{+}^{'} (a)$
- For absolute values: Rewrite and check one-sided derivatives
- If one-sided derivatives differ → NOT differentiable
Step 4: Check for vertical tangents
- Compute the one-sided difference quotients at the base input. Use finite equality for differentiability; same-sign unbounded rates for a continuous vertical-tangent example; opposite signs for a cusp of the type illustrated.
- If a defining quotient is unbounded, there is no ordinary finite derivative. Keep this conclusion distinct from simply seeing a formula undefined after substitution.
Step 5: If all tests pass
- Conclude differentiability only after establishing a finite two-sided difference-quotient limit, or applying a valid differentiability rule with its hypotheses. A visual checklist alone is not sufficient.

8. Know the limits of visual shortcuts

Oscillation supplies another failure case

Define $f (0) = 0$ and $f (x) = x \sin (\frac{1}{x})$ for nonzero inputs. Since $| f (x) | \leq | x |$ , the function is continuous at zero. However,

\frac{f (h) - f (0)}{h} = \sin (\frac{1}{h}), h \neq 0,

oscillates without a limit. Thus the function is continuous but not differentiable at zero, even though this behavior does not fit a simple corner/cusp/vertical-tangent sketch.

Differentiable does not mean the derivative is continuous

Define $g (0) = 0$ and $g (x) = x^{2} \sin (\frac{1}{x})$ for nonzero inputs. Its base-point quotient satisfies

| \frac{g (h) - g (0)}{h} | = | h \sin (\frac{1}{h}) | \leq | h | \to 0 .

Therefore $g^{'} (0) = 0$ . For nonzero inputs, $g^{'} (x) = 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})$ , which has no limit at zero. This is why a test involving only a nearby derivative formula cannot replace the defining quotient at the point.

These examples are extensions of the core topic. They clarify that differentiability is a precise finite-limit condition, not a synonym for an informal visual notion of “smooth.”

9. Recognition guides, strategies and common mistakes

Essential Tips

Memorize the theorem: Differentiable → Continuous (NOT the reverse!)
Check continuity first: If discontinuous, automatically not differentiable
Absolute value needs a check: If a differentiable inner function crosses zero with a nonzero derivative, its absolute value has a corner there. A zero alone does not guarantee a corner: $| x^{2} | = x^{2}$ is differentiable at zero, and even $| x^{3} |$ crosses with a flat derivative and remains differentiable there.
Fractional exponents: Watch for vertical tangents at $x = 0$ (e.g., $x^{\frac{1}{3}}$ )
One-sided derivatives: Must exist AND be equal for differentiability
Piecewise functions: Always check boundary points carefully

Quick Recognition Guide

Spotting Non-Differentiable Points:

If you see...	Likely problem	What to check
Absolute value	Corner	One-sided derivative limits at candidate zeros of the inner function
Piecewise function	Corner or discontinuity	Continuity and one-sided derivatives at boundaries
$x^{\frac{p}{q}}$ with $p < q$	Possible cusp, vertical or endpoint behavior; classify from the defining quotient.	Original domain, continuity and one-sided difference quotients; account for odd/even root domains.
Rational function	Discontinuity	Excluded inputs from the original denominator; check any separately defined extension
$x^{\frac{2}{3}}$ , $\sqrt[3]{x^{2}}$	Cusp	One-sided derivatives at critical point

Memory Tricks

Mnemonic for Non-Differentiable Cases:

"Don't Create Crazy Vertical tangents!"

Discontinuities
Corners
Cusps
Vertical tangents

Common Mistakes to Avoid

Mistake 1: Thinking continuous implies differentiable (FALSE!)
Mistake 2: Forgetting to check continuity before testing differentiability
Mistake 3: Only checking one-sided limit instead of one-sided derivatives
Mistake 4: Not checking BOTH continuity AND matching derivatives for piecewise functions
Mistake 5: Assuming smooth-looking graphs are differentiable everywhere (check corners!)
Mistake 6: Confusing vertical tangent with vertical asymptote (different concepts!)
Mistake 7: Assuming every zero or sign change inside an absolute value creates a corner. Compute the one-sided rates; a flat zero can remain differentiable.
Mistake 8: Forgetting that $f^{'} (a)$ must be a FINITE number to exist

10. Four practice problems with solutions

Decide whether a finite ordinary derivative exists at each stated input, and justify the answer.

1. Shifted absolute value

$f (x) = | x + 1 |$ at $x = - 1$ .

Check continuity and the one-sided rates

The function value and both function limits are zero, so it is continuous there. For nonzero input changes,

\frac{f (- 1 + h) - f (- 1)}{h} = \frac{| h |}{h} .

f_{-}^{'} (- 1) = - 1, f_{+}^{'} (- 1) = 1 .

The unequal finite rates form a corner; no derivative exists.

2. A real fractional power

$g (x) = x^{\frac{2}{5}}$ at zero, interpreted as $(\sqrt[5]{x})^{2}$ .

Check the cusp and real domain

The function is defined and continuous for every real input. Its derivative away from zero is $g^{'} (x) = \frac{2}{5 \sqrt[5]{x^{3}}}$ . At zero use the quotient itself:

\frac{g (h) - g (0)}{h} = \frac{h^{\frac{2}{5}}}{h} = \frac{1}{(\sqrt[5]{h})^{3}} .

It tends to $- \infty$ from negative inputs and $+ \infty$ from positive inputs. Hence zero is a cusp, with no finite derivative.

3. A continuous piecewise join

h (x) = {\begin{matrix} x^{2} + 1 & x < 1 \\ 3 x - 1 & x \geq 1 . \end{matrix}

Test differentiability at one. The function name here is $h$ ; use $u$ for an input change.

Check the defined value and both quotients

$h (1) = 2$ , and both one-sided function limits are two. The function is continuous. However,

h_{-}^{'} (1) = lim_{u \to 0^{-}} \frac{(1 + u)^{2} + 1 - 2}{u} = lim_{u \to 0^{-}} (2 + u) = 2,

h_{+}^{'} (1) = lim_{u \to 0^{+}} \frac{3 (1 + u) - 1 - 2}{u} = 3 .

The one-sided derivatives differ, so the continuous join is a corner and is not differentiable.

4. A square-root endpoint

$k (x) = \sqrt{x}$ at zero.

Check the endpoint convention and finite-rate requirement

The domain is $[0, \infty)$ , and the function is continuous from the right at zero. For positive input changes,

\frac{k (h) - k (0)}{h} = \frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}} \to + \infty .

There is no finite right-hand derivative, and no ordinary two-sided derivative at this endpoint. The graph has a vertical tangent from its available side. For $x > 0$ , the derivative formula $k^{'} (x) = \frac{1}{2 \sqrt{x}}$ is valid; it is not a finite derivative at zero.

11. Explain differentiability in an AP-style response

What the AP® Exam Expects:

Know the theorem: State that differentiability implies continuity
Check both conditions: For piecewise, verify continuity AND matching derivatives
Justify your answer: Explain WHY the function is/isn't differentiable
Use proper notation: Write $f_{-}^{'} (a)$ and $f_{+}^{'} (a)$ clearly
Show one-sided limits: Calculate left and right derivatives separately
Identify the case: State if it's a corner, cusp, vertical tangent, or discontinuity
Graph interpretation: Be able to identify non-differentiable points from graphs

Common FRQ Formats:

"Determine if f is differentiable at $x = a$ . Justify your answer."
"Find the values of $a$ and $b$ that make f both continuous and differentiable at $x = c$ "
"Explain why the derivative does not exist at $x = 2$ "
"Is f continuous at $x = 1$ ? Is f differentiable at $x = 1$ ? Justify."
"Identify all points where the function is not differentiable"
"Show that if $f^{'} (a)$ exists, then f is continuous at $x = a$ " (proof question)

12. Quick reference

Concept	Key Fact
Main Theorem	Differentiable → Continuous (NOT reverse!)
Contrapositive	Discontinuous → Not differentiable
4 Non-Diff Cases	Discontinuity, Corner, Cusp, Vertical tangent
One-Sided Test	$f^{'} (a)$ exists at an interior input exactly when $f_{-}^{'} (a) = f_{+}^{'} (a)$ and both values are finite.
Corner Example	$\| x \|$ at 0
Vertical Tangent Example	$x^{\frac{1}{3}}$ at 0
Cusp Example	$x^{\frac{2}{3}}$ at 0

13. Connections and next study steps

Topic 2.4 connects to:

Topic 2.1-2.2: Foundation for understanding what derivatives mean
Topic 2.5+: You need to know where derivatives exist before using rules
Unit 5: The Mean Value Theorem needs continuity on a closed interval and differentiability on its interior. Optimization also requires attention to endpoints and points where derivatives fail.
Later derivative tests: First and second derivatives provide sufficient tests for behavior when their hypotheses hold. Concavity is not defined merely by the existence of a second derivative.
Unit 6: A continuous integrand gives the standard differentiation conclusion in the Fundamental Theorem of Calculus. This theorem is distinct from the differentiability-implies-continuity result studied here.
Real-world: Many phenomena have points where instantaneous rates don't exist

Remember: Differentiability implies continuity at the same input. Continuity does not imply differentiability. Preserve the original domain, verify the base value, then check the defining one-sided rates. The ordinary derivative must be finite. For a piecewise join, both continuity and matching finite one-sided derivatives are required.

← Topic 2.3: estimating derivatives All Unit 2 study notes →Review continuity at a point →Review types of discontinuities →

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