2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

If a function is differentiable at a point, it must be continuous there, but a continuous function is not always differentiable. Derivatives fail to exist at corners, cusps, vertical tangents, and any discontinuity. For AP Calculus, justify non-differentiability with a specific reason instead of only naming the point.

Why This Matters for the AP Calculus Exam

This topic connects continuity from Unit 1 to the idea of the derivative, and it shows up in both multiple-choice and free-response work. You will be asked to explain why a derivative does or does not exist, read graphs of functions to spot non-differentiable points, and test piecewise functions algebraically at a join. The skill of presenting a clear difference quotient or one-sided derivative structure is important for clear exam work, since a correct conclusion without supporting reasoning often will not support a stronger score. Getting comfortable with this now also sets up graph analysis in Unit 5, where you move between a function, its first derivative, and its second derivative.

Key Takeaways

• Differentiable at a point means continuous at that point. The reverse is not guaranteed.
• If a point is not in the domain of $f$, it cannot be in the domain of $f'$.
• A derivative exists at a point only when the one-sided limits of the difference quotient agree.
• Common ways differentiability fails: corners, cusps, vertical tangents, and discontinuities (jump, removable, or infinite).
• For piecewise functions, check continuity at the join first, then check that the left and right derivatives match.
• The absolute value function $f(x)=|x|$ at $x=0$ and the cube root function $f(x)=\sqrt[3]{x}$ at $x=0$ are the classic examples of continuous but non-differentiable points.

Continuity and Differentiability

Most curves you meet in AP Calculus are both continuous and differentiable, but some points break one or both conditions. If you need a refresher on continuity, check this guide: Confirming Continuity over an Interval.

For a function to be differentiable at a point, it must have a derivative there. Visually, if you zoom in close enough, the graph looks like a straight line. That line does not have to be horizontal.

Look at the graph of $\cos(x)$. As you zoom into the point $(0,1)$, the curve starts to look like a line.

Zooming into graph of $\cos(x)$

For a function to be differentiable at a point $x=a$, the slope of the tangent line must approach the same value from the left and the right:

$\lim_{x\to a^{-}} f'(x) = \lim_{x\to a^{+}} f'(x) = f'(a)$

Here is the key relationship: if a function is differentiable at a point, then it is continuous there. The reverse does not hold. A function can be continuous at a point and still fail to be differentiable. A function is not differentiable at a point if it is discontinuous, has a sharp change in slope (a corner or cusp), or has a vertical tangent.

A useful consequence: if a point is not in the domain of $f$, then it is not in the domain of $f'$. You cannot have a derivative where the function itself does not exist.

Discontinuous Graphs

Unequal One-Sided Slopes

At most discontinuities, the slopes approaching from each side do not match.

For example, $g(x) = \frac{1}{x+2}$ has a domain of $(-\infty,-2) \cup (-2,\infty)$ because the denominator cannot be zero.

Graph of $g(x) = \frac{1}{x+2}$

Since this function is not continuous at $x=-2$ (it is not even defined there), it cannot be differentiable at $x=-2$.

Jump Discontinuity

The same reasoning applies to a jump discontinuity. The function value jumps from one level to another, so it is not continuous at that point and therefore not differentiable there.

Graph of a jump discontinuity

Removable Discontinuity

Removable discontinuities need a closer look. Consider $h(x) = \frac{x^2-4}{x-2}$. This simplifies to $x+2$ everywhere except $x=2$, where there is a hole.

Graph of a removable discontinuity

The derivative equals $1$ at every point except $x=2$. At $x=2$ the function is not defined, so the derivative does not exist there. Even though the slopes from both sides agree, the point is missing from the domain, so $h$ is not differentiable at $x=2$.

The takeaway: any kind of discontinuity at a point rules out differentiability at that point.

Vertical Tangents

Look at the graph of $f(x)=2\left(x+2\right)^{\frac{1}{3}}$.

Graph with a vertical tangent

What is the derivative at $x=-2$? Think about the slope of the tangent line there. As you approach the point, the slope grows without bound toward $\infty$, because the change in $y$ stays nonzero while the change in $x$ shrinks toward $0$. A vertical line has no defined slope, so the derivative at $x=-2$ does not exist, and the curve is not differentiable there.

This matches the classic example $f(x)=\sqrt[3]{x}$ at $x=0$, where the tangent line is vertical.

Corners and Cusps

A function is not differentiable at a corner or a cusp because the slope does not approach the same value from both sides. Sometimes a cusp also involves a vertical tangent.

Look at the graph of $g(x) = |x|$. This function has a corner at $x=0$. Approaching from the left, the slope is $-1$; approaching from the right, the slope is $+1$. Since the one-sided slopes do not match, $g$ is not differentiable at $x=0$, even though it is continuous there.

Graph of $g(x)=|x|$ with a corner at the origin

This is one of the two classic examples worth memorizing: $|x|$ at $x=0$ (a corner) and $\sqrt[3]{x}$ at $x=0$ (a vertical tangent).

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How to Use This on the AP Calculus Exam

MCQ

Many multiple-choice questions hand you a graph and ask where a function is not differentiable. Scan for:

• Breaks in the curve (jump, removable, or infinite discontinuities)
• Sharp corners where the slope flips suddenly
• Cusps where the curve comes to a point
• Vertical tangents where the slope blows up

Each of these spots fails differentiability, even if the function looks connected.

Free Response

When a free-response part gives a piecewise function and asks whether it is differentiable at the join, use a clear two-step structure:

1. Check continuity at the join by matching the two pieces' values.
2. Check that the left-hand derivative and right-hand derivative are equal.

Showing this structure clearly is important for full credit. A correct yes or no with no supporting work usually will not earn the point.

Problem Solving

Try this example. The function is

$f(x) = \begin{cases} \frac{8}{5}\sqrt{x+1}& x<3 \\ \frac{2}{5}x+2& x\geq3\end{cases}$

Given that $f(x)$ is continuous at $x=3$, is it differentiable there?

Check the left-hand derivative. Differentiate the piece for $x<3$:

$f'(x) = \frac{d}{dx}\left(\frac{8}{5}\sqrt{x+1}\right) = \frac{4}{5\sqrt{x+1}}$

So $\lim_{x\to 3^{-}} f'(x) = \frac{4}{5\sqrt{3+1}}= \frac{4}{5(2)} = \frac{2}{5}$.

Check the right-hand derivative. Differentiate the piece for $x\geq 3$:

$f'(x) = \frac{d}{dx}\left(\frac{2}{5}x + 2\right) = \frac{2}{5}$

So $\lim_{x\to 3^{+}} f'(x) = \frac{2}{5}$.

Compare. Both one-sided derivatives equal $\frac{2}{5}$, and the function is already continuous at $x=3$, so

$\lim_{x\to 3^{-}} f'(x) = \lim_{x\to 3^{+}} f'(x) = \frac{2}{5}$

The function is differentiable at $x=3$. The graph is smooth at that point, which confirms the result.

Common Trap

Continuity does not prove differentiability. A piecewise function can match perfectly at the join (so it is continuous) but still have a corner there if the slopes do not match. Always check both conditions.

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Common Misconceptions

• "Continuous means differentiable." Not true. $|x|$ is continuous everywhere but not differentiable at $x=0$.
• "If the slopes match, the function must be differentiable." Only if the function is also continuous and the point is in its domain. A removable discontinuity can have matching one-sided slopes yet still fail because the point is missing.
• "A vertical tangent means the derivative is a huge number." A vertical tangent means the derivative does not exist, because a vertical line has no defined slope.
• "The derivative can exist wherever I want it to." If a point is not in the domain of $f$, it cannot be in the domain of $f'$.
• "Corners are differentiable as long as the graph is connected." A corner has different left and right slopes, so the derivative does not exist there even though the graph is continuous.
• "Removable discontinuities are fine for derivatives." The function is not defined at that hole, so it is not differentiable there.

Related AP Calculus Guides

• Unit 2 Overview: Differentiation
• 2.1 Defining Average and Instantaneous Rates of Change at a Point
• 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
• 2.2 Defining the Derivative of a Function and Using Derivative Notation
• 2.3 Estimating Derivatives of a Function at a Point
• 2.7 Derivatives of cos x, sinx, e^x, and ln x