A Discontinuous Derivative

A function differentiable everywhere whose derivative alone is discontinuous. Zoom near the origin to watch the derivative oscillate.

The function itself settles smoothly at the origin. Its slope, though, never stops oscillating. Differentiable does not force the derivative to be continuous. Still, the derivative never skips an intermediate value.

For example, picture a ⚽ ball bouncing lower and faster each time it hits the floor. Its height calmly sinks to zero. Its bounce speed, though, keeps flickering to the very end.

Examples & Exercises

For $f(x) = x^2\sin(1/x)$ ( $x \ne 0$ ), $f(0) = 0$ , prove completely by the squeeze theorem that $f'(0) = 0$ .

\frac{f(h) - f(0)}{h} = h\sin\frac{1}{h}, \qquad \left|h\sin\frac{1}{h}\right| \le |h| \to 0

r = 0.1

h = 0.01

: the difference quotient

\approx -0.0051

is near 0, and the left panel shows the secant slope collapsing onto the horizontal (dashed) tangent.

Show proof

By the definition of the derivative, $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ ( $h \ne 0$ ). Since $f(0) = 0$ and $f(h) = h^2\sin(1/h)$ for $h \ne 0$ , the difference quotient simplifies to $\frac{h^2\sin(1/h) - 0}{h} = h\sin(1/h)$ .
Since $|\sin(1/h)| \le 1$ , we have $|h\sin(1/h)| \le |h|$ .
As $|h| \to 0$ ( $h \to 0$ ), the squeeze theorem gives $h\sin(1/h) \to 0$ — the limit of the difference quotient exists and $f'(0) = 0$ .
The oscillating factor $\sin(1/h)$ is completely damped by the factor $h$ .
In the scene, shrinking $h$ down to $0.0001$ shows the difference quotient converging to 0 numerically (alongside the bound $|h\sin(1/h)| \le |h|$ ).

Prove completely, via the product and chain rules, that $f'(x) = 2x\sin(1/x) - \cos(1/x)$ for $x \ne 0$ .

f'(x) = \frac{d}{dx}\left[x^2\sin\frac{1}{x}\right] = 2x\sin\frac{1}{x} - \cos\frac{1}{x} \qquad (x \ne 0)

r = 0.1

: the right panel plots exactly this formula — the

2x\sin(1/x)

term decays to 0 while the

-\cos(1/x)

term keeps oscillating with amplitude 1.

Show proof

Fix $x \ne 0$ . Since $f(x) = x^2 \cdot \sin(1/x)$ is a product of differentiable functions, apply the product rule: $f'(x) = 2x \cdot \sin(1/x) + x^2 \cdot \frac{d}{dx}\sin(1/x)$ .
By the chain rule, $\frac{d}{dx}\sin(1/x) = \cos(1/x) \cdot \frac{d}{dx}(1/x) = \cos(1/x) \cdot (-1/x^2)$ (since $1/x$ is differentiable at $x \ne 0$ ). Substituting, $f'(x) = 2x\sin(1/x) + x^2 \cdot \cos(1/x) \cdot (-1/x^2) = 2x\sin(1/x) - \cos(1/x)$ — proved.
Numerical check: $f'(0.5) = 2(0.5)\sin 2 - \cos 2 \approx 1.3254$ , matching the model. This formula is not defined at $x = 0$ — $f'(0)$ must be computed separately as the limit of the difference quotient, as in problem 1.

Prove completely, using the sequence $x_n = 1/(2\pi n)$ , that $f'$ is discontinuous at $x = 0$ .

x_n = \frac{1}{2\pi n} \to 0, \qquad f'(x_n) = 2x_n\sin(2\pi n) - \cos(2\pi n) = -1 \ne 0 = f'(0)

r = 0.01

h = 0.001

: zooming closer to 0, the right panel never stops oscillating, and the highlighted points

x_n = 1/(2\pi n)

keep landing on the dashed line

y = -1

Show proof

Set $x_n = 1/(2\pi n)$ ; then $x_n \ne 0$ and $x_n \to 0$ .
Plugging into problem 2's formula, $f'(x_n) = 2x_n\sin(1/x_n) - \cos(1/x_n) = 2x_n\sin(2\pi n) - \cos(2\pi n)$ .
Since $\sin(2\pi n) = 0$ and $\cos(2\pi n) = 1$ , we get $f'(x_n) = 0 - 1 = -1$ — constant for every $n$ .
Hence $f'(x_n) \to -1$ , but $f'(0) = 0$ by problem 1, so $f'(x_n) \not\to f'(0)$ even though $x_n \to 0$ . By the sequential criterion, $f'$ is not continuous at $0$ — $\lim_{x \to 0} f'(x)$ does not even exist (the never-ending oscillation in the right panel).
Remark: a discontinuous derivative still has the intermediate value property — by Darboux's theorem, $f'$ satisfies the IVP on intervals. Differentiability is strictly weaker than continuity of the derivative.