A Discontinuous Derivative

$f(x) = x^2\sin(1/x)$ ($f(0) = 0$) is differentiable everywhere with $f'(0) = 0$, yet the derivative $f'(x) = 2x\sin(1/x) - \cos(1/x)$ oscillates and is discontinuous at $0$. Differentiable does not imply continuously differentiable — though derivatives still keep the intermediate value property (Darboux).

Examples & Exercises

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

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

Show proof

1. 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 $\frac{h^2\sin(1/h) - 0}{h} = h\sin(1/h)$.
2. 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 $h$.
3. 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 $x \ne 0$.

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

Show proof

1. Fix $x \ne 0$. Since $f(x) = x^2 \cdot \sin(1/x)$ is a product of differentiable functions, apply the product rule: $f'(x) = 2x \cdot \sin(1/x) + x^2 \cdot \frac{d}{dx}\sin(1/x)$.
2. 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.
3. 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 $x = 0$.

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

Show proof

1. 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 $n$.
2. 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).
3. 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.

9. Differentiation