Analysis · Hierarchy

For a function of a real variable, the derivative $f^{\prime}(x)$ measures the slope of the graph at each point $x$. A differentiable function has a derivative at every point, but the derivative can oscillate wildly as $x$ changes causing the derivative to be discontinuous. A continuously differentiable function has a derivative at every point, and the derivative is also a continuous function of $x$.

continuously differentiable ⊂ differentiable

Definitions

differentiable

High school

Consider a function $f$ and a point $a$ in its domain. The function $f$ is differentiable at $a$ if the limit

\[f^{\prime}(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\]

exists. It is differentiable if this holds at every point of its domain.

Undergraduate (functions on open subsets of $\reals^n$)

Consider a function $f$ on an open subset of $\reals^n$ and a point $a$ in its domain. The function $f$ is differentiable at $a$ if there is a linear map $Df(a)$ with

\[\lim_{h \to 0} \frac{\norm{f(a + h) - f(a) - Df(a) h}}{\norm{h}} = 0.\]
Examples
  • $x \mapsto \sin x$
    $x \mapsto \sin x$ is differentiable at every $x$, with derivative $\cos x$.
  • $x \mapsto x \abs{x}$
    $x \mapsto x \abs{x}$ is differentiable at every $x$: away from $0$, the function equals $x^2$ or $-x^2$, and at $0$, the difference quotient $\frac{h \abs{h}}{h} = \abs{h}$ tends to $0$, so the derivative at $0$ is $0$.
  • $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$
    $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is differentiable at every $x$: away from $0$, the product and chain rules give $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$, and at $0$, the difference quotient $\frac{h^2 \sin(1/h)}{h} = h \sin(1/h)$ tends to $0$ because $\abs{h \sin(1/h)} \leq \abs{h}$, so $f^{\prime}(0) = 0$.
Non-examples
  • $f(x) = \abs{x}$
    $x \mapsto \abs{x}$ is not differentiable at $0$, because it has a corner there.
Properties
  • Every differentiable function is continuous.
  • Having all partial derivatives at $a$ does not imply differentiability at $a$. If the partial derivatives exist near $a$ and are continuous at $a$, then $f$ is differentiable at $a$.

continuously differentiable

High school

A function $f$ is called continuously differentiable if the derivative $f^{\prime}$ exists at each point in the domain of $f$ and the derivative is itself a continuous function.

Caveat. Having a derivative everywhere is not enough. The function given by $f(x) = x^2 \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ is differentiable everywhere, but its derivative is not continuous at $0$.
Undergraduate (functions on an open set $U \subseteq \reals^n$)

Consider a function $f : U \to \reals^m$ on an open set $U \subseteq \reals^n$. The function $f$ is continuously differentiable if every partial derivative of $f$ exists and is continuous on $U$.

Examples
Non-examples
  • $f(x) = \abs{x}$
    $f(x) = \abs{x}$ is not continuously differentiable, because $f$ has no derivative at $0$, so the derivative is not a continuous function on all of $\reals$.
  • $x \mapsto \abs{x}$ on $[-1, 1]$
    $x \mapsto \abs{x}$ is not continuously differentiable, because it is not differentiable at $0$.
  • $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$
    $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is not continuously differentiable, because $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$ has no limit as $x \to 0$.
  • $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$
    The function $f$ is not continuously differentiable, because its derivative is not continuous at $0$. At $0$, we have $f^{\prime}(0) = 0$ because \[\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} h \sin(1/h) = 0.\] Away from $0$, the product and chain rules give \[f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x),\] The points $x_n = 1 / (2 \pi n)$ tend to $0$, but $f^{\prime}(x_n) = 0 - \cos(2 \pi n) = -1$ for every $n$, so $f^{\prime}(x_n)$ does not tend to $f^{\prime}(0) = 0$. That is, \[f^{\prime}(0) \neq \lim_{x \to 0} f^{\prime}(x),\] because the limit does not exist. The discontinuity of $f^{\prime}$ at $0$ is not a jump discontinuity, at which the limits from the left and from the right exist but differ. Instead, $f^{\prime}$ oscillates: as $x \to 0$, the term $2x \sin(1/x)$ tends to $0$, but the term $-\cos(1/x)$ oscillates between $-1$ and $1$ faster and faster, so $f^{\prime}(x)$ has no limit from either side. This oscillation is the only way that a derivative can fail to be continuous, by Darboux’s theorem: every derivative has the intermediate value property, so a derivative cannot have a jump discontinuity, such as the jump of a step function.
  • The Cantor function on $[0, 1]$
  • $x \mapsto \sqrt{x}$ on $[0, 1]$
    $x \mapsto \sqrt{x}$ on $[0, 1]$ is not Lipschitz continuous or continuously differentiable on $[0, 1]$, because its slope is unbounded near $0$.
Properties

Examples

Example differentiable continuously differentiable Why
$x \mapsto \sin x$ ✓✓
$x \mapsto x \abs{x}$ ✓✓ $x \mapsto x \abs{x}$ shows that a continuously differentiable function can still have a derivative with a corner: the derivative $2 \abs{x}$ is continuous but not differentiable.
$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$ ✓✗ $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is the standard example of a differentiable function that is not continuously differentiable: the derivative exists at $0$, but the derivative $2x \sin(1/x) - \cos(1/x)$ oscillates between about $-1$ and $1$ near $0$, so the derivative is not continuous at $0$.
$f(x) = \abs{x}$ ✗✗

Notes

  • A derivative cannot jump: by Darboux’s theorem, every derivative has the intermediate value property, so a derivative that is discontinuous must be discontinuous by oscillating, as in $x^2 \sin(1/x)$.

See also

All distinctions · All definitions