Continuously differentiable function
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
Examples
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$f(x) = x^2$ on $\reals$
High school
$x \mapsto x^2$ on $\reals$ is continuously differentiable because its derivative $2x$ is continuous.
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$x \mapsto \sin x$
High school
$x \mapsto \sin x$ is continuously differentiable, because its derivative $\cos x$ is continuous.
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$x \mapsto x \abs{x}$
High school
$x \mapsto x \abs{x}$ is continuously differentiable, because its derivative $x \mapsto 2 \abs{x}$ is continuous.
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$x \mapsto x^2$ on $[0, 1]$
High school
$x \mapsto x^2$ on $[0, 1]$ is continuously differentiable because its derivative $2x$ is continuous.
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$x \mapsto \sin(1/x)$ on $(0, 1]$
Undergraduate
$x \mapsto \sin(1/x)$ is continuously differentiable, and so continuous, on $(0, 1]$.
Non-examples
Things that may be mistaken for examples.
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$f(x) = \abs{x}$
High school
$f(x) = \abs{x}$ is not continuously differentiable, because $f$ has no derivative at $0$, so the derivative is not a continuous function on all of $\reals$.
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$x \mapsto \abs{x}$ on $[-1, 1]$
High school
$x \mapsto \abs{x}$ is not continuously differentiable, because it is not differentiable at $0$.
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$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$
Undergraduate
$f(x) = x^2 \sin(1/x)$ 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 $x \to 0$.
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$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$
Undergraduate
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 $[0, 1]$ Undergraduate
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$x \mapsto \sqrt{x}$ on $[0, 1]$
Undergraduate
$x \mapsto \sqrt{x}$ on $[0, 1]$ is not Lipschitz continuous or continuously differentiable on $[0, 1]$, because its slope is unbounded near $0$.
Properties
- A continuously differentiable function is locally Lipschitz continuous, and it is Lipschitz continuous on every compact convex subset of its domain.
Distinctions involving this term
- Lipschitz continuous vs. continuously differentiable: Lipschitz continuity and continuous differentiability both rule out functions that are too steep or too wild, but in different ways. A Lipschitz continuous function has a bounded slope but may have corners. A continuously differentiable function has no corners, but its slope may be unbounded on an unbounded domain.
- Differentiable vs. continuously differentiable: 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 $x$.
- Types of continuity: There are many ways to describe the continuity of a function. On a compact interval, each is a special case of the next, from continuously differentiable to continuous.