Continuously differentiable function of a single real variable

In real analysis, given a subset <math>S\subseteq\mathbb{R}</math>, a real function <math>f:S\to\mathbb{R}</math> is said to be continuously differentiable at a point <math>x_{0}\in S</math> if the derivative of <math>f</math>, that is, <math>f'</math>, is continuous at <math>x_{0}</math>. Given a subset <math>U\subseteq S</math>, <math>f</math> is said to be continuously differentiable on <math>U</math> if it is continuously differentiable at every point in <math>U</math>. <math>f</math> is said to be continuously differentiable if it is continuously differentiable on its domain.

It is possible that <math>f</math> is differentiable at a point (that is, its derivative, <math>f'</math> exists at that point), but is not continuously differentiable at that point, that is, <math>f'</math> may have a discontinuity in its domain.

Differentiable but not continuously differentiable

Let <math>D</math> be the set of points at which a real function is differentiable, but not continuously differentiable.

D consisting only of isolated points

Let <math>f:\mathbb{R}\to\mathbb{R}</math> be such that <math>f(0)=0</math> and for all <math>x\in\mathbb{R}</math>, <math>x\neq0</math> implies <math>f(x)=x^{2}\operatorname{sin}(1/x)</math>. Then <math>f</math> is differentiable at <math>0</math>, but <math>f'</math> is discontinuous at <math>0</math>.

Since <math>-1\leq\operatorname{sin}(1/x)\leq 1</math> for all non-zero real <math>x</math>, then <math>-x^{2}\leq f(x)\leq x^{2}</math>, and hence <math>-|x|\leq\frac{f(x)-0}{x-0}\leq|x|</math>. Applying the squeeze theorem, <math>f'(0)=0</math>.

In every neighbourhood of <math>0</math>, the range of the derivative of <math>f</math> contains the interval <math>[-1,1]</math>. This implies the oscillation of <math>f'</math> at <math>0</math> is non-zero, hence <math>f'</math> is discontinuous at <math>0</math>.

The function <math>g:\mathbb{R}\to\mathbb{R}</math>, such that <math>g(x)=(x(1-x))^{2}\operatorname{sin}(\frac{1}{\pi x(1-x)})</math> for all <math>x\in(0,1)</math> and that <math>g(x)=0</math> for all <math>x\in(-\infty,0]\cup[1,\infty)</math>, is differentiable everywhere but not continuously differentiable at <math>0</math> and <math>1</math>.

D consisting of an accumulation point

The derivative of the composite function <math>f\circ f</math>, that is, <math>(f\circ f)'</math>, possesses discontinuities not only at <math>0</math>, but also at every deleted neighbourhood of <math>0</math>, while <math>f\circ f</math> is differentiable on <math>\mathbb{R}</math>.

Since <math>\{x\in\mathbb{R}|\operatorname{sin}(x)=0\}</math> is unbounded above and below, <math>\{x\in\mathbb{R}|\operatorname{sin}(1/x)=0\}</math> has an accumulation point at <math>0</math>. Furthermore, note that <math>f(0)=0</math>.

Since <math>f'(0)=0</math>, <math>(f\circ f)'(x)=0</math> for all <math>x\in\mathbb{R}</math> such that <math>f(x)=0</math>, by the chain rule. So the set of points at which <math>(f\circ f)'</math> is discontinuous is <math>\{x\in\mathbb{R}|\operatorname{sin}(1/x)=0\}\cup\{0\}</math>, which has an accumulation point at <math>0</math>.

General case

A subset of <math>\mathbb R</math> is the set of discontinuities of the derivative of a differentiable function if and only if it is a meagre <math>F_\sigma</math> set. In particular, there exist differentiable functions whose derivatives are discontinuous almost everywhere.

Derivatives of such functions are not Riemann integrable, regardless of whether they are bounded, and therefore the usual version of the fundamental theorem of calculus does not hold. They may fail to be Lebesgue integrable as well and therefore the fundamental theorem fails to hold for that integral. The Denjoy integral and the Henstock–Kurzweil integral were developed in part to address this problem. On a compact interval, every finite derivative is Denjoy integrable and Henstock–Kurzweil integrable, and these integrals recover the original function up to an additive constant.

Continuously differentiable, but not continuously differentiable in a neighbourhood

It is possible for a function's derivative to be continuous at a point, but also there to exist no neighbourhood of the point at which the derivative is continuous. In other words, the set of points at which a function's derivative is continuous need not be an open set, an example being the function <math>F:\mathbb{R}\to\mathbb{R}</math>, such that for all <math>x\in\mathbb{R}</math>, <math>F(x)=xf(f(x))</math> (where <math>f</math> is defined as above).

The derivative of <math>F</math>, that is, <math>F'</math>, is given by <math>F'(x)=f(f(x))+xf'(f(x))f'(x)</math>. Since <math>f</math> and <math>f'</math> are defined everywhere, so <math>F'</math> is also defined everywhere. However, at each root of <math>f</math>, <math>(f\circ f)'</math> is discontinuous, since <math>f'</math> is discontinuous at <math>0</math>. That is, <math>(f\circ f)'</math> has non-zero oscillation at each root of <math>f</math>. Since at each non-zero root of <math>f</math>, <math>f'</math> is non-zero, the oscillation of <math>xf'(f(x))f'(x)</math> is non-zero for non-zero roots of <math>f</math> (note that, although <math>(f\circ f)'</math> is zero at each root of <math>f</math>, it isn't close to <math>0</math> for values in any of the neighbourhoods of non-zero roots of <math>f</math>). However, when <math>x=0</math>, the oscillation of <math>xf'(f(x))f'(x)</math> is <math>0</math>. Since <math>(f\circ f)'</math> is <math>0</math> at each root of <math>f</math>, <math>F'</math> is <math>0</math> at each root of <math>f</math>, but the only root of <math>f</math> at which <math>F'</math> is continuous is <math>0</math>. Since the roots of <math>f</math> have an accumulation point at <math>0</math>, while <math>F'</math> being continuous at <math>0</math>, the set of points at which <math>F'</math> is continuous is non-open.

It is also possible that the set of points at which a function's derivative is discontinuous is a dense subset of the set of points at which the function is differentiable, as long as the set of points at which the derivative is continuous is a dense subset of the set of points at which the function is differentiable.

Possible discontinuities of a derivative

As a result of Darboux's theorem, a function's derivative must satisfy the intermediate value property. In other words, if a function is differentiable on a closed interval, its derivative must pass through every value between the images of the endpoints of the interval under the derivative.

What follows is that the only discontinuity a function's derivative can have is an essential discontinuity. If a function has a jump, removable, or infinite discontinuity, it does not satisfy the intermediate value property.

A necessary condition for non-zero continuous derivative

The inverse function theorem states that a real function whose derivative is continuous and is non-zero at a point is injective in a neighbourhood of that point.

If the derivative of a function at <math>x_{0}</math> is a non-zero real number, and the function's derivative is continuous at <math>x_{0}</math>, then there exists a non-empty open interval centered at <math>x_{0}</math> over which exactly one and only one of the following is true – the derivative is entirely positive (which implies the function is strictly increasing over the open interval), or the derivative is entirely negative (which implies the function is strictly decreasing over the open interval). In both cases, the function is injective in the open interval, which is a neighbourhood of <math>x_{0}</math>.