# Differentiability implies continuous derivative?

Gold Member
We know differentiability implies continuity, and in 2 independent variables cases both partial derivatives fx and fy must be continuous functions in order for the primary function f(x,y) to be defined as differentiable.

However in the case of 1 independent variable, is it possible for a function f(x) to be differentiable throughout an interval R but it's derivative f ' (x) is not continuous?

HallsofIvy
Homework Helper
Yes, it is. I gave an example, $f(x)= x^2 sin(x)$ if x is not 0, f(0)= 0, when you asked this question on another board.

lavinia
Gold Member
Yes, it is. I gave an example, $f(x)= x^2 sin(x)$ if x is not 0, f(0)= 0, when you asked this question on another board.

I think you meant ## x^{2}sin(1/x)## when x is not zero.

Svein
However in the case of 1 independent variable, is it possible for a function f(x) to be differentiable throughout an interval R but it's derivative f ' (x) is not continuous?
The simplest such function is f(x) = |x| on [-1, 1] (I know, there is a problem with f'(0), but let f'(0)=0 and see what happens).

A lot of functions with discontinuous derivatives can be found by starting with a discontinuous function g(x) and let $f(x) = \int_{0}^{x}g(t) dt$.

The simplest such function is f(x) = |x| on [-1, 1] (I know, there is a problem with f'(0), but let f'(0)=0 and see what happens).

A lot of functions with discontinuous derivatives can be found by starting with a discontinuous function g(x) and let $f(x) = \int_{0}^{x}g(t) dt$.

The function ##f(x)=|x|## is not differentiable at ##x=0##. For the second example, the function ##f## is usually not differentiable everywhere, one can only prove that it is differentiable almost everywhere. Fro the function ##f## in this example to be differentiable at all points, the function ##g## has to be very special.

The simplest example of the function differentiable everywhere with non-continuous derivative is probably the example by HallsofIvy (with the correction by lavina)

Svein
For the second example, the function ff is usually not differentiable everywhere, one can only prove that it is differentiable almost everywhere. Fro the function ff in this example to be differentiable at all points, the function gg has to be very special.
What happened to the old rule: "To derive with respect to the upper limit of an integral, insert the upper limit into the integrand".

What happened to the old rule: "To derive with respect to the upper limit of an integral, insert the upper limit into the integrand".
As quoted this rule work only for continuous integrands. In the general case it only guaranties equality almost everywhere.

lavinia
Gold Member
The simplest such function is f(x) = |x| on [-1, 1] (I know, there is a problem with f'(0), but let f'(0)=0 and see what happens).

A lot of functions with discontinuous derivatives can be found by starting with a discontinuous function g(x) and let $f(x) = \int_{0}^{x}g(t) dt$.

There are also functions that are continuous but nowhere differentiable or differentiable only at finitely many points.
There are functions that are differentiable except on a Cantor set.

Last edited:
jbunniii
Homework Helper
Gold Member
As others have pointed out, a differentiable function need not have a continuous derivative. However, the derivative ##f'## does satisfy the "intermediate value property," which means that if there are points ##x## and ##y## with ##f'(x) = a## and ##f'(y) = b##, then for any given ##c## between ##a## and ##b##, there is some point ##z## such that ##f'(z) = c##. This fact is known as Darboux's theorem and is interesting because it implies that ##f'## cannot have a step discontinuity. This does not contradict the example given by HallsOfIvy, in which ##f'## has an "oscillate to death" discontinuity, not a step discontinuity.

• Not anonymous
mathwonk