# Differentiability implies continuous derivative?

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1. Jan 28, 2015

### kelvin490

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?

2. Jan 28, 2015

### HallsofIvy

Staff Emeritus
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.

3. Jan 28, 2015

### lavinia

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

4. Jan 28, 2015

### Svein

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 $f(x) = \int_{0}^{x}g(t) dt$.

5. Jan 29, 2015

### Hawkeye18

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)

6. Jan 29, 2015

### Svein

What happened to the old rule: "To derive with respect to the upper limit of an integral, insert the upper limit into the integrand".

7. Jan 29, 2015

### Hawkeye18

As quoted this rule work only for continuous integrands. In the general case it only guaranties equality almost everywhere.

8. Jan 31, 2015

### lavinia

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: Jan 31, 2015
9. Feb 1, 2015

### jbunniii

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.

10. Feb 5, 2015

### mathwonk

It took me absurdly long to realize how trivial Darboux's theorem is: if f' > 0 at some point and < 0 at another point then f cannot be monotone in between (by the intermediate value theorem), so f has a local extremum where f' = 0.

by the way @Kelvin, the continuity of the partials is sufficient but not necessary for differentiability in several variables. the definition of differentiability is in terms of an approximating linear function.

11. Feb 5, 2015

### Svein

To maximize the confusion: Let I = [0, 1] and J be an open interval such that I ⊂ J. Then there exists a function that is C, equal to 1 on I and equal to 0 outside J.