Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question regarding the following exercise.

Let $f(x)= \begin{cases} x^n \mbox{ for } x \geq 0\\ 0 \mbox{ for } x<0 \end{cases}$

Show that the iterated derivatives $f^{(1)}$ through $f^{(n-1)}$ exist at all real numbers x, but the n-th iterated derivative at 0 does not.

I was able to prove that the n-th iterated derivative at 0 does not exist: at right of 0, its value is n! and left of 0, its value is 0.

Since the derivative left and right of 0 is different, the n-th iterated derivative is therefore non differentiable in 0.

But can someone show me how to prove that the first to the n-1 derivatives exist?

Thank you

share|cite|improve this question
Have you been taught the epsilon-delta definition of a differentiation? If so, just try to plug it in, using $\displaystyle \frac{x^n-y^n}{x-y} = \sum_{i=0}^{n-1}x^iy^{n-1-i}$ – Lord_Farin Oct 17 '12 at 19:10
No we did not yet – user43418 Oct 17 '12 at 19:10
up vote 1 down vote accepted

Consider the quotient $$\frac{f^{(n-1)}(h+0)-f^{(n-1)}(0)}{h}.$$

  • What if $h\to0$ from below (that is $h<0$)?
  • What if $h\to0$ from above (that is $h>0$)?

If you wish you might start with $n=1$.

Edit For $n=1$ we should look at $$\frac{f^{(0)}(h+0)-f^{(0)}(0)}{h}=\frac{f(h)-f(0)}{h},\tag{1}$$ with $$f(x)=\begin{cases} x \text{ for } x \geq 0\\ 0 \text{ for } x<0 \end{cases}\tag{2}$$ Now, substitute (2) into (1). What happens when $h\to0$ for $h<0$ and what happens when $h\to0$ for $h>0$?

Now go for $n=2$, $n=3$, etc. until you see and can explain the general picture.

share|cite|improve this answer
I don't understand – user43418 Oct 17 '12 at 20:32
@user43418 Is the edit easier? – AD. Oct 17 '12 at 21:48
If h>0, the quotient equals h and its limit is therefore 0 If h<0, the quotient equals 0 Since the limit left and right of 0 is the same, then $f^{(0)}$ exists. – user43418 Oct 17 '12 at 22:24
And now I suppose it is true for n-2 terms and show that it is true for n-1 right ? – user43418 Oct 17 '12 at 22:24
@user43418 No it is not $h$ for $h>0$. – AD. Oct 18 '12 at 5:25

You would do something very similar. The $k$th derivative would be given by $$f^{(k)}(x) = \begin{cases} (n)_kx^{n-k} \mbox{ for } x > 0\\ 0 \mbox{ for } x<0 \end{cases}$$ where $(n)_k = n(n-1)\cdots (n-k+1)$ is the falling factorial. There's no issues away from $0$, but as $x\rightarrow 0$ from the left and the right, do the limits coincide?

share|cite|improve this answer
yes I think they do – user43418 Oct 17 '12 at 19:32
I think they do too. But can you prove that they do? I would probably use Newton's quotient explicitly to show that the limits coincide. – EuYu Oct 17 '12 at 19:33
Can you prove it for me. I just did 10 problems of derivatives and this is my last question. Please :( – user43418 Oct 17 '12 at 19:34
That isn't how this site works. I'm sorry. I'd be happy to provide assistance if you show me some progress though. – EuYu Oct 17 '12 at 19:35
So I have to show it to be true for n=1 and then for n terms by induction ? – user43418 Oct 17 '12 at 20:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.