X not differentiable at x=0

1. Mar 24, 2012

Shawn Garsed

If the derivative of x^n equals nx^(n-1), then the derivative of x or x^1 equals x^0, but 0^0 is undefined. Does that mean x is not differentiable at zero?

2. Mar 24, 2012

Why do you think 0^0 is undefined? 0^0 = 1.

3. Mar 24, 2012

Shawn Garsed

That's what it says in my book. That may not be the best answer, but that's where my question comes from.

4. Mar 24, 2012

arildno

No.
But it means that that formula is only strictly valid for ALL n if you CHOOSE 0*0^(-1) (n=0) and 1*0^0 (n=1) in that particular context to mean 0 and 1, respectively.

Since those partial results are derivable by wholly independent ways, that local definition is utterly unproblematic.

5. Mar 24, 2012

I like Serena

As you can see here on wiki, your formula for the derivative of a power of x is indeed not generally valid.

Exceptions are specifically made for f(x)=c and for f(x)=x, in which cases f does have a proper derivative at x=0, but where the formula $nx^{n-1}$ is not properly defined.

6. Mar 24, 2012

Shawn Garsed

Thanks, I had a feeling that it had something to do with nx^(n-1) not being universally valid.

7. Mar 25, 2012

HallsofIvy

Staff Emeritus
When ever you see a "general formula" that appears to have a problem for specific values, it is best to go back to the basic formulas.

Here, to find the derivative of f(x)= x, we can write
$$\frac{f(x+h)- f(x)}{h}= \frac{x+h- x}{h}= \frac{h}{h}$$
Now, we can say that is equal to 1 as long as h is not 0!

Fortunately, for problems like this (and, after all, the derivative always involves a fraction in which both numerator and denominator go to 0) we have the very important limit property:
If f(x)= g(x) for all x except x= a, then $\lim_{x\to a} f(x)= \lim_{x\to a}g(x)$.
That allows us, as long as we are taking limits to handle things like "0/0".

8. Mar 25, 2012

Edgewood11

0^0= 1

9. Mar 25, 2012

HallsofIvy

Staff Emeritus
Please don't simply assert things (especially those others have already said were false) without giving some argument. What reason do you have to say that "0^0= 1"?

10. Mar 27, 2012

LikeMath

It seems to me that case should be eliminated from the general formula, so the formula is written like follows
If $f(x)=x^n$, $n\in \mathbb{N}$
then

$f^{\prime}(x)=\begin{cases} 1 & \text{ if } x= 0 \text{ and } n=1 \\ n x^{n-1} & \text{ otherwise.} \end{cases}$

11. Mar 27, 2012

I like Serena

@LikeMath: there is another special case for n=0.
However the formula also holds true for any $n \in \mathbb{R}$ if x>0.

12. Mar 27, 2012

hmm, sounds strange to me. I'm not a math expert like others on this forum, but from what I know from abstract algebra, if we take ordinary multiplication as a binary operation, then for any x we have x^0 = 1 (the identity element), or generally in a group we have g^0=e for any g in G.

I think your book says 0^0 is undefined because one can interpret it in two different ways and concludes that 0^0 = 0 and also 0^0=1. because:
1 - 0 times anything is equal to zero, hence 0^0 = 0
2 - x^0 = 1 for any x, hence 0^0=1
so your book considers 0^0 undefined because we can associate two different values to it (hence, it's not well-defined), but I think the first argument doesn't apply, because when we say 0^0 it means we're multiplying 0 zero times by itself, which means actually we don't multiply anything by 0. so I think the first case isn't relevant here or at least we can define 0^0=1, just like we define 0! = 1.

Do we really need to give any argument? Don't we define x^0 = 1 for any given x?

--------------------------------------------------------------------------------------

Well, after having spent some time on this, I realized that 0^0 might really be ill-defined, because I found no example in group theory that could explain 0^0=1, but still when I analyzed the function f(x)=x^x I found out that x^x = 1 as x approaches 0. Is it really ill-defined? or does it cause any contradiction if we define 0^0=1?

Last edited: Mar 27, 2012
13. Mar 27, 2012

I like Serena

R is not a group with multiplication, since 0 does not have an inverse.

You will need to specify a context for that definition, because contradictions are lurking around the corner.
EDIT: What should we do with 0^n for any given n?

14. Mar 27, 2012

That is exactly what I thought of. Please refresh. (I edited my previous post)

I can't say whether 0^0 is undefined or not, it seems that if we define 0^0=1 no contradiction would arise, but on the other hand as you rightly pointed out (and I had already thought of it) R is not a group under multiplication (R-{0} is), so my group theory argument is obviously wrong, but I'm not quite sure whether we could say 0^0=1 or not, because f(x)=x^x suggests us that it's possible to define 0^0=1.

15. Mar 27, 2012

D H

Staff Emeritus
The problem is by choosing different paths toward the origin, $\lim_{x,y\to 0} x^y$ can be made to have any value whatsoever. In other words, 0^0 is indeterminate (rather than undefined).

In many contexts such as the binomial expansion and the problem at hand, it is convenient to define 0^0 to be 1 as an abuse of notation. Extremely convenient, but still an abuse of notation.

16. Mar 27, 2012

I like Serena

For f(x)=0^x you might define 0^0=0.
For f(x)=x^0 you might define 0^0=1.

In other words, you would define 0^0 depending on which function f you're talking about.
The only reason you would do something like that, is to get out from under the burden to meticulously define f(0) in each case.

17. Mar 27, 2012

Yup. That's an insightful post.

You're right.

Actually I was thinking of a similar kind of abuse of notation that DH said. It's not immediately clear why we should define 0! = 1, but it turns out to be very useful in infinite series and binomial expansions, it would be a very useful abuse of notation, but still an abuse of notation. lol.
Thanks.

18. Mar 27, 2012

D H

Staff Emeritus
0! = 1 is not an abuse of notation.

19. Mar 27, 2012

I like Serena

Check out http://en.wikipedia.org/wiki/Empty_product.
As DH said, 0!=1 is well defined. ;)

20. Mar 27, 2012