1. ## differentiable function

Hi!

I must prove that the function f(x) is continuously differentiable function (just first derivate) on interval (-1,1).
$f(x)=exp(\frac{1}{x^2-1}) \ |x|<1 , \ 0$ for all other values.

I do that:

$f(x)=e^{\frac{1}{x^2-1}}=e^u \ [u(x)=\frac{1}{x^2-1}]$
$\frac{d}{dx}f(x)=\frac{df}{du}\frac{du}{dx}=-e^u \frac{2x}{(x^2-1)^2}=-\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}$

Is it ok?

2. Originally Posted by tom27
Is it ok?
If the question it is only related to $(-1,1)$ then, it is right (evidently $f'(x)$ is continuous in $(-1,1)$ ).

Fernando Revilla

3. The basic question is if the f(x) is a part of $C_0^1(\mathbb{R})$ space?

4. That's actually a different question from what you first asked. In your first post you said "continuously differentiable function (just first derivate) on interval (-1,1)". Now you are saying on all of $\mathbb{R}$. You had already answered the first question- f(x) is continuously differentiable on (-1, 1). Since f is defined to be a constant (0) for $x\le -1$ or $x\ge 1$, it is certainly continuously differentiable there. But now the question is about x= -1 and x= 1. What are the derivatives at those points? What are the derivatives on either side? Is the derivative continuous at x= -1 and x= 1?

5. Originally Posted by tom27
The basic question is if the f(x) is a part of $C_0^1(\mathbb{R})$ space?
In that case, you should prove $f'_{-}(1)=f'_{+}(-1)=0$ and $f'$ continuous at $x=\pm 1$ (for the rest is almost trivial)

Fernando Revilla

P.S. I suppose $C_0^1(\mathbb{R})$ means $C^1(\mathbb{R})$ (the standard notation for the set of continuosly differentiable functions on $\mathbb{R}$ ) .

Edited: Sorry, I didn't see HallsofIvy's post

6. How can I prove $f'_{-}(1)=f'_{+}(-1)=0$?

I tried with limits:

$\displaystyle \lim_{h \to 1}\frac{e^{\frac{1}{(x+h)^2-1}}-e^{\frac{1}{x^2-1}}}{h}$ and $\displaystyle \lim_{h \to -1}\frac{e^{\frac{1}{(-x-h)^2-1}}-e^{\frac{1}{x^2-1}}}{h}$

but this certainly is not the correct way.

7. While derivatives are not necessarily continuous, they do obey the "intermediate value property"- f'(x) takes on all values between f'(a) and f'(b) for x between a and b. In particular, that means that f'(c) exists if and only if $\displaystyle \lim_{x\to c^-} f'(x)= \lim_{x\to c^+}f'(x)$. That is, determine the derivatives for x< -1, and x>-1 and determine if the limits at x= -1 are the same. Do the same for x= 1.

By the way, you don't want to prove that " $f'_-(1)= f'_+(1)= 0$". You want to prove that $f'_-(1)= f'_+(1)$ and $f'_-(-1)= f'_+(-1)$. You don't need to use the limit definition of the derivative- but, of course, h goes to 0, not 1 and -1.

8. So $f'_-(-1)=f'_+(1)=0$ and i must calculate:

$f'_-(1)=\displaystyle \lim_{x \to 1} -\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}}$
$f'_+(-1)=\displaystyle \lim_{x \to -1} -\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}}$

but how can I do that? The exp function is not defined in those two points!

9. Fuction with exp is not define in those two points so limit does not exist. But wikipedia says: the function is smooth so is continuosly differentiable on $\mathbb{R}$. But I dont know how to prove that if one limit does not exist so $f'_-(1)\ne f'_+(1)$?