Questions about derivative and differentiation For a real-valued function $f$ defined on $\mathbb{R}$ or its subset,


*

*is it possible that it is
differentiable at one point and not
in one of its neighbourhoods except
the point itself?

*is it possible that it is
differentiable over an interval, but
its derivative over the interval is
not continuous?
I found on this link that for
$f(x) = x^2 \sin(1/x)$, when $x$ is not
0, the derivative is $2x \sin(1/x) -
    \cos(1/x)$ which does not have a limit
as x approaches 0, but the
derivative of $f$ does exist at 0:
$$\lim_{h \rightarrow 0} ( h^2
    \sin(1/h) - 0)/(h-0) = 0.$$
I was wondering how $f$ is
differentiable at 0? Doesn't it
require $f$ to be defined on 0?
Thanks and regards!
 A: Here is the answer to the third question
Let us take a look at the function $f(x) = x^2 \sin(\frac{1}{x})$.
The first question is "Is the function even in $C^{(0)}$"?
The answer is not yet since the function is ill-defined at the origin. However if we define $f(0) = 0$, then yes the function is in $C^0$. This can be seen from the fact that $\sin(\frac{1}{x})$ is bounded and hence the function is bounded above by $x^2$ and below by $-x^2$. So as we go towards $0$, the function is bounded by functions which themselves tend to $0$. And the limit is $0$ and thereby the function is continuous.
Now, the next question "Is the function differentiable everywhere?"
It is obvious that the function is differentiable everywhere except at $0$. At $0$, we need to pay little attention. If we were to blindly differentiate $f(x)$ using the conventional formulas, we get $g(x) = f'(x) = 2x \sin(\frac{1}{x}) + x^2 \times \frac{-1}{x^2} \cos(\frac{1}{x})$.
Now $g(x)$ is ill-defined for $x=0$. Further $\displaystyle \lim_{x \rightarrow 0} g(x)$ doesn't exist. This is what we get if we use the formula. So can we say that $f(x)$ is not differentiable at the origin. Well no! All we can say is $g(x)$ is discontinuous at $x=0$.
So what about the derivative at $x=0$? Well as I always prefer to do, get back to the definition of $f'(0)$.
$f'(0) = \displaystyle \lim_{\epsilon \rightarrow 0} \frac{f(\epsilon) - f(0)}{\epsilon} = \displaystyle \lim_{\epsilon \rightarrow 0} \frac{\epsilon^2 \sin(\frac{1}{\epsilon})}{\epsilon} = \displaystyle \lim_{\epsilon \rightarrow 0} \epsilon \sin(\frac{1}{\epsilon}) = 0$.
(Since $|\sin(\frac{1}{\epsilon})| \leq 1$ so it is bounded).
So we find that the function $f(x)$ has a derivative at the origin whereas the function $g(x) = f'(x)$, $\forall x \neq 0$ is not continuous or even well-defined at the origin.
So we have this function whose derivative exists everywhere but then $f(x) \notin C^{(1)}$ since the derivative is not continuous at the origin.
Look up Volterra's function as an answer to your second question.
A: *

*Let $f(x)=x^2$ if $x$ is rational, $f(x)=0$ if $x$ is irrational.  If $x\neq 0$, then $f$ is not continuous at $x$, and hence not differentiable at $x$.  If $h\neq0$ is rational, then $\frac{f(h)}{h}=h$, while if $h$ is irrational, then $\frac{f(h)}{h}=0$.  Therefore $f'(0)=\lim_{h\to0}\frac{f(h)}{h}=0$, and in particular it exists.   If you wanted the example to be continuous, take a continuous function that is nowhere differentiable and multiply by $x$ to get a continuous function differentiable only at $0$.

