continuity of the derivative under certain conditions I am working on this exercise in a book which asks to prove that $f$ is differentiable if $f$ is continuous and that $\lim \limits_{x\rightarrow x_0} f'(x)$ exists.
I know that this is easy to show using the Mean Value Theorem.
Now the problem asks: what can be said about the continuity of $f'$?
Well since $f$ is differentiable, I only have to show that $\lim \limits_{x\rightarrow x_0} f'(x)=f'(x_0)$. But ideas how to proceed elude me a bit. Any suggestions?
EDIT: Oh shoot...didn't realize how easy this was. The previous problem in this book showed how to prove half of it and I didn't realize till I read the comments below. Thanks guys!
 A: You can use l'Hospital by continuity of $\,f\,$ at $\,x_0\,$:
$$f'(x_0):=\lim_{x\to\ x_o}\frac{f(x)-f(x_0)}{x-x_0}\stackrel{\text{l'H}}=\lim_{x\to x_0}f'(x)$$
since you know the rightmost limit exists (finitely, I suppose), the limit defining the leftmost term exists (finitely), and this says $\,f'(x_0)\,$ exists.
A: You have proved that $f'(x_0)$ exists and that $f'(x)\rightarrow f'(x_0)$ as $x\rightarrow x_0$. This means exactly that $f'$ is continuous at $x_0$.
Existence of $f'(x_0)$. I assume that $f'$ exists in a neighborhood of $x_0$, excluded $x_0$ at most. Let $\ell$ be the limit of $f'$ in $x_0$. Choose $\varepsilon >0$ and $\delta>0$ such that: $$|x-x_0|<\delta \implies |f'(x)-\ell|< \varepsilon. $$
By the Mean Value Theorem, you have  $$|\dfrac{f(x)-f(x_0)}{x-x_0}-\ell|=|f'(t)-\ell|$$
for some $t$ beetween $x$ and $x_0$. Since $|t-x_0|<|x-x_0|<\delta$, it follows that  $$|\dfrac{f(x)-f(x_0)}{x-x_0}-\ell|<\varepsilon$$ for all $x$ such that $|x-x_0|<\delta$. This proves that $f'(x_0)$ exists and, at the same time, that $f'(x_0)=\ell$, by uniqueness the of the limit.
By hypothesis you have that $f'(x)\rightarrow \ell =f'(x_0)$ as $x\rightarrow x_0$ and this is a way to say that $f'$ is continuous in $x_0$.
A: (pppqqq's answer is correct, although I found the way it was written slightly confusing.  What follows amounts to the same proof.)
This is Theorem 5.31 in these notes.  I learned it from Spivak's Calculus.  Since I have not seen the result in other standard texts, I call it "a theorem of Spivak", although this is almost certainly not correct.  If someone can tell me something about the provenance of this result I will be grateful.
Remark: The way the question is phrased is a little strange to me.  If in fact you knew that $f$ was differentiable at $x_0$ as claimed, one can apply Darboux's Theorem (a.k.a. the Intermediate Value Theorem for Derivatives): this is in fact Theorem 5.30 in the aforelinked notes.  Namely, no derivative can have a "simple discontinuity".  However, in the setting of this problem I don't actually know how to prove that $f'(x_0)$ exists without showing at the same time that $f'(x_0) = \lim_{x \rightarrow x_0} f'(x)$.
