Confusion about non-derivable continuous functions I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.
And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.
I am almost certain I am getting something wrong here, but I can not even pin-point what.
 A: That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.
The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.
A: As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.
If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function 
$$f(x,y)=\begin{cases}(x^2+y^2)\sin(\frac{1}{\sqrt{x^2+y^2}}) &(x,y)\neq(0,0)\\0&(x,y)=(0,0)\end{cases}$$
at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.
A: Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:\mathbb{R}\to\mathbb {R}$ (like in your original post!) is continuous at any point $\left(a,f\left(a\right)\right)$ for which $$\lim\limits_{x\to a^-}f\left(x\right)=\lim\limits_{x\to a^+}f\left(x\right)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).
And finally, note that some functions can even be nowhere-continuous as well! Such as
$$f\left(x\right)=\left\{\begin{matrix}1, x\in\mathbb{Q}\\0,x\notin\mathbb {Q}\end{matrix}\right.$$
