differentiability check $$f(x)=\frac{1}{x-2}$$ 
number of points where $f$ is not differentiable?
I know that the domain of the function is $\mathbb{R}\setminus\{2\}$
and differentiability is checked only in the domain of the function
so according to me the answer should be 0,
But my teacher is saying that as the function is not continuous at $x=2$, it must be non-differentiable also.
please help by solving this confusion.
 A: Calculus as a topic deals with real functions. Whenever we talk about the behaviour of ANY function with respect to ANY point or interval, it is a basic step to first find the DOMAIN. The question in this case is fundamentally wrong. You can not question the behaviour of the function for any point OUTSIDE its domain. Simple example, is tan(x) differentiable at $\frac{\pi}{2}$? Acc to your sources, answer is no! But $\frac{d \tan(x)}{dx}$ = $sec^2x$. Why?
Because, we consider the domain only! 
EDIT: Your statement that a function is either continuous or discontinuous is also valid only in the domain.
A: In my experience, teachers tend to to consider 


*

*Non-continuity

*Discontinuity


as different ideas. In particular, many of us don't define discontinuity as the logical negation of continuity, since too many students would consider any point outside the domain of the given function as a point of discontinuity. The usual definition of discontinuity (at the point $x_0 \in E$ for the function $f \colon E \to \mathbb{R}$) is that either $\lim_{x \to x_0} f(x)$ does not exist in $\mathbb{R}$, or $f(x_0) \neq  \lim_{x \to x_0} f(x)$.
But there is a second interpretation that can be rephrased as follows: a function $f \colon E \to \mathbb{R}$ is discontinuous at the point $x_0 \in \overline{E}$ if $f$ cannot be defined at $x_0$ in such a way that this extension is continuous.
For these teachers, $x_0=2$ is a discontinuity point of $f(x)=\frac{1}{x-2}$, since no definition of $f(2)$ will ensure continuity. I think that, at least in Italy, this alternative definition is very popular among high-school instructors and high-school textbooks.
This said, I have never read a book in which singular points (i.e. points where a function is not differentiable) may fall outside the domain.
A: This is a very common kind of problem in calculus courses: there are multiple conventions for dealing with the basic notions of calculus. 
This is particularly the case when we deal with "functions" that are defined by expressions that are undefined for some real numbers, e.g. $f(x) = 1/x$. If you ask different people this exact question, you will get several different answers: "If $f(x) = 1/x$ continuous everywhere?". Some will say yes, because the function is continuous everywhere in its domain. Some will say no, because the function is not defined at $x = 0$. The case of $g(x) = x/x$ is even more complicated.
Rather than trying to analyze the sources for the different conventions, I want to just emphasize practical advice. Because the conventions for these things do differ from textbook to textbook and instructor to instructor, you want to do two things. First, understand the mathematics of what is going on, apart from the conventions of terminology. Second, ask your instructor whenever you encounter one of these conventions, because only your instructor can tell you what answer is going to be required on an exam.
