Assume has a finite derivative everywhere on except possibly at a point . If we have , then prove exists and is equal to .
You cannot hope to prove this unless you are given that f is continuous at c. Otherwise you can have a function that is continuous and differentiable everywhere in the interval apart from c but which takes an arbitrary value at c itself, or you could have a function that looks like tan(x) but with a small section round pi/2 removed. Then there is a huge discontinuity at c, and yet the derivative is the same both sides of c.
We must be allowed to assume that f is continuous on (a,b), otherwise the propostion would be clearly false (because in that case f(c) could be anything).
So, assuming f is continuous on (a,b), I suggest that you use the mean value theorem, applied to (a,c] and [c,b), respectively, to show that the left and the right derivative of f at c both exist and are both equal to A.
The question does not actually say that is continuous at . But it better had be, as noted by alunw and Failure. With that assumption, the result is true, but it seems to be quite a tricky piece of analysis, requiring a very careful proof.
We are told that . So given there exists such that whenever .
Now suppose that and . By the mean value theorem (as hinted by flyingsquirrel ) for some between and . But if and are both within distance from then so is . It follows that .
Next, let in that last inequality. By continuity of at , this gives . Since this holds whenever , and is an arbitrary positive number, it follows that is differentiable at , with derivative .
Good point. I should have said that y must be on the same side of c as x. Then I think the proof should work. But in fact your comment #5 above gives a shorter proof: once you know that f is continuous on the closed interval [c,x], you can apply the MVT on that interval and the result comes out straight away.