If f(x) is differentiable AT x= a, then f(x) is also continuous there. The converse is not true: f(x)= |x| is continuous at all x but not differentiable at x= 0.
The reason f(x) must be continuous in order to be differentiable is that
\lim_{h\rightarrow 0}\frac{f(a+ h)- f(a)}{h}
always has denominator going to 0. In order that the limit exist (necessary condition, not sufficient) the numerator must also go to 0. That is, we must have
\lim_{h\rightarrow 0} f(a+ h)= f(a)
precisely the condition that f be continuous at x= a.
Of course, if f(x) is differentiable on some interval, the derivative function may not be continuous on that interval itself. However, the derivative does always satisfy the "intermediate value property" (if f(a)= u and f(b)= v, then f(x) takes on every value between u and v at some point between a and b. Amoung other things, that means that for a differentiable function, if \lim_{x\rightarrow a^-} f'(x)= \lim_{x\rightarrow a^+}f'(x) then f is differentiable at x= a and has that common value as its derivative at a. In order not to be differentiable at x= a, those two limits must not be the same. That's why you can find the derivative, if it exists, of a piecewise continuous function, at the break points, by looking at the limits of the derivative on both sides of the break point.
And, since this has nothing to do with "Differential Equations", I am moving it to "Calculus and Analysis".
