I hope I read your question correctly as I edit my original answer considerably. To quote your original question To show that F(x) is differentiable ∀x∈(0,1) is it enough to say that as f(x) is continuous on (0,1) the derivative exists? The question is whether or not the fact the derivative function is continuous is important. The reason it's important, which isn't really clear from the Second Fundamental Theorem of Calculus, is if f isn't continuous, it can't have an antiderivative. That's what the first fundamental theorem of calculus says. Without the assumption of continuity of f,while it can be the case that the integral of f(x) is F(x) and F is continuous on [a,b],it does not follow that F(x) is differentiable with F'(x) = f(x). In other words,without the continuity assumption for f(x), the relationship with F is one sided and that means you can't simply evaluate the antiderivative at the endpoints to obtain the integral value of f on the interval.
It's important to note something else,though-the fundamental theorem of calculus (both parts) only guaruntees the existence of an antiderivative. Even if all the conditions are met, it does not follow that that antiderivative can be expressed in closed form i.e. as a formula of basic functions such as polynomials, exponentials, etc. For example, the function e^(-x^2) satisfies all the given conditions-yet it's antiderivative cannot be expressed in closed form!Even the clever trick everyone and his brother learns in basic calculus using polar coordinates to evaluate the integral-notice the result is a number,not a function.
That answer your question?