Is differentiation as a map discontinuous?

For a counterexample, take the sequence $$\frac {\sin nx} n$$ These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.

It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$, where $C([0,1])$ is is equipped with the supremum norm $\lVert f \rVert = \sup_{x \in [0,1]} \lvert f(x) \rvert$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm $$\lVert f \rVert^{(1)} = \lVert f \rVert + \lVert f' \rVert .$$ Then $A : (C^1([0,1]), \lVert - \rVert^{(1)}) \to (C([0,1]), \lVert - \rVert)$ is trivially continuous.

Edited:

$(C^1([0,1]), \lVert - \rVert^{(1)})$ is a Banach space. See Prove that $C^1([a,b])$ with the $C^1$- norm is a Banach Space. In contrast, $(C^1([0,1]), \lVert - \rVert)$ is not. You can generalize this to the sets $C^k([0,1])$ of $k$-times continuously differentiable functions. They are Banach spaces if equipped with $$\lVert f \rVert^{(k)} = \lVert f \rVert + \lVert f' \rVert + \ldots + \lVert f^{(k)} \rVert$$ and $$A : (C^{(k)}([0,1]), \lVert - \rVert^{(k)}) \to (C^{(k-1)}([0,1]), \lVert - \rVert^{[k-1)}), A(f) = f' ,$$ is continuous.