Does there exist a continuous and differentiable function which isn't smooth?

Question

As I understand, a smooth function is continuously differentiable.

But if I have a function which is continuous AND differentiable, I cannot automatically say that it is smooth. For it has to be so for all its differentials.

So I wonder, what function would be continuous and differentiable, but not continuously differentiable?

I cannot find the answer myself, as I do not clearly understand the difference between continuous AND differentiable, and continuously differentiable....

Context: I ask this because of an arc length contest. The function has to be continuous and differentiable on [0,1]. But does this automatically mean that I may always use the formula for an arc length, which has the condition that the function is smooth... (or, in another book, that it has a continuous derivative)?

Answer 1

EDIT: there is some confusion as to what is being asked here. I am answering "Does there exist a continuous and differentiable function which isn't smooth?" (mentioned in the title and the question), but I see that "what function would be continuous and differentiable, but not continuously differentiable?" is also asked. To this second question, I recommend looking at @Ian's comment below.

A good example is

$$ f(x) = \begin{cases} x^2 & : x \geq 0 \\ 0 & : x < 0 \end{cases} $$

You can check the right-hand limit of both the function and first derivative are $0$, but the second derivative is discontinuous.

Similarly, $$ f(x) = \begin{cases} x^n & : x \geq 0 \\ 0 & : x < 0 \end{cases} $$

provides an example where the function and first $n-1$ derivatives are continuous, but the $n$th derivative is not.

Answer 2

We say that $f$ is a function of class $C^k$ if it has $k$ continuous derivatives. You want to find a differentiable function which is not $C^1$, that is, a differentiable function with discontinuous derivative. Given $k \in \Bbb Z_{> 0}$, the function $$f(x) = \begin{cases} x^k\sin(1/x), & \text{if } x \neq 0 \\ 0, & \text{if }x=0\end{cases}$$ is $C^{k-1}$ but not $C^k$.