What is the form of the spectral derivative in the all-positive-frequency notation in DFT? The Discrete Fourier Transform (DFT) of a function $u:[0,2\pi] \to \mathbb R$ sampled over $N$ equidistant points $\theta_j = 2\pi j/N,\, j = 0, \dots, N-1,$ is defined by

$$
\tilde U_k = \frac1N \sum_{j=0}^{N-1} u_j e^{-ik\theta_j}\,, \qquad \text{where } k \in [k]_N \in \frac{\mathbb Z}{N\mathbb Z}\,.
$$

The indices $k$ could go from $-N/2+1$ to $N/2$ in one convention and from $0$ to $N-1$ in another. Since $\tilde U_{k + N} = \tilde U_k\,, \forall k \in \mathbb Z$, the negative frequencies can be pushed all the way over to the other side of the positive frequencies to get the all-positive convention. This means that the negative frequencies $k \in \{ -1, -2, \dots, -N/2 + 1\}$ in one convention map to the frequencies $k + N \in \{N-1, N-2, \dots, N/2 + 1\}$ in the other convention.

We know that the derivative of the function $u$ can be calculated as follows:

$$
u'(\theta) = \sum_{k=-N/2 + 1}^{N/2} ik\, \tilde U_k e^{ik\theta}\,.
$$

This would mean that in the all-positive convention,

$$
u'(\theta) = \sum_{k=0}^{N/2} ik\, \tilde U_k e^{ik\theta} + \sum_{k = N/2 + 1}^{N-1} i(k-N)\, \tilde U_k e^{ik\theta} \,.
$$

Is that correct? If yes, why is it that the spectral derivative requires the use of negative frequencies? What goes wrong in the claim that

$$
u'(\theta) = \sum_{k=0}^{N-1} ik\, \tilde U_k e^{ik\theta}\,?
$$
There is no requirement, it is a matter of choice.

Some conceptual clarity is required.  A sequence of sampled points does not have a derivative.  Derivatives come from continuous functions.

If you treat the inverse DFT definition as a continuous function, the result is a continuous interpolation function that will pass through all your sampled points.  Choosing different ranges (could be 6N to 7N-1 if you wanted) yields different interpolation functions each being differentiable.

This thread should give you some insight:

How to get Fourier coefficients to draw any shape using DFT?

The "simplest" or "most natural" range is using the negative frequencies, and if you have an even number of bins, the Nyquist bin should be split in half, one positive and one negative.