Based on Openheim's Signal and Systems book part 3.5 and with the same logic in here. We can conclude that if:
$$ f(x) \overset{FS}{\longleftrightarrow} a_k $$
then:
$$ \frac{d}{dx} f(x) \overset{FS}{\longleftrightarrow} jk\omega_0 \; a_k $$
But what about $a_0$?
Based on this then every $a_0$ of a function that can be integrated from a function with a Fourier series must be $0$.
That's obviously wrong since the Fourier Series of $f(x) = x$ (on any interval repeatedly) is attainable and the derivative of it has the $a_0 = 1$ there's same argument for any $x^{2n+1}$ and $x^{2n}$ functions.
Let’s say we’re dealing with sufficiently nice (say, continuously differentiable) $2\pi$-periodic functions on the interval $[-\pi,\pi]$ for simplicity.
Then by the fundamental theorem of calculus, by definition the $a_0$ corresponding to $f’(x)$ is
$$a_0 = \frac1\pi\int_{-\pi}^\pi f’(x)dx = f(\pi)-f(-\pi) $$
Which is zero since $f$ is periodic. So the relationship between Fourier series and derivatives indeed also holds for $k=0$ if $f$ is sufficiently nice.
In your case, $f(x) = x$ is not continuous when extended as a $2\pi$ periodic function, so in particular it’s not even differentiable and so the derivative property won’t hold in general.