Showing a set is compact using Arzelà–Ascoli theorem

Question

Let $f(x)$ be a bounded function on $[a,b]$ show that the set of functions

$F(x) = \int_{a}^{x} f(z) dz$ is compact but the set of functions

$G(x) = \frac{d}{dx} f(x)$ is not compact.

My trial

Let $\mathcal{F} = \{ F(x) |\int_{a}^{x} f(z) dz, x \in [a,b] \} $ Let $| x_1 - x_2 | < \delta$ and consider

$| F(x_2) - F(x_1) | \leq \int_{x_1}^{x_2} | f(z) | dz \leq M (b-a) = \epsilon \implies$

$\mathcal{F}$ is equicontinuous (1)

Now, we prove the uniform boundedness of $\mathcal{F}$ let $F \in \mathcal{F} $

$|F(x)| \leq \int_a^{x} |f(z)| dz \leq M(b-a) \forall F \in \mathcal{F} , x \in [a,b]$ (2)

From (1) , (2) and by Arzelà–Ascoli theorem of compactness the set $\mathcal{F}$ is compact.

I'm stuck at the second part of showing that the set of functions $G(x) = \frac{d}{dx} f(x)$ is not compact.

Thanks.