Therefore if $f(x)=\sin(\sqrt{x})$, then $f'(x)=\frac{\cos(\sqrt x)}{2\sqrt x}$ would be periodic, and this is impossible because $$\lim_{x\to \infty }f'(x)=0.$$
You can consider any two functions whose periods are incommensurable, for example $\sin(x)\sin(\pi x)$, or $\{x\}\{\sqrt 2 x\}$.
Edit: On better though, that may be false in general.
But since you are looking for a concrete example, consider this one: $\cos(x) \cos(\pi x)$. Observe that if $\cos(x)\cos(\pi x)=1$ then we have $\cos(x)=1$ and $\cos(\pi x)=1$ or $\cos(x)=-1$ and $\cos(\pi x)=-1$. In the first case, from the first equation we have $x=2\pi k$ for some integer $k$. From the second equation we have $x=2m$ for some integer $m$. So we have $\pi k = m$ for integers $k$ and $m$. If $k\ne 0$ then $\pi = m/k \in \Bbb Q$, a contradiction. So $k=0$ and then $x=0$. A similar argument shows that the second case can't happen.
This proves that $\cos(x)\cos(\pi x)$ take the value $1$ only once, hence cannot be periodic.
Best Answer
An other way to prove it is to remark that if this function is $T-$ periodic, the derivate would be $T-$ periodic too. Indeed,
$$f'(x+T)=\lim_{h\to 0}\frac{f(x+T+h)-f(x+T)}{h}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x).$$
Therefore if $f(x)=\sin(\sqrt{x})$, then $f'(x)=\frac{\cos(\sqrt x)}{2\sqrt x}$ would be periodic, and this is impossible because $$\lim_{x\to \infty }f'(x)=0.$$