Prove that $\sin(\sqrt x)$ not periodic

$\sin\sqrt x$ is not a periodic function. How can one prove this?

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 $$\lim_{x\to \infty }f'(x)=0.$$

Assume that it is. Then $\sin{\sqrt{x}}=\sin{\sqrt{x+p}}=$ for all $x$ and some $p$. Solve for $p$.

Take $f(x)=\sin \sqrt{x}$. If $f(x+T)=f(x)$ for all $x$, and $T \neq 0$, then $f(T)=f(0)=0$. So $T$ is a root of $f$. So $T=n^2 \pi^2$ for some integer $n$.

If $f(x+T)=f(x)$ for all $x$. Then $f(2T)=f(T)=0$. Recall we assumed $T \neq 0$. That implies that there exists two distinct roots of $f$, $x_1$ and $x_2$ such that $\frac{x_2}{x_1}=2$. That means,

$$\frac{m^2 \pi^2}{n^2 \pi^2}=\frac{m^2}{n^2}=\left(\frac{m}{n} \right)^2=2$$

For some integers $n$ and $m$. This is impossible since $\sqrt{2}$ is irrational.