Show that the sequence of derivatives $(h_n')$ diverges for every $x\in\mathbf{R}$, where $h_n(x)=\frac{\sin nx}{\sqrt{n}}$

Question

I am self-learning Real Analysis from the text, Understanding Analysis by Stephen Abbott.

[Abbott 6.3.4] Let

$$h_n(x) = \frac{\sin nx}{\sqrt{n}}$$

Show that $h_n(x) \to 0$ uniformly on $\mathbf{R}$ but that the sequence of derivatives $(h_n')$ diverges for every $x \in \mathbf{R}$.

Whilst it is an easy exercise to show that $(h_n)$ converges uniformly on $\mathbf{R}$ to the constantly zero function $h(x) = 0$, I don't quite know how to prove the latter part of the question.

If we fix $n\in\mathbf{N}$, since the trigonometric function $\sin (x)$ is everywhere differentiable, by the familiar rules of differentiation:

$$h_n'(x) = \sqrt{n} \cos {nx}$$

Now, $h'(x) = 0$. Doesn't $h_n'(\pi/2)$ converge pointwise to $h'(\pi/2)$? Any hints/clues would be extremely helpful.