Show that $f'$ is unbounded on $(0,1]$ but that $f$ is nevertheless uniformly continuous on $(0,1].$

The question is this. Let $f(x) = \sqrt{x}, \forall x \geq 0.$ Show that $f'$ is unbounded on $(0,1]$ but that $f$ is nevertheless uniformly continuous on $(0,1].$ Compare with Theorem 19.6.

The theorem 19.6 is this.

Theorem : Let $f$ be a continuous function on an interval $I$ [$I$ may be bounded or unbounded.] Let $I'$ be the interval obtained by removing from $I$ any endpoints that happen to be in $I.$ If $f$ is differentiable on $I'$ and if $f'$ is bounded on $I',$ then $f$ is uniformly continuous on $I.$

Can anyone explain this??