$f$ bounded but $f'$ isn't

Question

Is there a bounded function $f$ that holomorphic on the open unit disc but $f'$ isn't bounded?

I think first $f$ shouldn't be analytic outside the unit disc then we can't use Cauchy's inequality, because $$|f'(z)|\le\frac{\max|f|}{R}$$ When $z$ near the bound of unit disc $R$ should be small so $|f'(z)|$ can be large

Answer 1

$(z+1)\log(z+1)$ is bounded on the open unit disk $|z|\lt1$. Its derivative, $1+\log(z+1)$ is not.

Answer 2

Does this work? $$\sum_n \frac{x^{2^n}}{2^n}$$

Answer 3

Any branch of $\sqrt{1+z}$ is bounded in the unit disk, but the derivative $\frac{1}{2\sqrt{1+z}}$ is not.