If $f$ is holomorphic on a compact, is $||\frac{df}{dz}||$ controlled by $||f||$? Let $U\subset \mathbb{C}$ an open set and  $f:U\rightarrow \mathbb{C}$ a holomorphic function which is injective onto its image. Suppose that $f$ is uniformly bounded on every compact subset of $U$. Then also the derivatives of $f$ will be bounded.
I would like to go a little further: is it true that if $\frac 1 c <||f'||<c$ with $c\in \mathbb{R},c>1$ close enough to $1$, then also $||f''||$ will be close to $1$? What I mean is: for every $K\subset U$ and for every $\epsilon>0$ does there exist a $\delta>0$ such that if $\frac{1}{1+\delta}<||f'(x)||<1+\delta$ for every $x \in K$ then $\frac{1}{1+\epsilon}<||f''(x)||<1+\epsilon$ for every $x\in K$?
I could only think about Cauchy integral formula, but it seems useless in this case.
Do you have other ideas?

$f$ being holomorphic on a compact set does not make sense in general. You should rather consider $f:U\rightarrow {\Bbb C}$ holomorphic on an open non-empty subset of the complex plane. If uniformly bounded then $f'$
(and any higher order derivative)  is uniformly bounded on any given compact subset of $U$. For example when using a Cauchy formula at $z$ you  need $f$ being holomorphic in a (closed) $r$-neighborhood of $z$. Not quite sure what you aim at for the second part but it seems that ( $1+\delta  \times z$) on the unit disk is a counter-example?
