Let $f$ be an analytic function in $X$ such that $|f(z)|\leq Me^{c|z|}$ for all $z \in \mathbb{C}.$ Use the Cauchy formula for derivatives to prove that $|f'(z)| \leq ceMe^{c|z|}$
I'm stumped. My first thoughts were to use the fact that if $X$ is a subspace of $\mathbb{C},$ $\bar{S}(\alpha,\rho)$ is in $X$, $f$ analytic in X and $|f(z)| \leq M$ for all $\alpha$ in the boundary of $S(\alpha,\rho)$, then $$|f^{(n)}(\alpha)| \leq M\frac {n!}{\rho}$$
but I'm not quite sure how to use this.
Any help would be greatly appreciated!
We can write from Cauchy's Integral Formula
$$f'(z)=\frac{1}{2\pi i}\oint_C\frac{f(z')}{(z'-z)^2}\,dz' \tag 1$$
Now, we apply $(1)$ by choosing the contour $C$ to be a circle of radius $r$ with center at $z$. Then, we have the following set of upper bounds for $f'$.
$$\begin{align} |f'(z)|&\le\frac{1}{2\pi }\oint_C\frac{|f(z')|}{|z'-z|^2}\,d|z'|\\\\ &\le\frac{1}{2\pi}\oint_C\frac{Me^{c|z'|}}{|z-z'|^2}\,d|z'|\\\\ &\le \frac{1}{2\pi}\int_0^{2\pi}\frac{Me^{c|z+re^{i\theta}|}}{r^2}\,r\,d\theta\\\\ &\le \frac{Me^{c|z|}e^{cr}}{r} \tag 2 \end{align}$$
Note that the value for $r$ for which the right-hand side of $(2)$ is a minimum is $r=1/c$. Thus, for all real-valued $r$, the right-hand side is at least
$$\frac{Me^{c|z|}e^{c\times \frac1c}}{\frac1c}=ceMe^{c|z|}$$
Therefore, we have
$$|f'(z)|\le ceMe^{c|z|}$$
as was to be shown!