Deriving upper bound for derivative of analytical function

$f$ is an analytical function on the unit disk $\mathbb{D}$ for which there is a $C > 0$ such that

$\forall z \in \mathbb{D} \colon |f(z)| \leq \dfrac{C}{1-|z|}$.

I know have to prove that

$\forall z \in \mathbb{D} \colon |f'(z)| \leq \dfrac{4C}{(1-|z|)^2}$.

I tried using the Cauchy integral formula but I can't find a fitting circle to get this result.

The way to do this is to apply the Cauchy bound in a disk $D_1$ centered at fixed $z, |z|=r$ and of radius $R-r, r<R<1$ and then hope we find an $R$ satisfying the above constraint that leads to what we require.

We know by hypothesis (on the $R$ disc centered at the origin and maximum modulus, noting that the disc $D_1$ is included in the $R$ disc centered at the origin) that on $D_1, |f(w)| \leq \dfrac{C}{1-R}$, so by Cauchy $|f'(z)| \leq \dfrac{C}{(R-r)(1-R)}$.

Maximizing the denominator to give the best estimate leads (quadratic in $R$ or just make the terms equal) to $R=\frac{1+r}{2}$ which satisfies the constraint $r<R<1$ and gives us precisely what we need:

$|f'(z)| \leq \dfrac{4C}{(1-r)^2}=\dfrac{4C}{(1-|z|)^2}$