Stirling's formula for derivatives of the Gamma function Let $s = \sigma + it$ be a complex number such that $|\arg s| \leq \pi-\varepsilon$. Fix $\sigma > 0.$ For $t \neq 0$, Stirling's formula gives
$$\Gamma(\sigma+it) = \sqrt{2\pi}(it)^{\sigma-\tfrac{1}{2}}e^{-\frac{\pi}{2}|t|}\left(\frac{|t|}{e}\right)^{it}\left\{1+O\left(\frac{1}{|t|}\right)\right\}.$$
Let $k$ be a large positive integer. Is it possible to prove a similar expression for the derivatives $\Gamma^{(k)}(\sigma + it)$, where $\Gamma^{(k)}(\sigma+it)$ denotes the $k$-th derivative with respect to $t$. In particular, can one show that $\Gamma^{(k)}(\sigma + it)$ has exponential decay as $t \to \infty$?
Edit: A previous version of the question defined $\Gamma^{(k)}(z)$ to mean the complex derivative. What I am really after is derivative with respect to $t$.

Rewrite it in the form $$\Gamma(s)=F(s)(1+O(1/|t|))=F(s)G(s)=F(s)+H(s)$$
with $F(s)$ analytic non-vanishing in the region and $G(s)=\Gamma(s)/F(s)=O(1/|t|)$.
The differentiation
$$\Gamma^{(k)}(s)=\sum_{m=0}^k {k\choose m}F^{m}(s)G^{(k-m)}(s)= F^{(k)}(s)+O(1/|t|)\sum_{m=0}^{k-1} {k\choose m}F^{m}(s)$$ $$ = F^{(k)}+O(1/|t|) F(s+kr)$$
works because the Cauchy integral formula gives that for a fixed $r$
$G^{(m)}(s)=O(\sup_{|z-s|=r} |G(z)|)= O(1/|t|)$ and $F^{(m)}(s)=O(\sup_{|z-s|=r} |F^{(m-1)}(z)|)=O(F^{(m-1)}(s+r))$, where the $O$ constants depend on $r$.
