Suppose I have an Euler product absolutely convergent for $\sigma >1$ $$\mathcal Z_\mathbf z(s)=\prod _p\left (1-\frac {1}{2p^{s}}\left (\frac {1}{p^z}+\frac {1}{p^{z'}}\right )\right );$$ note $$\mathcal Z_\mathbf 0(s)=\frac {1}{\zeta (s)}.\hspace {10mm}(1)$$ I want to say something about the derivatives $$ \left (\frac {d}{d\mathbf z}\right )^\mathbf D\Bigg |_{\mathbf z=0}\mathcal Z_\mathbf z(s) = \left (\frac {d}{dz}\right )^D \left (\frac {d}{dz'}\right )^{D'} \Bigg |_{z=z'=0}\mathcal Z_{z,z'}(s) $$ and ideally link them to the derivatives of $\mathcal Z_\mathbf 0(s)$. Is that possible?
One thing I'm trying is: Define for squarefree $n$ $$a_\mathbf z(n):=\frac {1}{2^{\omega (n)}}\prod _{p|n}\left (\frac {1}{p^z}+\frac {1}{p^{z'}}\right )=\frac {1}{2^{\omega (n)}}\sum _{ml=n}\frac {1}{m^zl^{z'}}.$$ Then $$\mathcal Z_\mathbf z(s)=\sum _{n=1}^\infty \frac {\mu (n)a_\mathbf z(n)}{n^s}=\sum _{n=1}^\infty \frac {\mu (n)}{n^s2^{\omega (n)}}\sum _{ml=n}\frac {1}{m^zl^{z'}}$$ so $$\mathcal Z_\mathbf 0(s)=\sum _{n=1}^\infty \frac {\mu (n)d(n)}{n^s2^{\omega (n)}}=\sum _{n=1}^\infty \frac {\mu (n)}{n^s}=\frac {1}{\zeta (s)}.$$ (This is obviously the same as (1) but I can differentiate the expressions in the derivations, and I guess I was hoping that matching $\log $ powers have equal coefficients, or something like that.)
Proposed method: Begin with Faà di Bruno's formula for the high-order derivatives of a composition of two functions: $$ \frac{d^n}{dz^n} h\bigl( g(z) \bigr) = \sum_{k=1}^n h^{(k)}(g(z)) B_{n,k}\bigl( g'(z),g''(z),\dots,g^{(n)}(z) \bigr), $$ where the incomplete exponential Bell polynomials are defined to be $$ B_{n,k}(x_1,\dots,x_n) = \sum_{\substack{0\le m_1,m_2,\dots,m_n\le n \\ m_1+2m_2+\cdots+nm_n = n \\ m_1+m_2+\cdots+m_n=k}} \frac{n!}{m_1!m_2!\cdots m_n!} \biggl( \frac{x_1}{1!} \biggr)^{m_1} \biggl( \frac{x_2}{2!} \biggr)^{m_2} \cdots \biggl( \frac{x_n}{n!} \biggr)^{m_n}, $$ When applied with $h(u)=e^u$, this becomes $$ \frac{d^n}{dz^n} e^{g(z)} = \sum_{k=1}^n e^{g(t)} B_{n,k}\bigl( g'(z),g''(z),\dots,g^{(n)}(z) \bigr) = e^{g(z)} B_n\bigl( g'(z),g''(z),\dots,g^{(n)}(z) \bigr). $$ where the complete exponential Bell polynomial is \begin{align*} B_n(x_1,\dots,x_n) &= \sum_{k=1}^n B_{n,k}(x_1,\dots,x_n) \\ &= \sum_{\substack{0\le m_1,m_2,\dots,m_n\le n \\ m_1+2m_2+\cdots+nm_n = n}} \frac{n!}{m_1!m_2!\cdots m_n!} \biggl( \frac{x_1}{1!} \biggr)^{m_1} \biggl( \frac{x_2}{2!} \biggr)^{m_2} \cdots \biggl( \frac{x_n}{n!} \biggr)^{m_n}. \end{align*} You can apply this with $g(\mathbf z) = \log \mathcal Z_\mathbf z(s)$ and then set $\mathbf z=0$; and you can also apply it with $g(s) = \log \mathcal Z_\mathbf 0(s)$. Compare the answers and see what happens....