First of all, I hope you all are healthy.
Now, the context here is basic differential geometry for general relativity (therefore index notation is applied heavily) and $g$ is the metric tensor. The aim of this question is to prove this result (I'm writing a little book):
$$\Gamma^{j}\hspace{0.3mm}_{ji} = \frac{1}{\sqrt{|g|}}\partial_{i}\sqrt{|g|} \tag{1}$$
In order to prove $(1)$ we must to prove another result:
$$\frac{\partial \mathfrak{g}}{\partial g_{ab}} = A^{ab} \tag{2}$$
i.e. the derivative of metric determinant are given by $A^{ab}$, the elements of cofactor matrix of $g_{ab}$.
My Solution:
\begin{equation} \mathfrak{g} \equiv \mathrm{det}(g) = \sum_{(a \hspace{0.1mm} fixed)}\sum^{n}_{b = 0} g_{ab}A^{ab} \end{equation}
Then,
\begin{equation} \frac{\partial}{\partial g_{ab}}\big(\mathfrak{g} \big)=\frac{\partial}{\partial g_{ab}}\Bigg( \sum_{(a \hspace{0.1mm} fixed)}\sum^{n}_{b = 0} g_{ab}A^{ab}\Bigg) \end{equation}
Therefore, by the chain rule:
$$\frac{\partial}{\partial g_{ab}}\Bigg( \sum_{(a \hspace{0.1mm} fixed)}\sum^{n}_{b = 0} g_{ab}A^{ab}\Bigg) = \sum_{(a \hspace{0.1mm} fixed)}\sum^{n}_{b = 0} \Bigg( \frac{\partial g_{ab}}{\partial g_{ab}} A^{ab} + g_{ab}\frac{\partial A^{ab}}{\partial g_{ab}}\Bigg) =$$
$$ = \sum_{(a \hspace{0.1mm} fixed)}\sum^{n}_{b = 0} \Bigg( A^{ab} + g_{ab} \cdot 0\Bigg) \implies$$
\begin{equation} \frac{\partial \mathfrak{g}}{\partial g_{ab}} = A^{ab} \end{equation}
My question:
Since $A^{ab}$ are the components of $A(g)$ (due to the fact that $A(g)$ is the cofactor matrix of $g$) why $$\frac{\partial A^{ab}}{\partial g_{ab}} = 0? \tag{3}$$
Remember that you're thinking of the determinant of the matrix as a function of the $n^2$ entries $x_{ij}$ of the matrix. The cofactor $A^{\alpha\beta}$ is a function of the $(n-1)^2$ variables $x_{ij}$ with $i\ne\alpha$ and $j\ne\beta$. Thus, it is independent (in your notation) in particular of $g_{\alpha\beta}$.