(29)
$$ \frac{1}{\sqrt{-g}} \frac{\partial \sqrt{-g}}{\partial x_{\sigma}} =
\frac{1}{2} \frac{\log (-g)}{\partial x_{\sigma}} = \frac{1}{2}
g^{\mu\nu} \frac{\partial g_{\mu\nu}}{\partial x_{\sigma}} = -
\frac{1}{2} g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x_{\sigma}}
$$
Again, since _g__{μν} _g_^{νσ} = δ^{ν}_{μ} , we have, by
differentiation,
$$ g_{\mu\sigma} dg^{\nu\sigma} = -g^{\nu\sigma} dg_{\mu\sigma} $$
or
$$ g_{\mu\sigma} \frac{\partial g^{\nu\sigma}}{\partial x_{\lambda}} = -
g^{\nu\sigma} \frac{\partial g_{\mu\sigma}}{\partial x_{\lambda}} $$
By mixed multiplication with _g_^{στ} and _g__{νλ} respectively we
obtain (changing the mode of writing the indices).
(31)
_dg_^{μν} = -_g_^{μα} _g_^{νβ} _dg__{αβ}
∂_g_^{μν}/∂_x__{σ} = -_g_^{μα} _g_^{νβ} _dg__{αβ}
and
(32)
_dg__{μν} = -_g__{μα} _g__{νβ} _dg_^{αβ}
∂_g__{μν}/∂_x__{σ} = -_g__{μα} _g__{νβ} ∂_g_^{αβ}/∂_x__{σ}.
The expression (31) allows a transformation which we shall often use;
according to (21)
(33)
$$ \frac{\partial g_{\alpha\beta}}{\partial x_{\sigma}} =
\begin{bmatrix}\alpha & & \sigma\ & \beta &\end{bmatrix} +
\begin{bmatrix}\beta & & \sigma\ \alpha&\end{bmatrix} $$
If we substitute this in the second of the formula (31), we get,
remembering (23),
(34)
$$ \frac{\partial g^{\mu\nu}}{\partial x_{\sigma}} = - ( g^{\mu\tau}
\begin{Bmatrix}\tau & & \sigma\ \nu&\end{Bmatrix} + g^{\nu\tau}
\begin{Bmatrix}\tau & & \sigma\ \mu&\end{Bmatrix} ) $$