Computation of the second derivative of the Jacobian of the change of the coordinates I am trying to understand how to derive the equality $1.135$ of the book "A course in minimal surfaces" by Colding and Minicozzi.

I derive
$$(g^{ij}g_{ij}')' = (g^{ij})' g_{ij}' + g^{ij} g_{ij}''$$
and derive twice $(g^{ij}g_{ij})'' = (\delta_j^i)''$ to obtain
$$(g^{ij})' g_{ij}' + (g^{ij})'' g_{ij} = - (g^{ij})' g_{ij}' - g^{ij} g_{ij}'',$$
but this does not help me to obtain the equality desired. I would like to understand how I can do it.
Thanks in advance!
$\textbf{P.S.:}$ I know that the authors considering an ortohonormal frame, but my attempt does not give the expression desired even in this frame.
 A: Deriving $(1.133)$,
\begin{align*}
2 \frac{d^2 \nu}{dt^2} \bigg\vert_{t=0} &= \sum_{i,j=1}^n \left( (g^{ij})'(0) g_{ij}'(0) + g^{ij}(0) g_{ij}''(0) \right) \nu(0) + \left( \sum_{i,j=1}^n g^{ij}(0) g_{ij}'(0) \right) \frac{d \nu}{dt} \bigg\vert_{t=0}\\
&= \sum_{i,j=1}^n \left( (g^{ij})'(0) g_{ij}'(0) + g^{ij}(0) g_{ij}''(0) \right) \nu(0) + \frac{1}{2} \left( \sum_{i,j=1}^n g^{ij}(0) g_{ij}'(0) \right)^2 \nu(0)\\
&= \sum_{i,j=1}^n \left( (g^{ij})'(0) g_{ij}'(0) + g^{ij}(0) g_{ij}''(0) \right) + \frac{1}{2} \left( \sum_{i,j=1}^n g^{ij}(0) g_{ij}'(0) \right)^2 \ (1)
\end{align*}
We will omit the point $0$ from here to not charge the notation.
$$(g^{ij})' = - g_{ij}' \ (2)$$
and
$$g_{ij}' = 4 \langle A(F_{x_i}, F_{x_j}), F_t \rangle = 4 ||F_t|| \langle A(F_{x_i}, F_{x_j}), N \rangle \ (3)$$
in normal coordinates.
$(3)$ implies that $(g_{ij}')$ is diagonalizable. This and $(2)$ imply that $((g^{ij})')$ is diagonalizable. If $D$ is the diagonal matrix of $(g_{ij}')$, then
\begin{align*}
\sum_{i,j=1}^n (g^{ij})' g_{ij}' &= \text{tr} \ ((G^{-1})'G')\\
&= \text{tr} \ ((Q^{-1}(-D)Q)(Q^{-1}DQ))\\
&= \text{tr} \ (Q^{-1}(-D)DQ)\\
&= \text{tr} \ ((-D)DQQ^{-1})\\
&= - \text{tr} \ (D^2)\\
&= - \text{tr} \ ((G')^2) = - \sum_{i,j=1}^n (g_{ij}')^2,
\end{align*}
where we used this result of linear algebra in the fourth equality.
Substituting this in $(1)$, we obtain $(1.135)$.
