If we allow two domains $\Omega, \bar{\Omega}\in \mathbb{R}^3$, allow $\mathbf{\Theta}: \Omega \to \mathbf{E}^3$ and $\mathbf{\bar \Theta}: \bar \Omega \to \mathbf{E}^3$ to be two $C^1$-diffeomorphisms (where $\mathbf{E}^3$ is a 3D Euclidean space) and define $\mathbf{g_i}\equiv \partial_i\mathbf{\Theta}$, $\mathbf{\bar g_i}\equiv \bar\partial_i\mathbf{\bar \Theta}$, each representing the same point $\hat x= \mathbf{\Theta}(x)=\mathbf{\bar \Theta}(\bar x)\in \mathbf{E}^3$, where all $\mathbf{g_i}, \mathbf{\bar g_i}$ are linearly independent, then:
$$ \mathbf{g_i}=(\partial_i\chi^j)\mathbf{\bar g_j} $$
Where $\chi^j \equiv (\mathbf{\bar \Theta}^{-1}\circ \mathbf{\Theta})_j$
Though the text says a 'simple' calculation gives this, I've been trying to work it out and nothing comes to mind (it's one of 'those' days).
I mean, it seems like a straightforward application of the chain rule, but I am receiving oddly complicated results.
Thanks for any help.
I realized how simple it was after a while.
We have that: $$ \partial_i{\bf \bar \Theta}(\bar x(x))=\partial_i{\bf \Theta}(x) $$
And, knowing that: $$ \partial _i {\bf \Theta}(x)={\bf g}_i $$
Then, using the chain rule and the fact that $\chi= {\bf \bar x} = \mathbf{\bar \Theta}^{-1}\circ \mathbf{\Theta}$ $$ {\bf g}_i=\frac{\partial {\bf \Theta}}{\partial \chi}{\partial_i\chi}=\left[\frac{\partial\Theta_k}{\partial \chi^j}\right]_{jk}[\partial_i\chi_l]_l=\left[\frac{\partial\Theta_k}{\partial \bar x_j}\right]_{jk}[\partial_i\chi_l]_l $$
This the, gives us, as expected: $$ {\bf g}_i=\left[\sum_j\frac{\partial \Theta_k}{\partial \bar x_j}\partial_i\chi^j\right]_k=(\partial_i\chi^j){\bf \bar g}_j $$