Note that $B = A[x, x_1, \ldots, x_n]/(xh - 1, f_1, \ldots, f_n)$ is a relative global complete intersection over $A$, see Algebra, Definition \ref{algebra-definition-relative-global-complete-intersection}. By Algebra, Lemma \ref{algebra-lemma-relative-global-complete-intersection} we see that $B$ is flat over $A$. \medskip\noindent Write $P' = A[x, x_1, \ldots, x_n]$ and $P = P'/(xh - 1) = A[x_1, \ldots, x_n, 1/g]$. Then we have $P' \to P \to B$. By More on Algebra, Lemma \ref{more-algebra-lemma-relative-global-complete-intersection-koszul} we see that $xh - 1, f_1, \ldots, f_n$ is a Koszul regular sequence in $P'$. Since $xh - 1$ is a Koszul regular sequence of length one in $P'$ (by the same lemma for example) we conclude that  $f_1, \ldots, f_n$ is a Koszul regular sequence in $P$ by More on Algebra, Lemma \ref{more-algebra-lemma-truncate-koszul-regular}. \medskip\noindent Let $g_i \in P \otimes_A B$ be the image of $x_i \otimes 1 - 1 \otimes x_i$. Let us use the short hand $y_i = x_i \otimes 1$ and $z_i = 1 \otimes x_i$ in $A[x_1, \ldots, x_n] \otimes_A A[x_1, \ldots, x_n]$ so that $g_i$ is the image of $y_i - z_i$. For a polynomial $f \in A[x_1, \ldots, x_n]$ we write $f(y) = f \otimes 1$ and $f(z) = 1 \otimes f$ in the above tensor product. Then we have $$ P \otimes_A B/(g_1, \ldots, g_n) = \frac{A[y_1, \ldots, y_n, z_1, \ldots, z_n, \frac{1}{h(y)h(z)}]} {(f_1(z), \ldots, f_n(z), y_1 - z_1, \ldots, y_n - z_n)} $$ which is clearly isomorphic to $B$. Hence by the same arguments as above we find that $f_1(z), \ldots, f_n(z), y_1 - z_1, \ldots, y_n - z_n$ is a Koszul regular sequence in $A[y_1, \ldots, y_n, z_1, \ldots, z_n, \frac{1}{h(y)h(z)}]$. The sequence $f_1(z), \ldots, f_n(z)$ is a Koszul regular in $A[y_1, \ldots, y_n, z_1, \ldots, z_n, \frac{1}{h(y)h(z)}]$ by flatness of the map $$ P \longrightarrow A[y_1, \ldots, y_n, z_1, \ldots, z_n, \textstyle{\frac{1}{h(y)h(z)}}],\quad x_i \longmapsto z_i $$ and More on Algebra, Lemma \ref{more-algebra-lemma-koszul-regular-flat-base-change}. By More on Algebra, Lemma \ref{more-algebra-lemma-truncate-koszul-regular} we conclude that $g_1, \ldots, g_n$ is a regular sequence in $P \otimes_A B$. \medskip\noindent At this point we have verified all the assumptions of Lemma \ref{lemma-tate} above with $P$, $f_1, \ldots, f_n$, and $g_i \in P \otimes_A B$ as above. In particular the annihilator $J$ of $I$ is freely generated by one element $\delta$ over $B$. Set $f_{ij} = \partial f_i/\partial x_j \in A[x_1, \ldots, x_n]$. An elementary computation shows that we can write $$ f_i(y) = f_i(z_1 + g_1, \ldots, z_n + g_n) = f_i(z) + \sum\nolimits_j f_{ij}(z) g_j + \sum\nolimits_{j, j'} F_{ijj'}g_jg_{j'} $$ for some $F_{ijj'} \in A[y_1, \ldots, y_n, z_1, \ldots, z_n]$. Taking the image in $P \otimes_A B$ the terms $f_i(z)$ map to zero and we obtain $$ f_i \otimes 1 = \sum\nolimits_j \left(1 \otimes f_{ij} + \sum\nolimits_{j'} F_{ijj'}g_{j'}\right)g_j $$ Thus we conclude from Lemma \ref{lemma-tate} that $\delta = \det(g_{ij})$ with $g_{ij} = 1 \otimes f_{ij} + \sum_{j'} F_{ijj'}g_{j'}$. Since $g_{j'}$ maps to zero in $B$, we conclude that the image of $\det(\partial f_i/\partial x_j)$ in $B$ generates the Noether different of $B$ over $A$.