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-rw-r--r--main.tex3
1 files changed, 2 insertions, 1 deletions
diff --git a/main.tex b/main.tex
index d0184d5..a458623 100644
--- a/main.tex
+++ b/main.tex
@@ -313,7 +313,8 @@ which implicitly gives $\eta[\epsilon]$ and $F_\e[\epsilon]=F[\eta[\epsilon],\ep
cannot be solved explicitly, we can make use of the inverse function theorem.
First, denote by $\eta^{-1}[\eta]$ the inverse functional of $\eta$ implied by
\eqref{eq:implicit.eta}, which gives the function $\epsilon_\X$ corresponding
-to each solution of \eqref{eq:implicit.eta} it receives. Now, we use the inverse function
+to each solution of \eqref{eq:implicit.eta} it receives. This we can immediately identify from \eqref{eq:implicit.eta} as $\eta^{-1}[\eta](x)=-2/b(\delta F_\o[\eta]/\delta\eta(x))$.
+Now, we use the inverse function
theorem to relate the functional reciprocal of the derivative of $\eta[\epsilon]$ with respect to $\epsilon_\X$ to the derivative of $\eta^{-1}[\eta]$ with respect to $\eta$, yielding
\begin{equation}
\begin{aligned}