Computing the Frechet derivative of the inverse endomorphism. For every $n$, compute the $n$-th Frechet derivative of 
$$\begin{array}{crcl}
f:\!\!\! & GL(R)\!\!\! & \to\!\!\! & GL(R) \\
  & u\!\!\! & \mapsto\!\!\! & u^{-1} \\
\end{array}$$
I know answer just for $n=1$
it is :
$(Df(u))(h)=-u^{-1}\cdot h\cdot u^{-1}$

Use the Neumann series $(I-T)^{-1}=I+T+T^2+T^3+...$ for $\|T\|<1$:
\begin{align}
(A+X)^{-1}
&=A^{-1}(I+XA^{-1})^{-1}
=A^{-1}A^{-1}\sum_{k=0}^\infty(-XA^{-1})^k\\
&=A^{-1}+\sum_{k=1}^\infty\frac{1}{k!} \left((-1)^kk!\,A^{-1}(XA^{-1})^k\right)
\end{align}
which tells you all about the derivative maps.
First derivative is $X\mapsto-A^{-1}XA^{-1}$. 
Second, after polarization, $(X,Y)\mapsto A^{-1}XA^{-1}YA^{-1}+A^{-1}YA^{-1}XA^{-1}$
and so on, product of the direction matrices separated and bracketed by $A^{-1}$, sum over all permutations, minus sign for odd derivatives.

The second derivative in direction $X$ and $Y$ is the first derivative in direction Y of the first derivative $-A^{-1}XA^{-1}$ in direction $X$. Thus the first derivative formula has, by the product rule, to be applied to both factors $A^{-1}$. Thus the second derivative is
$$
-(-A^{-1}YA^{-1})XA^{-1}-A^{-1}X(-A^{-1}YA^{-1})=A^{-1}XA^{-1}YA^{-1}+A^{-1}YA^{-1}XA^{-1}
$$
For the third derivative in an additional direction $Z$, one has to add three terms for each summand, for each position where  $A^{-1}$ gets replaced with $(-A^{-1}ZA^{-1})$, giving 6 terms, one for each permutation of $(X,Y,Z)$.

If you are given a quadratic map that is already in the form of a bilinear map $Q(X)$, as $B(X,X)$, then polarization results in $\tfrac12(B(X,Y)+B(Y,X))$. In general you have to compute one of
$$
B(X,Y)=\tfrac14 (Q(X+Y)-Q(X-Y))=\tfrac12(Q(X+Y)-Q(X)-Q(Y))
$$
