Calculating the derivative $\frac{{\partial {{\bf{X}}^{{\rm{ - }}1}}}}{{\partial {\bf{X}}}}$ How to calculate the derivation $\frac{{\partial {{\bf{X}}^{{\rm{ - }}1}}}}{{\partial {\bf{X}}}}$,where ${\bf{X}}$ is square matrix.Thanks a lot for your help! 
 A: If $F(X) = X^{-1}$, the differential of $f$ in $X$ is the linear function given by
$$DF(X)H = - X^{-1}HX^{-1}$$
Proof:
$$F(X+H) - F(X) - DF(X)H = (X+H)^{-1} - X^{-1} + X^{-1}HX^{-1}$$
$$= -(X+H)^{-1}HX^{-1} + X^{-1}HX^{-1} = (X^{-1}-(X+H)^{-1})HX^{-1},$$
so
$$\|F(X+H) - F(X) - DF(X)H\|\le\|X^{-1}-(X+H)^{-1}\|\|H\|\|X^{-1}\| = \|X^{-1}-(X+H)^{-1}\|O(\|H\|).$$
And
$$\lim_{H\to 0}\frac{\|F(X+H) - F(X) - DF(X)H\|}{\|H\|} = 0$$
by the continuity of $F$.
A: First note that $\frac{\partial F}{\partial X}$ is a 4th order tensor. 
Let $F=X^{-1},\,$ then working out the problem in index notation yields
$$\eqalign{
 dF_{ij} &= -F_{ik}\,dX_{kl}\,F_{lj} \cr\cr
\frac{\partial F_{ij}}{\partial X_{kl}} &= -F_{ik}\,F_{lj} \cr\cr
}$$
A: Let $\mathrm F (\mathrm X) := \mathrm X^{-1}$. Hence,
$$\mathrm F (\mathrm X + h \mathrm V) = (\mathrm X + h \mathrm V)^{-1} = \left( \mathrm X ( \mathrm I + h \mathrm X^{-1} \mathrm V) \right)^{-1} \approx ( \mathrm I - h \mathrm X^{-1} \mathrm V) \mathrm X^{-1} = \mathrm F (\mathrm X) - h \mathrm X^{-1} \mathrm V \mathrm X^{-1}$$
Thus, the directional derivative of $\mathrm F$ in the direction of $\mathrm V$ at $\mathrm X$ is $\color{blue}{- \mathrm X^{-1} \mathrm V \mathrm X^{-1}}$. Vectorizing,
$$\mbox{vec} (- \mathrm X^{-1} \mathrm V \mathrm X^{-1}) = - \left( \mathrm X^{-\top} \otimes \mathrm X^{-1} \right) \mbox{vec} (\mathrm V)$$
