Derivative of matrix inverse from the definition How to compute (directly from the definition) the derivative of matrix-valued function $M^{-1}(t)$ with respect to $t$ and recover the standard result $-M^{-1}(t)\frac{dM}{dt}M^{-1}(t)$? 
A similar question has been asked on this site before involving this computation several times before, but without the restriction the proof come directly from the definition. In this case, I know how to give a proof using the fact that $M^{-1}(t)M(t) = I$, and applying the product rule. However, I would like to give an argument directly from the definition if possible.
A few hours of tinkering have led me nowhere fast - the issue is that the known formula is in terms of the derivative of $M$, and all methods I know of relating $M$ to $M^{-1}$, such as through the adjugate formula seem to be too ugly to recover the formula in question. Either I'm missing something, or this problem is difficult without the slick product-rule approach.
Since this is homework (of course, why else would such an arbitrary restriction be imposed on an otherwise fine argument?), a full solution is probably not necessary. 
 A: $$\left( \mathrm M (t + \mathrm d t) \right)^{-1} = \left( \mathrm M (t) + \dot{\mathrm M} (t) \,\mathrm d t \right)^{-1} = \cdots = \mathrm M^{-1} (t) \color{blue}{- \mathrm M^{-1} (t) \, \dot{\mathrm M} (t) \, \mathrm M^{-1} (t)} \,\mathrm d t$$
A: Let $M \, : \, t \in I \, \rightarrow \, M(t) \in \mathrm{GL}_{n}(\mathbb{R})$ be a differentiable matrix-valued function defined on an open interval $I$ of $\mathbb{R}$ and $\mathrm{Inv} \, : \, M \in \mathrm{GL}_{n}(\mathbb{R}) \, \rightarrow \, M^{-1}$ be the matrix inverse. You are interested in the derivative of:
$$ f = \mathrm{Inv} \circ M $$
It follows from the chain rule that for $t \in I$:
$$ f'(t) = \mathrm{D}_{M(t)} \mathrm{Inv} \cdot M'(t) $$
where $D_{M} \mathrm{Inv} \cdot H$ denotes the value at $H$ of the differential $D_{M} \mathrm{Inv}$ (differential of $\mathrm{Inv}$ at $M$). 
You can show that (see this post, for example):
$$ \mathrm{D}_{M} \mathrm{Inv} \cdot H = - M^{-1} H M^{-1}. $$
As a result:

$$ \forall t \in I, \; f'(t) = - M(t)^{-1} M'(t) M(t)^{-1}. $$

