Calculating the differential of the inverse of matrix exp? Let $A(t)$ and $B(t)$ be two matrix-valued smooth function satisfying the equation, $B(t) = e^{A(t)}$. I need to express $\frac{dA(t)}{dt}$ in terms of $B(t)$. I know that there is a formula of Wilcox, namely
$$ \frac{d}{dt} e^{A(t)} = \int_0^1 e^{s A(t) } \frac{dA(t)}{dt}  e^{(1-s) A(t) } ds.$$
But I need something of the opposite direction. Does anyone know if there is such a formula or a general method to calculate that?
 A: $\log(U)$, the principal $\log$ is defined when $U$ has no $\leq 0$ eigenvalues. Note that $B=e^A$ implies that $A=\log(B)$ only when, for every eigenvalue $\lambda$ of $A$, $im(\lambda)\in(-\pi,\pi)$. 
Assume that $A(t)=\log(B(t))$; then (*) $A'(t)=\int_0^1(u(B-I_n)+I_n)^{-1}B'(t)(u(B-I_n)+I_n)^{-1}du$.
EDIT. That follows is a proof of the previous formula (*).
Step 1. (**)  $\log(A)=\int_0^1(A-I)(u(A-I)+I)^{-1}du$. Proof. Since the principal logarithm $\log()$ is $C^{\infty}$ over $\mathbb{C}\setminus(-\infty,0]$, it suffices to prove the result when $A$ is diagonalizable (by density and continuity) and even only when $A$ is diagonal, that is easy.
Step 2. The derivative (with respect to $t$) of (**) is $(\log(A))'=\int_0^1A'(u(A-I)+I)^{-1}du-\int_0^1(A-I)(u(A-I)+I)^{-1}uA'(u(A-I)+I)^{-1}du=$
$\int_0^1 ZA'(u(A-I)+I)^{-1}du$ where $Z=I-u(A-I)(u(A-I)+I)^{-1}=(u(A-I)+I-u(A-I))(u(A-I)+I)^{-1}=(u(A-I)+I)^{-1}$
and we are done.
A: Recall that $$ e^A = I + A + \frac{A^2}{2} +\cdots $$
If $A=A(0)+A'(0)t + \frac{A''(0)t^2}{2} + \cdots $, then \begin{align} e^A =& I
+ (A(0)+ A'(0)t) + \frac{ A(0)^2 + t(A(0)A'(0)+A'(0)A(0) ) }{2} +
\cdots \\& + \frac{ A(0)^n + t \sum_{i=1}^n A(0)\cdots\underbrace{
A'(0)}_{=i-th}\cdots A(0) }{n!} +\cdots + O(t^2) \end{align}
Then $$ B'(0)=\frac{d}{dt} e^A = \sum_{n=1}^\infty \sum_{i=1}^n
\frac{A(0)\cdots\underbrace{ A'(0)}_{=i-th}\cdots A(0) }{n!} $$
