Let $\rho:(a,b)\to \operatorname{ISO}\left(\Bbb{R}^n\right)$ be differentiable. I want to prove that \begin{align}\frac{d}{dt}\left[\rho(t)\right]^{-1}=-\rho(t)^{-1}\rho'(t)\rho(t)^{-1}\end{align} $\operatorname{ISO}\left(\Bbb{R}^n\right)$ is a space of continuous linear maps from $\Bbb{R}^n$ to $\Bbb{R}^n$. I have proven before that
\begin{align}f:\operatorname{ISO}\left(\Bbb{R}^n\right)\to \operatorname{ISO}\left(\Bbb{R}^n\right)\end{align} \begin{align}u\mapsto u^{-1}\end{align}is differentiable and \begin{align}f'(u)(h)=-u^{-1}h u^{-1}\end{align} Along the line, I used Neumann's Lemma but I got stuck when proving this.
Here is my work:
I tried by using definition:
Let $h\in(a,b),$ then \begin{align}\rho(t+h)-\rho(t)&=\rho(t+h)^{-1}-\rho(t)^{-1}\\&=\left[\rho(t+h)-\rho(t)+\rho(t)\right]^{-1}-\rho(t)^{-1}\\&=\left[\left[\left(\rho(t+h)-\rho(t)\right)\left(\rho(t)\right)+I\right]^{-1}-I\right]\rho(t)^{-1}\end{align} I'm thinking of applying Neumann's Lemma but I can't move on. I want someone to help see my fault. Otherwise, alternative methods will be highly regarded. Thanks!
Idea: $$\forall t\in(a,b): \left[\rho(t)\right]^{-1}\rho(t) = I\ \hbox{(constant)},$$ so $$ 0 = \frac{d}{dt}(\left[\rho(t)\right]^{-1}\rho(t)) = (\frac{d}{dt}\left[\rho(t)\right]^{-1})\rho(t) + \rho(t)^{-1}\frac{d}{dt}\rho(t). $$