How to obtain $\frac{d}{dt}(XY^{-1})=\dot{X}Y^{-1}-XY^{-1}\dot{Y}Y^{-1}$

Question

I am reading the following paper:

http://www.sciencedirect.com/science/article/pii/S0005109815004239

My question is the following:

I think:

$$\frac{d}{dt}(XY^{-1})=\dot{X}Y^{-1}-XY^{-2}\dot{Y}=\dot{X}Y^{-1}-XY^{-1}Y^{-1}\dot{Y}$$

If I want to get the term in the red box, then $Y^{-1}\dot{Y}=\dot{Y}Y^{-1}$. However, it seems that it may not be correct even though $X,Y$ are in the general linear group.

Answer 1

From $Y Y^{-1} = I$ one gets $0 = \frac{d}{dt} (YY^{-1}) = Y' Y^{-1} + Y (Y^{-1})'$, hence $(Y^{-1})' = -Y^{-1}Y'Y^{-1}$.

Then $\frac{d}{dt}(XY^{-1} ) = X'Y^{-1} + X (Y^{-1})' = ...$