Associativity of ore-polynomials

Question

I'm fairly new to ore-polynomials. Until now I thought the rule $\frac{d}{dt}x(t)=\dot{x}(t)+x(t)\frac{d}{dt}$ resp. $x(t)\frac{d}{dt}=\frac{d}{dt}x(t)-\dot{x}(t)$ would do the trick in right- or left-shifting the $\frac{d}{dt}$ operator in equations. Now I was wondering what the result of the following expression would be:

$-\dot{x}(t)\frac{d}{dt}\frac{1}{\dot{x}(t)}=?$

I've assumed associativity is fulfilled, but that does not seem to be the case:

1. $-\dot{x}(t)\left(\frac{d}{dt}\frac{1}{\dot{x}(t)}\right)=-\dot{x}(t)\left(\frac{1}{\ddot{x}(t)} + \frac{1}{\dot{x}(t)}\frac{d}{dt}\right)=-\frac{d}{dt}-\frac{\dot{x}(t)}{\ddot{x}(t)}$

2. $-\left(\dot{x}(t)\frac{d}{dt}\right)\frac{1}{\dot{x}(t)}=-\left(\frac{d}{dt}\dot{x}(t)-\ddot{x}(t)\right)\frac{1}{\dot{x}(t)}=-\frac{d}{dt}+\frac{\ddot{x}(t)}{\dot{x}(t)}$

What am I missing? What ist the correct result?

Answer 1

the first calculation is wrong: $\frac{\partial}{\partial t}\frac{1}{\dot{x}(t)}=-\frac{\ddot{x}(t)}{\dot{x}^2(t)}$ and so it should instead be

1. $-\dot{x}(t)\left(\frac{d}{dt}\frac{1}{\dot{x}(t)}\right)=-\dot{x}(t)\left(-\frac{\ddot{x}(t)}{\dot{x}^2(t)} + \frac{1}{\dot{x}(t)}\frac{d}{dt}\right)=-\frac{d}{dt}+\frac{\ddot{x}(t)}{\dot{x}(t)}$ which is equal to the result of the second calculation.