Evaluating: $\frac{d}{d\tau}(\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu})=0$
$=\eta_{\mu\nu}\ddot{x}^\mu\dot{x}^\nu+\eta_{\mu\nu}\dot{x}^\mu\ddot{x}^\nu$
I understand that we can relabel $\mu$ and $\nu$ in the second term
$=\eta_{\mu\nu}\ddot{x}^\mu\dot{x}^\nu+\eta_{\mu\nu}\dot{x}^\nu\ddot{x}^\mu=2\eta_{\mu\nu}\ddot{x}^\mu\dot{x}^\nu=0$ $\eta_{\mu\nu}\ddot{x}^\mu\dot{x}^\nu=0$
Does this imply that $\ddot{x}^\mu$ and $\dot{x}^\nu$ commute? If so why?