Differential Equations: Time-Reversal Symmetry In Strogatz's Nonlinear Dynamics and Chaos (page: 163-164), he claims that any mechanical system of the form mx" = F(x) is T-R symmetric. He considers a system that has T-R symmetry to be one that is invariant under the change of variable

$ t -&gt; -t $

First, I see that if we reverse time along a trajectory, that the first derivative would inherit a negative sign. Second, I'm not positive I see why the second derivative, x", would remain unchanged. Is this due to the following?

$ x''= d^2x/dt^2 $  ... which since t is squared would yield $t^2$ = $(-t)^2$ under the mapping of t -> -t. 

However, it seems a bit odd that you can manipulate the $dx/dt$ notation like that. If someone could show why $dt^2$ is able to obey the above I would appreciate it. Namely, why is $d(t)^2$ considered a product or function where we can first compute $t^2$ then apply d? 

Thanks!
This is the chain rule. If you have a function 
$$
f(t)
$$
and you compose with the function $g(t)=-t$, then 
$$
\frac{d}{dt}f(-t)=\frac{d}{dt}f(g(t))=g'(t)f'(g(t))=-f'(-t)\\
\implies \frac{d^2}{dt^2}f(-t)=\frac{d}{dt}-f'(-t)=-\frac{d}{dt}f'(-t)=f''(-t)
$$
Due to the power rule, if the first derivative is odd such that $F(-x)=-F(x)$, than the second derivative is even. That is why it is invariant to the transformation $t\to-t$