The following function is defined for $x \in (-\infty,\infty)$ \begin{equation} f(x) = \begin{cases} x & x<1 \\ 1 & x \geq 1 \end{cases} \end{equation}0 The function is obviously continuous. Now, let us assume the composite function $f(y(t))$. We will then have \begin{equation} f(y(t)) = \begin{cases} y(t) & y(t)<1 \\ 1 & y(t) \geq 1 \end{cases} \end{equation} I am interested in defining the derivatives $df/dt$ and $d^2f/dt^2$. The first derivative would be \begin{equation} \frac{df(y(t))}{dt} = \begin{cases} \frac{dy(t)}{dt} & y(t)<1 \\ 0 & y(t) \geq 1 \end{cases} \end{equation} which was justified in another post (link missing for the moment). The idea is because the intervals are strictly defined (i.e. no single point is undefined), not other considerations are required (like delta function or a prescribed jump).
Next, to differentiate $df/dt$, that may now be discontinuous, could I still say the following: \begin{equation} \frac{d^2f(y(t))}{dt^2} = \begin{cases} \frac{d^2y(t)}{dt^2} & y(t)<1 \\ 0 & y(t) \geq 1 \end{cases} \end{equation} Or do I miss some terms (or even NDE) that come-up at stencils $t$ where $y(t) = 1$?
The second derivative is correct. I think this may be an interesting undergrad question if some additional information is available to make f differentiable twice everywhere.