Even if the function equals a constant, it may depend on the input variable and its output may happen to equal a constant for all values of that particular variable - but not produce the same output for another variable.
Let v(t) = t^p / p!
Consider the function (d/dt)^p [ v(t) ] = 1
If we replace the input variable with (1-t), keeping the function itself (as a p-th derivative with respect to t) unchanged, we get
(d/dt)^p [ v(1-t) ] = { 1 if p even, -1 if p odd}
because each differentiation of v(1-t) = (1-t)^p / p! introduces a multiplicative factor of (-1)
If we take an additional derivative with respect to t, as in the case of (d/dt)^(p+1) [ v(t) ], we get a function that is zero for all values of (1-t). The function happens to be equal to its cousin (d/dt)^(p+1) [ v(1-t) ] = 0, only because (d/dt)^p [ v(t) ] and (d/dt)^p [ v(1-t) ] are both constants.
It would be incorrect and misleading to assert that because (d/dt)^(p+1) [ v(t) ] = 0, therefore (d/dt)^(p+1) [ v(1-t) ] = 0. Even if the function equals zero everywhere, we cannot simply substitute one input variable with another.
For example, if we had tried (d/dt)^(p+1) [ v(1-t^2) ], the function is not identically zero. It is a parabolic curve with squiggles crossing the x-axis between -1<t<1, as the graph for p=8 shows below.
