I believe they are equal, according to SymPy's rule that
(g'(t)).diff(g(t)) == 0:
f(g(t)).diff(t, g(t)) == (f'(g(t))*g'(t)).diff(g(t)) == f''(g(t))*g'(t)
f(g(t)).diff(g(t), t) == f'(g(t)).diff(t) == f''(g(t))*g(t)
You can also verify this with SymPy:
>>> f(g(t)).diff(t).diff(g(t))
Derivative(f(g(t)), (g(t), 2))*Derivative(g(t), t)
>>> f(g(t)).diff(g(t)).diff(t)
Derivative(f(g(t)), (g(t), 2))*Derivative(g(t), t)
Aaron Meurer
> --
> You received this message because you are subscribed to the Google Groups
> "sympy" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> email to
sympy+un...@googlegroups.com.
> To post to this group, send email to
sy...@googlegroups.com.
> Visit this group at
https://groups.google.com/group/sympy.
> To view this discussion on the web visit
>
https://groups.google.com/d/msgid/sympy/73c688ce-d447-4767-8ded-373346b822b7%40googlegroups.com.
> For more options, visit
https://groups.google.com/d/optout.