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


On Tue, Aug 28, 2018 at 11:55 PM, Chris Smith <[email protected]> wrote:
> SymPy allows derivative wrt non-Symbols. Under the current assumptions,
> `g(t).diff(g(t),t) == g(t).diff(t, g(t)) == 0`. Can anyone give an example
> where `f(g(t)).diff(t, g(t))` would not equal `f(g(t)).diff(g(t), t)`?
>
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