The original motivation for derivatives wrt functions and derivatives
of functions was to support the Euler-Lagrange equations
https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation. In
Euler-Lagrange, dL(t, q(t), q'(t))/dq'(t) means consider L as a
function of three variables, L(x, y, z) and take dL/dz evaluated at
z=q'(t).

See also https://github.com/sympy/sympy/issues/15048

Aaron Meurer

On Wed, Aug 29, 2018 at 11:52 AM, Robert Dougherty-Bliss
<[email protected]> wrote:
> Why do we assume that g'(t).diff(g(t)) == 0?
>
> Here's a question on math.SE about derivatives w.r.t. functions:
> https://math.stackexchange.com/questions/954073
>
> Does the assumption work with the accepted answer there?
>
>
> On Wednesday, August 29, 2018 at 2:12:07 AM UTC-4, Aaron Meurer wrote:
>>
>> 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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