使用 sympy 关于时间的导数

Question

I looking for a way to declare a variable as a function of time, to then perform the time derivative. ie

import sympy as sp
from sympy import cos
from sympy import sin

t = sp.symbols('t')
x(t) = sp.symbols('x(t)')
f = cos(x(t))*sin(x(t))
df = sp.diff(f, t)

However, this code generates the following error:

    x(t) = sp.symbols('x(t)')
    ^
SyntaxError: cannot assign to function call

Answer 1

Well why not use Function like so

import sympy as sp
from sympy import sin, cos, Function

t = sp.symbols('t')
x = Function('x')
f = cos(x(t))*sin(x(t))
df = sp.diff(f, t)
df

Answer 2

I found a solution to this problem by using the "Function" function, as follows:

t = sp.symbols('t')
x = sp.Function('x')(t)
f = cos(x)*sin(x)
df = sp.diff(f, t)

The result yields the following expression:

-sin(x(t))**2*Derivative(x(t), t) + cos(x(t))**2*Derivative(x(t), t)