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I have the following equation, like this:

y = 3x2 + x

Then, I want to differentiate the both side w.r.t the variable t with sympy. I try to implement it in the following code in JupyterNotebook:

>>> import sympy as sp
>>> x, y, t = sp.symbols('x y t', real=True)
>>> eq = sp.Eq(y, 3 * x **2 + x)
>>>
>>> expr1 = eq.lhs
>>> expr1

>>> expr1.diff(t)
0
>>>
>>> expr2 = eq.rhs
>>> expr2
3^2+
>>> expr2.diff(t)
0

As the result, sympy will treat the symbol x and y as a constant. However, the ideal result I want should be the same as the result derived manually like this:

y = 3x2 + x

d/dt (y) = d/dt (3x2 + x)

dy/dt = 6 • x • dx/dt + 1 • dx/dt

dy/dt = (6x + 1) • dx/dt

How can I do the derivative operation on a expression with a specific symbol which is not a free symbol in the expression?

Jet C.
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2 Answers2

2

You should declare x and y as functions rather than symbols e.g.:

In [8]: x, y = symbols('x, y', cls=Function)

In [9]: t = symbols('t')

In [10]: eq = Eq(y(t), 3*x(t)**2 + x(t))

In [11]: eq
Out[11]: 
          2          
y(t) = 3⋅x (t) + x(t)

In [12]: Eq(eq.lhs.diff(t), eq.rhs.diff(t))
Out[12]: 
d                 d          d       
──(y(t)) = 6⋅x(t)⋅──(x(t)) + ──(x(t))
dt                dt         dt

https://docs.sympy.org/latest/modules/core.html#sympy.core.function.Function

Oscar Benjamin
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Alternatively, the idiff function was made for this purpose but it works with expressions like f(x, y) and can return the value of dy/dx. So first make your Eq and expression and then calculate the desired derivative:

>>> from sympy import idiff
>>> e = eq.rewrite(Add)
>>> dydx = idiff(e, y, x); dydx
6*x + 1

Note, too, that even in your equation (if you write it explicitly in terms of functions of t) you do not need to isolate y(t) -- you can differentiate and solve for it:

>>> from sympy.abc import t
>>> x,y=map(Function,'xy')
>>> eq = x(t)*(y(t)**2 - y(t) + 1)
>>> yp=y(t).diff(t); Eq(yp, solve(eq.diff(t),yp)[0])
Eq(Derivative(y(t), t), (-y(t)**2 + y(t) - 1)*Derivative(x(t), t)/((2*y(t) - 1)*x(t)))
smichr
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  • Can you explain more for the second part? From the line `from sympy.abc import t`. – Jet C. Mar 18 '22 at 11:13
  • You can always solve for `dy/dt` even if you can't isolate `y(t)` in an expression. So if you just have a relationship between `x(t)` and `y(t)`, `f(x(t),y(t)) = 0`, the value of `y(t)` will be given by `dydt = lambda f: Eq(y(t).diff(t), solve(f.diff(t), y(t).diff(t))[0])`. – smichr Mar 18 '22 at 11:19