如何对向量求sympy的导数？

Question

I want to know if it is possible in sympy to take derivatives of polynomials and expressions using vector notation. For example, if I have an expression as a function of two coordinates, x1 and x2, can I just make one call to diff(x) , where x is a vector of x1 and x2 , or do I need to make two separate diff calls to x1 and x2 , and stack them in a matrix?

This illustrates what works, versus what I want to work ideally:

import sympy
from sympy.matrices import Matrix

# I understand that this is possible: 
x1 = sympy.symbol.symbols('x1')  
x2 = sympy.symbol.symbols('x2')

expr_1 = x1**2+x2
poly_1 = sympy.poly(expr_1, x1, x2)

print Matrix([[poly_1.diff(x1)],[poly_1.diff(x2)]])

# but is something like this also possible?
x1 = sympy.symbol.symbols('x1')  
x2 = sympy.symbol.symbols('x2')
x_vec = Matrix([[x1],[x2]])

expr_1 = x1**2+x2
poly_1 = sympy.poly(expr_1, x1, x2)

# taking derivative with respect to a vector
poly_1.diff(x_vec)

# should ideally return same as first example: 
'''
Matrix([
[Poly(2*x1, x1, x2, domain='ZZ')],
[   Poly(1, x1, x2, domain='ZZ')]])
'''
# but it fails :(

Is there a way to take derivatives with respect to vectors in sympy ?

Thank you.

Answer 1

Perhaps you are thinking of the jacobian :

>>> Matrix([Poly(x**2+y,x,y)]).jacobian([x, y])
Matrix([[Poly(2*x, x, y, domain='ZZ'), Poly(1, x, y, domain='ZZ')]])

And that [x, y] argument can be Matrix([x, y]) or its transpose, too.