I ran the following code

```
using SymPy
@syms x y
f(x,y) = cos(x - y)
df(x,y) = sympy.diff(f(x,y),x)
df(0,pi/2) # it is supposed to be 1, but an error occurs
```

Please help me to resolve this problem.

Thank you.

I ran the following code

```
using SymPy
@syms x y
f(x,y) = cos(x - y)
df(x,y) = sympy.diff(f(x,y),x)
df(0,pi/2) # it is supposed to be 1, but an error occurs
```

Please help me to resolve this problem.

Thank you.

Do one of the following:

- remove the arguments from the definition of
`df`

, so write`df = sympy.diff(f(x,y),x)`

instead of`df(x,y) = sympy.diff(f(x,y),x)`

- call
`df(x,y)(0,pi/2)`

instead of`df(0, pi/2)`

The reason it fails is because you are creating a *Julia function* that takes two arguments `x, y`

and evaluates the expression on the right hand side. So `df(0, 1)`

would try to do `sympy.diff(f(0,1),0)`

which is not what you want I suppose.

Instead, SymPy already returns an object from `diff`

which can be called with the remaining variables.

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