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**arbolis** If $\displaystyle \vec F=\vec \nabla \varphi$, show that $\displaystyle \vec \nabla \times \vec F=\vec 0$. $\displaystyle \vec F$ is a vector field and $\displaystyle \varphi$ is a scalar field.

I know it's an obvious question but I can't remember how to formally prove it.

I know it's true if $\displaystyle \frac{\partial ^2 \varphi}{\partial y \partial z} =\frac{\partial ^2 \varphi }{\partial z \partial y} ,\frac{\partial ^2 \varphi}{\partial x \partial z} =\frac{\partial ^2 \varphi }{\partial z \partial x}$ and $\displaystyle \frac{\partial ^2 \varphi}{\partial x \partial y} =\frac{\partial ^2 \varphi }{\partial y \partial x} $ in Cartesian coordinates, I've showed it for $\displaystyle \varphi (x,y,z)=f(x)g(y)h(z)$ but not for a general form of $\displaystyle \varphi$. I think I'm over-complicating this.

Could you give me a tip?