Prove that $\text{rot(grad}f)=0$
$X$ is a vector field on $\mathbb{R^n}$, $\omega$ it's a form
$X\mapsto \omega \mapsto d\omega \mapsto *(d\omega)=\text{rot}X$
(This is a problem from do Carmo's differential forms.)
$\text{grad}f=\sum _{i=1}^n \frac{\partial f}{\partial x^i} \frac{\partial }{\partial x^i} \mapsto \sum _{i=1}^n \frac{\partial f}{\partial x^i}dx^i$
now if I apply the exterior derivative
$\bigg(\frac{\partial^2 f}{\partial x^1\partial x^1}dx^1+...+\frac{\partial^2 f}{\partial x^n\partial x^1}dx^n \bigg)\wedge dx^1+...+ \bigg(\frac{\partial^2 f}{\partial x^1\partial x^n}dx^1+...+\frac{\partial^2 f}{\partial x^n\partial x^n}dx^n \bigg)\wedge dx^n=$
$= \Bigg[\bigg(- \frac{\partial^2 f}{\partial x^2\partial x^1}+\frac{\partial^2 f}{\partial x^1\partial x^2}\bigg)dx^1\wedge dx^2 + \bigg(- \frac{\partial^2f}{\partial x^3\partial x^1}+\frac{\partial^2 f}{\partial x^1\partial x^3}\bigg)dx^1\wedge dx^3+...+ \bigg(- \frac{\partial^2 f}{\partial x^n\partial x^1}+\frac{\partial^2 f}{\partial x^1\partial x^3}\bigg)dx^1\wedge dx^n\Bigg]+ +\Bigg[\bigg(- \frac{\partial^2 f}{\partial x^3\partial x^2}+\frac{\partial^2 f}{\partial x^2\partial x^3}\bigg)dx^2\wedge dx^3+...+\bigg(- \frac{\partial^2 f}{\partial x^n\partial x^2}+\frac{\partial^2 f}{\partial x^2\partial x^n}\bigg)dx^2\wedge dx^n\Bigg]+...+\Bigg[ \bigg( - \frac{\partial^2f}{\partial x^n\partial x^{n-1}}+\frac{\partial^2 f}{\partial x^{n-1}\partial x^n} \bigg)dx^{n-1}\wedge dx^n\Bigg]\mapsto 0=\text{rot(grad}f)$
Is my solution correct ?