Can i extract the exterior derivative of a vector field from the matrix?

Question

Suppose i want to extract the exterior derivative $d\phi$ of a 0-form $\phi$. I can use the classical definition of gradient, together with the musical isomorphism, to get: $$(d\phi)^\sharp=\nabla\phi=\begin{bmatrix}\frac{\partial\phi}{\partial x^1}\\\vdots\\\frac{\partial\phi}{\partial x^n}\end{bmatrix}$$ Or: $$d\phi=(\nabla\phi)^T=\begin{bmatrix}\frac{\partial\phi}{\partial x^1}&\cdots&\frac{\partial\phi}{\partial x^n}\end{bmatrix}$$ Can i apply a similar strategy to get the exterior derivative of a vector field $F$? i.e.: $$dF=(\nabla F)^T=\begin{bmatrix}\frac{\partial F_{x_1}}{\partial x^1}&\cdots&\frac{\partial F_{x_n}}{\partial x^1}\\\vdots&\ddots&\vdots\\\frac{\partial F_{x_1}}{\partial x^n}&\cdots&\frac{\partial F_{x_n}}{\partial x^n}\end{bmatrix}^T$$