Generally \begin{align} &\texttt{The 4-gradient contravariant vector operator} \nonumber\ & \qquad \qquad \qquad \partial^{\mu}\boldsymbol{\equiv}\dfrac{\partial }{\partial x_{\mu}}\boldsymbol{=}\left(\dfrac{\partial }{\partial x^{0}}\,,\boldsymbol{-\nabla}\right) \tag{01a}\label{01a}\ &\texttt{The 4-gradient covariant vector operator} \nonumber\ &\qquad \qquad \qquad \partial_{\mu}\boldsymbol{\equiv}\dfrac{\partial }{\partial x^{\mu}}\boldsymbol{=}\left(\dfrac{\partial }{\partial x^{0}}\,,\boldsymbol{+\nabla}\right) \tag{01b}\label{01b} \end{align} The 4-divergence of a 4-vector $A^{\mu}\boldsymbol{=}\left(A^{0},\mathbf{A}\right),A_{\mu}\boldsymbol{=}\left(A^{0},\boldsymbol{-}\mathbf{A}\right) $ is the invariant \begin{equation} \partial^{\mu}A_{\mu}\boldsymbol{=}\partial_{\mu}A^{\mu}\boldsymbol{=}\dfrac{\partial A^{0} }{\partial x^{0}}\boldsymbol{+\nabla\cdot}\mathbf{A} \tag{02}\label{02} \end{equation}
$=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=$
Note : May be in the future you ask yourself if the converse of \eqref{02} is true. No, it's not : if $A^{\mu}\boldsymbol{=}\left(A^{0},\mathbf{A}\right)$ is a 4-dimensional quantity then \begin{equation} \partial_{\mu}A^{\mu}\boldsymbol{=}\texttt{invariant}\quad \boldsymbol{=!\ne!\Rightarrow}\quad A^{\mu}\boldsymbol{=}\texttt{contravariant Lorentz 4-vector} \tag{03}\label{03} \end{equation}
$-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!$
Counter example
Consider that $Y^{\mu}\left(\mathbf{x},t\right)\boldsymbol{=}\left[!!\left[ Y^{0}\left(\mathbf{x},t\right),\mathbf{Y}\left(\mathbf{x},t\right)\right]!!\right]$ is a (contravariant) 4-vector function of the space-time coordinates $\left(\mathbf{x},t\right)$. 616