Suppose that $\mathbf{T}$ is a tensor field of order $n\ge1$ and also introduce curvilinear coordinates $q^i$ and the corresponding basis by the relations
\begin{align*} \mathbf{x}&=\chi(\boldsymbol{q}) \\ \mathbf{g}_i&=\frac{\partial\chi}{\partial q^i} \end{align*}
then one can usually construct a reciprocal basis $\mathbf{g}^i$ having the property
\begin{align*} \mathbf{g}_i\cdot\mathbf{g}^j=\delta_i^j \end{align*}
let us make the following definitions for gradient, divergence and curl
\begin{align*} \text{grad}\mathbf{T}\cdot\mathbf{n}&=\lim_{\alpha\to0}\frac{\mathbf{T}(\mathbf{x}+\alpha\mathbf{n})-\mathbf{T}(\mathbf{x})}{\alpha} \\ \text{div}\mathbf{T}&=\lim_{|\Omega|\to0}\frac{1}{|\Omega|}\int_{\partial\Omega}\mathbf{T}\cdot\mathbf{n}dA \\ \text{curl}\mathbf{T}\cdot\mathbf{n}&=\lim_{|\Gamma|\to0}\frac{1}{|\Gamma|}\int_{\partial\Gamma}\mathbf{T}\cdot\mathbf{t}dL \end{align*}
then I am interested to prove the following relations
\begin{align*} \text{grad}\mathbf{T}&=\frac{\partial\mathbf{T}}{\partial q^i}\otimes \mathbf{g}^i\\ \text{div}\mathbf{T}&=\frac{\partial\mathbf{T}}{\partial q^i}\cdot\mathbf{g}^i \\ \text{curl}\mathbf{T}&=\mathbf{g}^i\times\frac{\partial\mathbf{T}}{\partial q^i} \end{align*}
where $\otimes,\cdot,\times$ are tensor product, contraction and cross product.
Any hint or help is appreciated.