If $A_{ij}$ is grade $2$ tensor, show that :
$$\frac{\partial(A_{ij})}{\partial x_k}$$
is a grade 3 tensor.
Solution :
$$\frac{\partial(A_{ij})}{\partial x_k}= \frac{\partial(A_{ij})}{\partial x_1} + \frac{\partial(A_{ij})}{\partial x_2} + \frac{\partial(A_{ij})}{\partial x_3} = A_{ij,k}$$
which has $3$ free pointers, hence it's a grade $3$ tensor.
Is my approach correct and as thorough as it could be ?
It all has to do with transformation properties. When there is a change in coordinates the tensors are supposed to change in a very particular way.
Suppose there is a change from coordinates $x^j$ to coordinates $(x')^{j}$.
Then a rank 0 tensor field will be a scalar which doesn't change.
$$ \phi' =\phi $$
A rank 1 tensor field transforms in the following way,
$$ (A')^{\mu} = \frac{\partial (x')^{\mu}}{\partial x^\alpha} A^\alpha$$
A rank two tensor field transforms like the product of two rank 1 fields,
$$ (A')^\mu (B')^\nu = \frac{\partial (x')^{\nu}}{\partial x^\beta}\frac{\partial (x')^{\mu}}{\partial x^\alpha} A^\alpha B^\beta$$
$$ (T')^{\mu\nu} = \frac{\partial (x')^{\nu}}{\partial x^\beta}\frac{\partial (x')^{\mu}}{\partial x^\alpha} T^{\alpha\beta}$$
These are the rules for contravariant indices (the raised ones). For covariant indices (sub-scripted indices)the rank 1 tensor transforms slightly differently.
$$ (A')_{\mu} = \frac{\partial x^\alpha}{\partial (x')^{\mu}} A_\alpha$$
You must show that your object transforms as a rank three covariant tensor.