 Solution

Under the same Cartesian change, a vector field has $u_i^{\prime }(x^{\prime })=Q_{ip}{u}_p(x)$ with $x=Q^T{x}^{\prime }$. The chain rule gives $\partial /\partial x_j^{\prime }=Q_{jq}\partial /\partial x_q$. Since $Q$ is constant,This is exactly the order-two tensor component transformation law. Thus the Cartesian vector gradient is a Cartesian second-rank tensor, as in the velocity gradient tensor when $u$ is a velocity field.

The restriction to Cartesian transformations matters. In a general position-dependent basis, differentiating the basis contributes extra terms, so the tensorial derivative is the covariant derivative, not an arbitrary array of partial derivatives.