= Solution
Under the same Cartesian change, a <vector field> has $u'_i(x')=Q_{ip}u_p(x)$ with $x=Q^Tx'$. The <chain rule> gives $\partial/\partial x'_j=Q_{jq}\partial/\partial x_q$. Since $Q$ is constant,
$$
\boxed{\frac{\partial u'_i}{\partial x'_j}=Q_{ip}Q_{jq}\frac{\partial u_p}{\partial x_q}.}
$$
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.
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