= Solution
Label a trajectory by $\mathbf a$ and write its <flow map> as $\mathbf X(\mathbf a,t)$, with $\partial_t\mathbf X=\mathbf u(\mathbf X,t)$. An infinitesimal <material line> joining neighboring labels has tangent $\ell_i=F_{ij}\,d a_j$, where $F_{ij}=\partial X_i/\partial a_j$ is the <deformation gradient>. Differentiating at fixed labels gives
$$
\frac{D\ell_i}{Dt}=\partial_j u_i\,\ell_j,
\qquad
\boxed{\partial_t\boldsymbol\ell+\mathbf u\cdot\nabla\boldsymbol\ell=(\boldsymbol\ell\cdot\nabla)\mathbf u.}
$$
This identity holds for an unnormalized <material line> tangent even in a compressible flow. <Incompressibility> also makes an advected <magnetic field> obey the same stretching law. A unit tangent would instead require a projection removing its change of length.
Use the source's <velocity gradient> convention $G_{ij}=\partial_i u_j$. For a <outer product> $A_{ij}=\ell_i\ell_j$, the <product rule> yields
$$
\frac{DA_{ij}}{Dt}=(\partial_k u_i)A_{kj}+A_{ik}(\partial_k u_j)
=(G^T\mathsf A+\mathsf A G)_{ij}.
$$
The same law holds for a sum or locally transported average of such <outer products>. Thus the <dyadic transport of material line elements> motivates
$$
\boxed{\overset{\triangledown}{\mathsf A}=\partial_t\mathsf A+\mathbf u\cdot\nabla\mathsf A-\mathsf A G-G^T\mathsf A.}
$$
This <upper-convected derivative> vanishes for a symmetric <tensor> transported solely by stretching of <material lines>. It subtracts both advection and geometric deformation, allowing a <constitutive equation> to isolate intrinsic relaxation or production. More generally $\mathsf A=F\mathsf A_{\rm ref}F^T$ gives $\overset{\triangledown}{\mathsf A}=F\dot{\mathsf A}_{\rm ref}F^T$. Under a time-dependent rigid rotation $Q$, its value transforms as $Q\overset{\triangledown}{\mathsf A}Q^T$, making it an <objective time derivative>. This is a suitable derivative for this transport type; symmetry alone does not make it the unique possible <objective time derivative>.
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