 Solution

Regard $f$ as a scalar field evaluated on a specified worldline $x^{\lambda }(\tau )$. The ordinary multivariable chain rule givesThis is the directional derivative along the four-velocity. With $x^0=ct$, it reads $df/d\tau =\gamma ({\partial }_{t}f+\mathbf{v}\cdot \nabla f)$. The spatial term is present because the worldline moves through the field; it would be lost by identifying a total derivative along the particle with a partial derivative at a fixed spatial position. For a scalar the contracted expression has the same value in all coordinate systems. For a tensor field the analogous covariant derivative also transports its indices.