 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-3/3c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

[F \times (\nabla \times G)]_{i} = F_{j} \partial_{i} G_{j} - F_{j} \partial_{j} G_{i}

. Similarly

[G \times (\nabla \times F)]_{i} = G_{j} \partial_{i} F_{j} - G_{j} \partial_{j} F_{i}

. Add the two directional-derivative terms to cancel the last terms, leaving

F_{j} \partial_{i} G_{j} + G_{j} \partial_{i} F_{j} = \partial_{i} (F_{j} G_{j}) .

The product rule has therefore proved the gradient of a dot product identity

\nabla (F \cdot G) = (F \cdot \nabla) G + (G \cdot \nabla) F + F \times (\nabla \times G) + G \times (\nabla \times F) .

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