I can see that $\nabla \times (f^2 \nabla f)=\nabla f^2 \times \nabla f+f^2\nabla \times \nabla f$ where the second term is $0$ but I don't have any idea about the first term...What should I do with the $f^2$ here?
2026-05-15 16:01:39.1778860899
How can I prove that $\nabla \times ( f^2 \nabla f)=0$?
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1
You can do the same thing you would if you encountered the ordinary derivative of $f(x)^2$, or more generally, $f(x) g(x)$, that is, apply product rule.
In this case, it would simply be: $$ \nabla (fg) = f\nabla g + g\nabla f $$
The first term will then be evaluated to zero, because $A\times A = 0$, for any $A$.
=].