Prove that, for $n, l \in \mathbb{N}$ the identity $\vec\nabla \times (f^n \vec\nabla(f^l)) = \textbf{ $\vec 0$} $

Question

a) Let $f$ and $g$ be two smooth scalar fields. Prove the following identity:

\begin{equation} \vec\nabla \times (f \vec\nabla g) + \vec\nabla \times (g \vec\nabla f) = \textbf{$\vec 0$} \end{equation}

b) Prove that, for $n, l \in \mathbb{N}$ the identity $\vec\nabla \times (f^n \vec\nabla(f^l)) = \textbf{ $\vec 0$} $

Answer 1

Another way to see (a) is

$\nabla \times (f\nabla g )+ \nabla \times (g\nabla f) = \nabla f \times \nabla g + f \nabla \times (\nabla g) + \nabla g \times \nabla f + g \nabla \times (\nabla f)$ $=\nabla f \times \nabla g + \nabla g \times \nabla f = \nabla f \times \nabla g - \nabla f \times \nabla g = 0, \tag{1}$

where we have used the identity

$\nabla \times (fX) = \nabla f \times X + f \nabla \times X \tag{2}$

which may be found in this wikipedia entry. Also, we used

$\nabla \times \nabla h = 0 \tag{3}$

for any smooth function $h$.

As for (b), we have

$\nabla \times (f^n \nabla f^l) = \nabla f^n \times \nabla f^l + f^n \nabla \times \nabla f^l = \nabla f^n \times \nabla f^l \tag{4}$

since $\nabla \times \nabla f^l = 0$; again (2) is also used here. Now for $m \in \Bbb N$,

$\nabla f^m = mf^{m -1} \nabla f, \tag{5}$

which is in fact the chain rule; again, see the linked wiki page or simply look at components in a Cartesian coordinate system $x_1, x_2, x_3$:

$(\nabla f^m)_i = \dfrac{\partial f^m}{\partial x^i} = mf^{m -1} \dfrac{\partial f}{\partial x_i} = mf^{m -1}(\nabla f)_i; \tag{6}$

using (5) in (4):

$\nabla \times (f^n \nabla f^l) = \nabla f^n \times \nabla f^l = nf^{n -1}\nabla f \times lf^{l - 1} \nabla f = nl f^{n + l - 2}\nabla f \times \nabla f = 0, \tag{7}$

since $X \times X = 0$ for any vector $X$.

Answer 2

I used the identities to prove a)

\begin{equation} \vec\nabla \times (f \vec\nabla g) + \vec\nabla \times (g \vec\nabla f) \end{equation} \begin{equation} = \vec\nabla \times (f\vec\nabla g +g \vec\nabla f) \end{equation} \begin{equation} = \vec\nabla \times (\vec\nabla (fg)) = \textbf{$\vec 0$} \end{equation}

I know for b) I will need to use the chain rule but I'm confused about how to do it