Show that $$\nabla\cdot (\nabla f\times \nabla h)=0,$$ where $f = f(x,y,z)$ and $h = h(x,y,z)$.
I have tried but I just keep getting a mess that I cannot simplify. I also need to show that
$$\nabla \cdot (\nabla f \times r) = 0$$
using the first result.
Thanks in advance for any help
We use $\nabla f =(f_x,f_y,f_z)$ and $\nabla h=(h_x,h_y,h_z)$. For the cross product we have $(a,b,c) \times (u,v,w)=\hat{i} (bw-cv)+\hat{j}(cu-aw)+\hat{k} (av-bu) $, alternatively written this is expressed $(a,b,c) \times (x,y,z) = (bw-cv,cu-aw,av-bu)$ but this is exactly the same. Here $a,b,c,u,v,w$ are not meant to mean anything special but are just to illustrate the point.
Using this information we calculate $(\nabla f\times \nabla h)$: $$(\nabla f\times \nabla h)=\hat{i}(f_yh_z-f_zh_y)+\hat{j}(f_zh_x-f_xh_z)+\hat{k}(f_xh_y-f_yh_x)$$ Now for the dot product $\nabla \cdot (\nabla f\times \nabla h)$ we repeatedly use the product rule to obtain: $$\nabla \cdot (\nabla f\times \nabla h)= \frac{\partial}{\partial x}(f_yh_z-f_zh_y)+\frac{\partial}{\partial y}(f_zh_x-f_xh_z)+\frac{\partial}{\partial z}(f_xh_y-f_yh_x)$$ $$=(f_{yx}h_z+f_yh_{zx}-f_{zx}h_y-f_zh_{yx})+(f_{zy}h_x+f_zh_{xy}-f_{xy}h_z - f_xh_{zy})+(f_{xz}h_y+f_xh_{yz}-f_{yz}h_x-f_yh_{xz})$$ Now $f_{yz}=f_{zy}$ etcetera assuming that $f$ and $h$ are twice continuous differentiable. Upon close inspection we see that all the terms cancel to give $$\nabla \cdot (\nabla f\times \nabla h)=0$$
If we consider $r$ to be the radial vector then this is irrotational (from vector calculus). Then it is a result from vector calculus that there exists function $\phi$ such that $r=\nabla \phi$. Then $$\nabla\cdot (\nabla f\times r)=\nabla\cdot (\nabla f\times \nabla \phi)=0$$ by the previous result.