Show that $\nabla\cdot (\nabla f\times \nabla h)=0$ 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
 A: 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.
A: While we're here having fun, you can prove this easily enough with geometric calculus.  Instead of using the cross product, we use the wedge product instead and Hodge duality.
$$\nabla \cdot (\nabla f \times \nabla h) = \nabla \cdot [-i (\nabla f \wedge \nabla h)] = -i [\nabla \wedge (\nabla f \wedge \nabla h)]$$
Using the product rule, we get
$$\nabla \cdot (\nabla f \times \nabla h) = -i [(\nabla \wedge \nabla f) \wedge \nabla h - \nabla f \wedge (\nabla \wedge \nabla h)]$$
But $\nabla \wedge \nabla g = 0$ for any scalar field $g$, so the result is zero.
You might be thinking this is exotic and strange.  It's not.  To show you this, let's move the $i$ back into our products to get something straight out of vector calculus.
$$-i [(\nabla \wedge \nabla f) \wedge \nabla h - \nabla f \wedge (\nabla \wedge \nabla h)] = (\nabla \times \nabla f) \cdot \nabla h - \nabla f \cdot (\nabla \times \nabla h)$$
You might be accustomed to seeing $\nabla \times \nabla g = 0$, which yields the same conclusion as above.
Now then, note that the vector $r = \nabla \frac{1}{2} |r|^2$.
A: $$\nabla f\times\nabla h\\=\left(\left(\hat{x}\frac{\partial}{\partial x}+\hat{y}\frac{\partial}{\partial y}+\hat{z}\frac{\partial}{\partial z}\right)f\right)\times\left(\left(\hat{x}\frac{\partial}{\partial x}+\hat{y}\frac{\partial}{\partial y}+\hat{z}\frac{\partial}{\partial z}\right)h\right)\\=\left(\hat{x}\frac{\partial f}{\partial x}+\hat{y}\frac{\partial f}{\partial y}+\hat{z}\frac{\partial f}{\partial z}\right)\times\left(\hat{x}\frac{\partial h}{\partial x}+\hat{y}\frac{\partial h}{\partial y}+\hat{z}\frac{\partial h}{\partial z}\right)\\=\begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}&\frac{\partial f}{\partial z}\\\frac{\partial h}{\partial x}&\frac{\partial h}{\partial y}&\frac{\partial h}{\partial z}\end{vmatrix}\\=\left(\hat{x}\frac{\partial f}{\partial y}\frac{\partial h}{\partial z}+\hat{y}\frac{\partial f}{\partial z}\frac{\partial h}{\partial x}+\hat{z}\frac{\partial f}{\partial x}\frac{\partial h}{\partial y}\right)-\left(\hat{x}\frac{\partial f}{\partial z}\frac{\partial h}{\partial y}+\hat{y}\frac{\partial f}{\partial x}\frac{\partial h}{\partial z}+\hat{z}\frac{\partial f}{\partial y}\frac{\partial h}{\partial x}\right)$$$$=\left(\hat{x}\tfrac{\partial f}{\partial y}\tfrac{\partial h}{\partial z}+\hat{y}\tfrac{\partial f}{\partial z}\tfrac{\partial h}{\partial x}+\hat{z}\tfrac{\partial f}{\partial x}\tfrac{\partial h}{\partial y}\right)-\left(\hat{x}(\tfrac{\partial f}{\partial z}\tfrac{\partial h}{\partial y})(\tfrac{\partial y}{\partial z}\tfrac{\partial z}{\partial y})+\hat{y}(\tfrac{\partial f}{\partial x}\tfrac{\partial h}{\partial z})(\tfrac{\partial z}{\partial x}\tfrac{\partial x}{\partial z})+\hat{z}(\tfrac{\partial f}{\partial y}\tfrac{\partial h}{\partial x})(\tfrac{\partial x}{\partial y}\tfrac{\partial y}{\partial x})\right)$$$$=\left(\hat{x}\frac{\partial f}{\partial y}\frac{\partial h}{\partial z}+\hat{y}\frac{\partial f}{\partial z}\frac{\partial h}{\partial x}+\hat{z}\frac{\partial f}{\partial x}\frac{\partial h}{\partial y}\right)-\left(\hat{x}\frac{\partial f}{\partial y}\frac{\partial h}{\partial z}+\hat{y}\frac{\partial f}{\partial z}\frac{\partial h}{\partial x}+\hat{z}\frac{\partial f}{\partial x}\frac{\partial h}{\partial y}\right)=\vec{0}$$
Hence
$$\nabla\bullet\nabla f\times\nabla h=\nabla\bullet\vec{0}=0$$
We have furthermore
$$\nabla\times(\nabla f\times\nabla h)=\nabla\times\vec{0}=\vec{0}$$
Then concerned to $\nabla\bullet\nabla f\times\vec{r}=0$ , $\vec{r}\times\nabla h=\vec{0}$ is REQUIRED, that means $\vec{r}//\nabla h$ !! :))
A: Suppose everything is smooth: consider an arbitrary smooth simply connected domain $D\subset \mathbb{R}^3$, divergence theorem reads:
$$
\iiint_D \nabla \cdot (\nabla f\times \nabla h) \,dV = \iint_{\partial D} \nabla f\times \nabla h\cdot d\mathbf{S} = \iint_{\partial D} \nabla f\times \nabla h\cdot\mathbf{n} \,dS
$$
Since everything is smooth, we can use triple product to switch positions:
$$
\iint_{\partial D} \nabla f\times \nabla h\cdot\mathbf{n} \,dS = \iint_{\partial D} \nabla h\times \mathbf{n}\cdot \nabla f\, dS = 
\iiint_D\Big(\nabla h\cdot \nabla\times(\nabla f) - \nabla f\cdot \nabla \times(\nabla h)\Big)dV = 0
$$
by the curl of an irrotational vector field is always zero. For $D$ is arbitrary, then 
$$
\nabla \cdot (\nabla f\times \nabla h) = 0.
$$ 
Looks like cheated though, but the intuition behind is that:

The cross product of two irrotational vector fields: $\mathbf{v} = \nabla h\times\nabla f$, must be perpendicular to both on any surface they span: $\mathbf{v}\perp \nabla h$ and $\mathbf{v}\perp \nabla f$, hence it is a solenoidal (rotational or harmonic) vector fields, and have zero divergence.

A: Introducing the Levi-Civita symbol, we have
\begin{align*}
\nabla \cdot (\nabla f \times \nabla h)
&= \epsilon^{ijk} \frac{\partial}{\partial x^{i}} \left( \frac{\partial^2 f}{\partial x^{j}} \frac{\partial h}{\partial x^{k}} \right) \\
&= \epsilon^{ijk} \left( \frac{\partial f}{\partial x^{i} \partial x^{j}} \frac{\partial h}{\partial x^{k}} + \frac{\partial f}{\partial x^{j}} \frac{\partial h^2}{\partial x^{k} \partial x^{i}} \right)
\end{align*}
But since $\epsilon^{ijk}$ is anti-symmetric and $\partial^2 / \partial x^{i}\partial x^{j}$ is symmetric,
$$ \epsilon^{ijk} \frac{\partial f^2}{\partial x^{i} \partial x^{j}} \frac{\partial h}{\partial x^{k}}
= -\epsilon^{jik} \frac{\partial f^2}{\partial x^{j} \partial x^{i}} \frac{\partial h}{\partial x^{k}}, $$ 
this quantity vanishes as it is identical to its negative. Similar argument applies to the second term, yielding
$$ \nabla \cdot (\nabla f \times \nabla h) = 0. $$
Finally, note that $\mathrm{r} = \nabla (\frac{1}{2} r^2) $. Then the second identity follows.
I guess this solution is equivalent to Muphrid's answer, but I am not sure as I know nothing on the Hodge dual (which replaces Levi-Civita pseudotensor in context-free description).
