Taking the cross product of a cross product? Proving an identity that involves gradients and vectors? Problem 20:

Solution:

I am having difficulty understanding how the boxed is not equal to 0. The derivative of 1 is equal to 0.
 A: The boxed entity is a differential operator that is to be applied to $\mathbf{r}$:
$$
\frac{\partial\mathbf{r}}{\partial x}=\mathbf{i}\qquad
\frac{\partial\mathbf{r}}{\partial y}=\mathbf{j}\qquad
\frac{\partial\mathbf{r}}{\partial z}=\mathbf{k}
$$

After seeing your comments, I see that what might be bothering you is that the product rule in this situation is
$$
(\nabla\cdot\mathbf{u})\mathbf{v}=(\mathbf{u}\cdot\nabla)\mathbf{v}+\mathbf{v}(\nabla\cdot\mathbf{u})
$$
and since $\mathbf{u}$ is constant, $\nabla\cdot\mathbf{u}=0$.
A: Let's back up.  What you're really having trouble with is understanding
$$\nabla \times (a \times r) = (\nabla \cdot r) a - (\nabla \cdot a) r$$
The notation could be considered unclear.  The product rule tells us that both $a$ and $r$ must be differentiated here; there are actually four terms.  One way to denote this is to use overdots; I'll consider only the second term:
$$(\nabla \cdot a) r = (\dot \nabla \cdot \dot a) r + (\dot \nabla \cdot a) \dot r$$
The overdot means "if $\nabla$ has a dot over it, differentiate only that which also has a dot over it".
Let's look at the first term; it's just a divergence.
$$(\dot \nabla \cdot \dot a) r =r ( \mathrm{div} \, a) = 0$$
Because $a$ is a constant vector.
The second term is more clearly written as a directional derivative.
$$(\dot \nabla \cdot a) \dot r = (a \cdot \dot \nabla) \dot r = a$$
This is the boxed term.  It's a known identity that the directional derivative of the position vector is just the direction the derivative is taken in.
Unfortunately, without any overdots or other decorations to denote what is to be differentiated, it becomes very unclear whether to apply the product rule or not.  I would always regard a statement like $(\nabla \cdot a) r$ to differentiate only the vector $a$, for instance, but the authors of this problem did not choose to do that.
