Proving $\nabla\cos(k\|x\|)=-k\sin(k\|x\|)\cdot \frac{x}{\|x\|}$ in $\mathcal{D}(\mathbb{R}^3)$ Using the chain rule, it is easy to show that for all $x\in\mathbb{R}^3$:
$\nabla\cos(k\|x\|)=-k\sin(k\|x\|)\cdot \frac{x}{\|x\|}.\qquad(\star)$
However, how do we verify this equality in $\mathcal{D}'(\mathbb{R}^3)$? I would normally proceed by saying that for all $\phi\in C^\infty_c(\mathbb{R}^3)$, we have
$$\langle\nabla\cos(k\|x\|),\phi\rangle=-\langle\cos(k\|x\|),\nabla\phi\rangle,$$
although I suppose that this is not what is meant by verifying $(\star)$ in $\mathcal{D}'(\mathbb{R}^3)$, since $\cos(k\|x\|)\in C^\infty(\mathbb{R}^3)$ and is thus a function rather than a functional which acts on $\phi$.
 A: By definition of the derivative of a distribution
$$\langle \nabla \cos(k\|x\|), \phi \rangle = - \langle \cos(k\|x\|), \nabla\phi \rangle$$
Here $\cos(k\|x\|) \in C(\mathbb R^3)$ so
$$\langle \cos(k\|x\|), \nabla\phi \rangle = \int_{\mathbb R^3} \cos(k\|x\|) \, \nabla\phi(x) \, dx$$
We split the integral into two pieces:
$$\int_{\mathbb R^3} = \int_{\|x\| < \epsilon} + \int_{\|x\| > \epsilon}$$
On $\|x\| < \epsilon$ the integrand $\cos(k\|x\|) \, \nabla\phi(x)$ is bounded so the integral will be of order $\operatorname{volume}(\|x\|<\epsilon) \sim \epsilon^3$ and therefore tends to $0$ as $\epsilon \to 0$.
On $\|x\| > \epsilon$ we have
$\cos(k\|x\|) \, \nabla\phi(x) = \nabla \{\cos(k\|x\|) \, \phi(x) \} - \nabla\{\cos(k\|x\|)\} \, \phi(x)$
so
$$\begin{align}
\int_{\|x\| > \epsilon} \cos(k\|x\|) \, \nabla\phi(x) \, dx 
& = \int_{\|x\| > \epsilon} \nabla \{\cos(k\|x\|) \, \phi(x) \} \, dx
- \int_{\|x\| > \epsilon} \nabla\{\cos(k\|x\|)\} \, \phi(x) \, dx \\
& = \oint_{\|x\| = \epsilon} \cos(k\|x\|) \, \phi(x) \, n \, dS
- \int_{\|x\| > \epsilon} -\sin(k\|x\|) \, k\frac{x}{\|x\|} \, \phi(x) \, dx
\end{align}$$
where $d\Omega$ is the solid angle measure and $n$ is the outward directed unit normal ($n = x/\|x\|$).
Again the first integral clearly tends to $0$ as $\epsilon \to 0$; in this case the integral is of order $\operatorname{area}(\|x\|=\epsilon) \sim \epsilon^2$.
Thus we have
$$\langle \nabla \cos(k\|x\|), \phi \rangle = \lim_{\epsilon \to 0}
\int_{\|x\| > \epsilon} \sin(k\|x\|) \, k\frac{x}{\|x\|} \, \phi(x) \, dx$$
But
$$\sin(k\|x\|) \, k\frac{x}{\|x\|} = k^2 \frac{\sin(k\|x\|)}{k\|x\|} \, x$$
which is boundad as $x\to 0$ and thus locally integrable, so we can actually write
$$\langle \nabla \cos(k\|x\|), \phi \rangle = 
\int_{\mathbb R^3} \sin(k\|x\|) \, k\frac{x}{\|x\|} \, \phi(x) \, dx
= \left\langle k \sin(k\|x\|) \, \frac{x}{\|x\|} , \phi \right\rangle$$
i.e.
$$\nabla \cos(k\|x\|) = k \sin(k\|x\|) \, \frac{x}{\|x\|}$$
as a distribution.
