Verifying a solution from Strauss' "Partial Differential Equations - An Introduction", 2nd edition In section 9.2 on page 241, question #12 is given as follows:
"Solve the three-dimensional wave equation in $\{r\ne0,t>0\}$ with zero initial conditions and with the limiting condition
\begin{equation*}
\lim_{r\to 0}4\pi r^{2}u_{r}(r,t) = g(t).
\end{equation*}
Assume that $g(0) = g^{\prime}(0) = g^{\prime\prime}(0) = 0$.
The text also provides the following solution:
\begin{equation*}
u = \begin{cases} -\frac{1}{4\pi r}g\left(t-\frac{r}{c}\right) &\mbox{if } t \ge \frac{r}{c} \\
0 & \mbox{if } 0\le t\le \frac{r}{c}. \end{cases}
\end{equation*}
For simplicity, let $c=1$.  Now,
\begin{alignat*}{2}
u_{r} &= \frac{4\pi r^{2}g^{\prime}(t-r) + 8\pi rg(t-r)}{16\pi^{2}r^{4}} \\
&= \frac{g^{\prime}(t-r)}{4\pi r^{2}} + \frac{g(t-r)}{2\pi r^{3}} \Rightarrow \\
u_{rr} &= \frac{-4\pi r^{2}g^{\prime\prime}(t-r)-8\pi rg^{\prime}(t-r)}{16\pi^{2}r^{4}} - \left[ \frac{2\pi r^{3}g^{\prime}(t-r)+6\pi r^{2}g(t-r)}{4\pi^{2}r^{6}}\right] \\
&= -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} - \frac{g^{\prime}(t-r)}{\pi r^{3}} - \frac{3g(t-r)}{2\pi r^{4}}.
\end{alignat*}
However, 
\begin{alignat*}{2}
u_{rr} + \frac{2}{r}u_{r} &= -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} - \frac{g^{\prime}(t-r)}{2\pi r^{3}} - \frac{g(t-r)}{2\pi r^{4}} \\
&\ne -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} \\
&= u_{tt}.
\end{alignat*}
So,
\begin{equation*}
u_{tt} \ne u_{rr} + \frac{2}{r}u_{r} = \Delta u \Rightarrow
\end{equation*}
$u$ does not solve the wave equation in 3D.  What am I missing in my reasoning or calculation that can make the given solution for $u$ correct? 

I prefer to write   $u_{rr}+\frac{2}{r}u_r$ as  $\frac{1}{r^2}(r^2 u_r)_r$, which has more meaning ($r^2 u_r$ is the flux through a sphere of radius $r$) and is easier to calculate with. Given
$$u=-\frac{1}{4\pi} r^{-1} g(t-r/c)$$
we get
$$u_r= \frac{1}{4\pi} r^{-2} g(t-r/c)+\frac{1}{4\pi c} r^{-1} g'(t-r/c)$$
hence
$$r^2u_r= \frac{1}{4\pi}   g(t-r/c)+\frac{1}{4\pi c} r  g'(t-r/c)$$
Take one more derivative: 
$$\begin{split}(r^2u_r)_r&=  -\frac{1}{4\pi c}   g'(t-r/c)+\frac{1}{4\pi c}    g'(t-r/c)-\frac{1}{4\pi c^2}  r  g''(t-r/c) \\ & = -\frac{1}{4\pi c^2}  r  g''(t-r/c)\end{split}$$
Finally, 
$$\frac{1}{r^2}(r^2 u_r)_r=-\frac{1}{4\pi rc^2}     g''(t-r/c)$$
No quotient rule at all. It has its uses, but when you can avoid it by using the negative power of a variable, do that. 
