Laplacian Problem I'm going through past exams for study and I've come across a question I can never seem to solve.
Question: Given r = (x,y,z) and $r = \sqrt{x^2 + y^2 +z^2}$. Find $\nabla^2{\cosh(r)}$
I gave it a go and attempted to solve it directly, but the method was very tedious, and in an exam format would take too long. I know there is a way to solve using vector identities (nabla identities), but I'm not sure where to start.
Any help would be appreciated.
Thanks
 A: $r = \sqrt{\mathbf{r} \cdot \mathbf{r}}$, so by the chain rule,
$$ \nabla \cosh{r} = (\nabla r) \frac{\partial}{\partial r} \cosh{r} = (\nabla r) \sinh{r}, $$
and then
$$ \nabla^2 \cosh{r} = \nabla \cdot \left( (\nabla r) \sinh{r} \right) = \lvert \nabla r \rvert^2 \cosh{r} + (\nabla^2 r)\sinh{r}. $$
Now,
$$ \nabla r = \nabla \sqrt{\mathbf{r} \cdot \mathbf{r}} = \frac{\mathbf{r}}{(\mathbf{r} \cdot \mathbf{r})^{3/2}} $$
and then
$$ \nabla \cdot \frac{\mathbf{r}}{(\mathbf{r} \cdot \mathbf{r})^{3/2}} = \frac{3}{r^{3/2}} - \frac{3 \mathbf{r} \cdot \mathbf{r} }{r^{5/2}}=0, $$
so we find
$$ \nabla \cosh{r} = \frac{1}{r^2}\cosh{r}. $$
(Of course, the easiest way to do this is to use the Laplacian in spherical coordinates, $ \nabla^2 f = r^{-2}\partial_r r^2\partial_r f + \dotsb $.)
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mbox{If a function}\ \mrm{f}\ \mbox{depends solely on}\ r = \verts{\vec{r}},\quad 
\nabla\mrm{f}\pars{r} = \mrm{f}'\pars{r}\,{\vec{r} \over r} =
{\mrm{f}'\pars{r} \over r}\,\vec{r}
\\[5mm]
&\mbox{Note that}\ \partiald{\mrm{f}\pars{r}}{x} =
\totald{\mrm{f}\pars{r}}{r}\,\partiald{r}{x} =
\totald{\mrm{f}\pars{r}}{r}\,{1 \over 2r}\,\,\partiald{r^{2}}{x} =
\totald{\mrm{f}\pars{r}}{r}\,{x \over r}\quad
\pars{\substack{\mbox{Similarly for}\\[1mm] \ds{y}\ \mbox{and}\ z}}
\\[1cm]
&\mbox{In order to evaluate}\ \nabla^{2}\mrm{f}\pars{r} \stackrel{\mbox{def.}}{=} \nabla\cdot\bracks{\nabla\mrm{f}\pars{r}} =
\nabla\cdot\bracks{{\mrm{f}'\pars{r} \over r}\,\vec{r}}\
\mbox{I'll use the identity}
\\ & \nabla\cdot\pars{\varphi\vec{A}} =
\pars{\nabla\varphi}\cdot\vec{A} + \varphi\nabla\cdot\vec{A}.
\end{align}

\begin{align}
\nabla^{2}\mrm{f}\pars{r} & =
\braces{\nabla\bracks{\mrm{f}'\pars{r} \over r}}\cdot\vec{r} + {\mrm{f}'\pars{r} \over r}\,\ \overbrace{\nabla\cdot\vec{r}}^{\ds{=\ 3}} =
{\vec{r} \over r}\,\totald{}{r}\bracks{\mrm{f}'\pars{r} \over r}\cdot\vec{r} +
{3\,\mrm{f}'\pars{r} \over r}
\\[5mm] & =
r\,{\mrm{f}''\pars{r}r - \mrm{f}'\pars{r} \over r^{2}} +
{3\,\mrm{f}'\pars{r} \over r} =
\bbx{\mrm{f}''\pars{r} + {2\,\mrm{f}'\pars{r} \over r}}
\end{align}


With $\ds{\mrm{f}\pars{r} \equiv \cosh\pars{r}}$:

$$
\nabla^{2}\cosh\pars{r} = \bbx{\cosh\pars{r} + {2\sinh\pars{r} \over r}} 
$$
