Catenoid is a minimal surface i want to show that the catenoid is a minimal surface. I have given 
$f:I \times (0,2\pi)\longrightarrow \mathbb{R}^3$ with $f(r,\phi)=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\r\end{array} \right)$.
I know that:
$f$ is minimal surface $\Longleftrightarrow$ $\Delta f=0$.
$f$ is given in polar coordinates so i have to calculate the following: 
$\Delta f= \frac{\partial^2f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}\frac{\partial^2f}{\partial \phi^2}$
$\frac{\partial f}{\partial r}= \left( \begin{array}{c}\sinh(r) \;\cos(\phi)\\\sinh(r) \;\sin(\phi)\\1\end{array} \right)$ , $\frac{\partial^2f}{\partial r^2}=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\0\end{array} \right)$
$\frac{\partial f}{\partial \phi}=\left( \begin{array}{c}-\cosh(r) \;\sin(\phi)\\\cosh(r) \;\cos(\phi)\\0\end{array} \right)$ ,  $\frac{\partial^2 f}{\partial \phi^2}=\left( \begin{array}{c}-\cosh(r) \;\cos(\phi)\\-\cosh(r) \;\sin(\phi)\\0\end{array} \right)$.
But when I put all together I can not show that $\Delta f$ is 0. What did I do wrong?
Can someone help me please?
Thanks in advance.
 A: I think you are a little bit confused about the harmonic characterization of (conformally immersed) minimal surfaces.  
We know that every regular $2$-dimensional surface can be described locally in isothermal coordinates (i.e. for neighborhood of the surface, there's a coordinate map that preserves angles aka is conformal aka has 1st fundamental form satisfying $E=G$, $F=0$).  
So we can cover the surface by a family of coordinate maps $$\vec{x_\alpha}(u,v) = \big(x_1(u,v), x_2(u,v), x_3(u,v)\big)$$
with each $\vec{x_\alpha}$ conformally mapping an open subset of $\mathbb{R}^2$ to $\mathbb{R}^3$.
The harmonic characterization says that the surface is minimal iff for each $\vec{x}_\alpha$ in such a family, the coordinates $x_i(u,v)$ are harmonic functions with respect to the coordinates (u,v).  I would advise going back to look at the proof of this characterization for clarification, and thinking about geometrically what it means to be conformal (preserve angles).  
I think where you have been misled is in thinking of this as a polar parametrization and using the so-called "polar form of the laplacian." 
In your case you have a conformal coordinate map describing the entire catenoid in coordinates $r, \varphi$.  What does $$\big(\cosh(r)\cos(\varphi)\big)_{rr} + \cosh(r)\cos(\varphi)\big)_{\varphi \varphi}$$ look like?
A: Theorem: If $(u,v) \to f(u,v)$ is an isothermal parametrisation, then $$f_{uu}+f_{vv} = 2 E H\mathbf{N}$$
where $\mathbf{N}$ is the principal normal to the surface. It is clear from this that a) we must check the parametrisation is isothermal, and b) that the 'Laplacian' is not the usual $\Delta u = u_{xx} + u_{yy}$ (and the coordinates you use are not 'polars'. They are just abstract coordinates).  
The definition of isothermal is that the first fundamental form takes the form 
$$\pmatrix{\lambda^{2} & 0 \\ 0 & \lambda^{2}}$$
(I leave it to you to check this). Your coordinates are $(r,\phi)$, so we look at 
$$f_{rr} = \pmatrix{\cosh(r) \cos (\phi) \\ \cosh(r) \sin(\phi) \\ 0}$$
and 
$$F_{\phi \phi} = \pmatrix{\cosh(r) (-\cos(\phi)) \\ \cosh(r) (-\sin (\phi)) \\ 0}$$
from which we see that the catenoid is minimal.
A: 
This not a complete answer but that's too long for me to post it in comment.

Preliminary for differential geometry of surfaces
\begin{align*}
  \mathbf{x}(u,v)
  &= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\
  \mathbf{x}_u &= \frac{\partial \mathbf{x}}{\partial u} \\
  \mathbf{x}_v &= \frac{\partial \mathbf{x}}{\partial v} \\
  \mathbf{N} &=
  \frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_v|}
  \tag{unit normal vector} \\
\end{align*}
First fundamental form
$$\mathbb{I}=
\begin{pmatrix} E & F \\ F & G \end{pmatrix}= 
\begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} $$
Second fundamental form
$$\mathbb{II}=
 \begin{pmatrix} e & f \\ f & g \end{pmatrix}=
-\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
 \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix}$$
Metric
$$ds^2=E\, du^2+2F\, du\, dv+G\, dv^2$$
Element of area
$$dA=|\det \mathbb{I}| \, du \, dv
=|\mathbf{x}_u \times \mathbf{x}_v| \, du\, dv
=\sqrt{EG-F^2} \, du\, dv$$
Principal curvatures
Let $\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}$
where $\mathbb{A}=
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$.
Now
\begin{align*}
   \begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
   \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} &=
   \mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
   \begin{pmatrix} \mathbf{x}_{u} & \mathbf{x}_{v} \end{pmatrix} \\
  -\begin{pmatrix} e & f \\ f & g \end{pmatrix} &=
   \mathbb{A} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \\
   \mathbb{A} &=
  -\begin{pmatrix} e & f \\ f & g \end{pmatrix}
   \begin{pmatrix} E & F \\ F & G \end{pmatrix}^{-1}
\end{align*}
The principal curvatures $k_{1}, k_{2}$ are the eigenvalues of $-\mathbb{A}$.
Mean curvature
$$H=\frac{k_{1}+k_{2}}{2}
=-\frac{1}{2} \operatorname{tr} \mathbb{A}
=\frac{eG-2fF+gE}{2(EG-F^2)}$$
Gaussian curvature
$$K=k_{1} k_{2}
=(-1)^{2} \det \mathbb{A}
=\frac{eg-f^2}{EG-F^2}$$

For minimal surface, $$H=0$$

