linearzation of curvature in spherical coordinates Curvature in polar coordinates (two dimensions) is
\begin{equation}
\kappa = \frac{\rho^2 + 2 \rho'(\theta)^2 -\rho \rho''(\theta)}{\left(\rho^2 + \rho'(\theta)\right)^{\frac{3}{2}}}
\end{equation}
when we assume $\frac{\rho'(\theta)}{\rho} \ll1$, above expression is written as
\begin{align}
\kappa = \frac{1}{\rho} + 2\frac{\rho'(\theta)^2}{\rho^2} - \frac{\rho''(\theta)}{\rho^2}
\end{align}
what is the corresponding expression for the mean curvature in three dimensions in spherical coordinates?
Typos EDIT:
\begin{equation}
\kappa = \frac{\rho^2 + 2 \rho'(\theta)^2 -\rho \rho''(\theta)}{\left(\rho^2 + \rho'(\theta)^2\right)^{\frac{3}{2}}}
\end{equation}
\begin{align}
\kappa = \frac{1}{\rho} - \frac{\rho''(\theta)}{\rho^2}
\end{align}
 A: Since the surface is defined by $r=f(\theta,\phi)$, we compute an inward normal (not normalised)
$$
\mathbf{N}=\nabla (f(\theta,\phi)-r)=-\mathbf{e}_r+\frac1r f_\theta\,\mathbf{e}_\theta+\frac1{r\sin\theta}f_\phi\,\mathbf{e}_\phi
$$
so the mean curvature $H$ is
\begin{align*}
2H&=-\nabla\cdot\left(\frac{\mathbf{N}}{\left\lvert\mathbf{N}\right\rvert}\right)\\
&=\frac{\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N})-2\nabla\cdot\mathbf{N}}{2(\mathbf{N}\cdot\mathbf{N})^{3/2}}
\end{align*}
(sign chosen so the mean curvature of a sphere is positive).
But if $\lvert N\rvert\approx 1$ (i.e., $\frac{f_\theta^2}{r^2}+\frac{f_\phi^2}{r^2\sin^2\theta}\ll 1$), then
$$
H\approx\frac14\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N})-\frac12\nabla\cdot\mathbf{N}
$$
We calculate
\begin{align*}
-\nabla\cdot\mathbf{N}
&=\frac1{r^2}\frac{\partial(r^2)}{\partial r}+\frac1{r\sin\theta}\frac{\partial(-\frac1r f_\theta\sin\theta)}{\partial\theta}+\frac1{r\sin\theta}\frac{\partial(-\frac1{r\sin\theta}f_\phi)}{\partial\phi}\\
&=\frac2r-\frac{f_{\theta\theta}\sin\theta+f_\theta\cos\theta}{r^2\sin\theta}-\frac{f_{\phi\phi}}{r^2\sin^2\theta}\\
&\approx\frac2r-\frac{f_{\theta\theta}}{r^2}-\frac{f_{\phi\phi}}{r^2\sin^2\theta}\\
\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N})
&=\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N}-1)\approx 0
\end{align*}
since every term in $\mathbf{N}\cdot\nabla(\mathbf{N}\cdot\mathbf{N}-1)$ has a factor of $f_\theta$ or $f_\phi$ which are assumed to be negligible.  So we end up with
$$
2H\approx\frac2{f}-\frac{f_{\theta\theta}}{f^2}-\frac{f_{\phi\phi}}{f^2\sin^2\theta}
$$
or abusing notation,
$$
H\approx\frac1r+\frac12\left(-\frac{r_{\theta\theta}}{r^2}-\frac{r_{\phi\phi}}{r^2\sin^2\theta}\right)
$$
which you should recognise big parenthesis term is basically the Laplacian of $r$ in the spherical variables.

You can also derive this using a more general result: If $f\colon M^2\to\mathbb{R}^3$ is an embedding and we have a normal vector field given by $\varphi N$, $N$ is the Gauss map and $\varphi$ is some smooth function, then the first variation of the principal curvatures are
$$
\delta^{(1)}\kappa_i=\operatorname{Hess}(\varphi)(e_i,e_i)+\kappa_i^2 \varphi
$$
(exercise in computing using moving frames, for example) and hence
$$
\delta^{(1)} H=-\frac12\Delta\varphi+(2H^2-K)\varphi.
$$
(Sign convention: the Laplacian $\Delta:=-\operatorname{Tr}_{I}\operatorname{Hess}$ is minus div grad.)
In our case, $M=S^2$ and we start with the round sphere of radius $r_0:=f(\theta_0,\phi_0)$ for our (local) calculation.  If $\nabla f$ is small, then we are making small perturbation of the round sphere $r=r_0$ with $\varphi=r_0-f$ (note $N$ points inwards, to get principal curvatures positive for our starting sphere) and so the first-order approximation we get is
$$
H\approx H_0+\delta^{(1)} H=\frac1{r_0}-\frac12\Delta^{\mathbb{S}^2(r_0)}\varphi=\frac1{r_0}+\frac1{2r_0^2}\Delta^{\mathbb{S}^2} f.
$$
where the Laplacian on standard unit sphere $\mathbb{S}^2$ is
$$
\Delta^{\mathbb{S}^2}=-\frac{\partial^2}{\partial\theta^2}-\frac1{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}.
$$
Of course, there is nothing special about the suffix $0$ here except it is the starting point of our calculation, so dropping it altogether recovers the previous formula.
Similarly, for smooth $n$-dimensional hypersurfaces in $\mathbb{R}^{n+1}$ that are described by "$C^1$-approximately-constant $\log r$", we get
$$
nH\approx\frac{n}r+\frac1{r^2}\Delta^{\mathbb{S}^n}r.
$$
generalising both examples $n=1,2$.
A: Please note that the following are in polar / cylindrical coordinates ( but not spherical).
We can derive curvatures on the basis of differential lengths/ triangles and curvature as follows. $\angle AOB$ is negligible in comparison to $\angle BOX ,$ and so on. $PB$ is differential arc of meridian. Primes are differentiation w.r.t $\theta. $ Using $r$ instead of $\rho$ for polar coordinate radius vector.
Two dimensional
$(x,y)$ or $( r,\theta) $ are here labeled as $(z,r). OA = r $

$$ k_1 = \frac{d \phi}{ds} = \frac{d \psi+ d \theta}{ds}= \frac {d ( \tan^{-1}\dfrac{r}{r'})+ d \theta}{ds}=$$
$$\frac{r^2+2r^{'2}-r r{''}}{r^2+r^{'2}}\cdot\frac{1}{\sqrt{r^2+r^{'2}}} \tag 1 $$
Three dimensional
$Z$ is the axis of symmetry and $r$ is the radius in cylindrical coordinates.
If H is mean curvature of a surface of revolution in particular, then $ 2 H = k_1+k_2; $
Arc $PB $ on meridian is extended up to Z-axis as $PQ.$ The minor curvature
$k_2=\dfrac{1}{PQ}=\dfrac{1}{R_2} $ can be represented as in the above meridional diagram.
$$k_2=\frac{\cos \phi}{r}=\frac{\cos {(\theta+\psi)}}{r}=\frac{\cos \theta \cos \psi- \sin \theta \sin \psi}{r}$$
$$=\frac{\cos \theta \;r'- \sin \theta\;r}{r \sqrt{r^2+r^{'2}}}\tag 2 $$
$$ 2H=\frac{(r^2+2r^{'2}-r r{''})/({{r^2+r^{'2}}} )+{(\cos \theta \;r'/r- \sin \theta\;)}}{\sqrt{r^2+r^{'2}}} ,\tag 3 $$
Linearizations can next  be  made on this.
$$ H=\frac{(1- r{''}/r- \sin \theta\;)}{2r}.\tag 4 $$
