Gradient in cylindrical coordinate using covariant derivative

The $nabla$ in differential geometry is NOT the same $nabla$ that you learn about in your vector calculus courses. In a typical vector calculus course, when one considers a function $f:Bbb{R}^ntoBbb{R}$ and introduces $nabla f$ as the gradient vector field, this is what I will henceforth refer to as $text{grad}(f)$. This is a vector field i.e a tensor field of type $(1,0)$. In differential geometry, $nabla f$ is a $(0,1)$ tensor field, i.e a covector field. By definition, begin{align} nabla f&:= df=sum_{i=1}^nfrac{partial f}{partial x^i},dx^itag{$*$} end{align} Even from your first equation, you can see that because $f$ is a smooth function, it is a $(0,0)$ tensor field, so that $r=s=0$. So, if $r$ and $s$ are $0$, there shouldn't even be any $Gamma$ symbols. Just look at the formula you wrote: begin{align} nabla_{p}f&=frac{partial f}{partial y^p} end{align} (slightly sloppy notation, but this is exactly what $(*)$ says).

From here, if you want to recover the familiar vector-calculus formula, then it's not the $Gamma$'s which matter, but rather the metric tensor itself, for which you need to know how one can convert between vectors and covectors using the musical isomorphism. By definition, if $g$ refers to the metric tensor on our manifold then begin{align} text{grad}(f)&:= g^{sharp}(df) &=g^{sharp}left(sum_{i=1}^nfrac{partial f}{partial x^i},dx^iright) &=sum_{i=1}^nfrac{partial f}{partial x^i}g^{sharp}(dx^i) &=sum_{i,j=1}^nfrac{partial f}{partial x^i}g^{ij}frac{partial }{partial x^j}tag{$**$} end{align} Here, $g_{ij}:=gleft(frac{partial}{partial x^i},frac{partial}{partial x^j}right)$, and $(g^{ij})$ is the inverse matrix of $(g_{ij})$.

Now, $frac{partial}{partial x^j}$ is the $j^{th}$ coordinate basis-vector, but it is not necessarily normalized. If we let $mathbf{e}_j$ denote the normalized version, then the relationship is that begin{align} frac{partial}{partial x^j}&=left|frac{partial}{partial x^j}right|cdot mathbf{e}_j =sqrt{g_{jj}}mathbf{e}_jtag{$***$} end{align} (i.e the length of a vector is the square root of its inner/dot-product with itself). Plugging $(***)$ into $(**)$ yields begin{align} text{grad}(f)&=sum_{i,j=1}^ng^{ij}sqrt{g_{jj}}frac{partial f}{partial x^i},mathbf{e}_j. end{align} This is the general formula for the gradient vector field of a smooth function on any (Pseudo)-Riemannian manifold $(M,g)$.

In the case where $M=Bbb{R}^3$ and $g$ is the standard metric tensor field on $Bbb{R}^3$, the components relative to the cylindrical coordinates are begin{align} [g_{ij}]&= begin{pmatrix} g_{rr}& g_{rphi}&g_{rz} g_{phi r}&g_{phiphi}&g_{phi z} g_{zr}&g_{zphi}&g_{zz} end{pmatrix} = begin{pmatrix} 1&0&0 0& r^2 & 0 0&0&1 end{pmatrix} end{align} Since the matrix is diagonal, the inverse matrix is simply the matrix whose entries are reciprocals. So, plugging this into the above expression, it actually simplifies a lot: begin{align} text{grad}(f)&=sum_{i,j=1}^ng^{ij}sqrt{g_{jj}}frac{partial f}{partial x^i}mathbf{e}_j &=sum_{i=1}^nfrac{1}{g_{ii}}sqrt{g_{ii}}frac{partial f}{partial x^i},mathbf{e}_itag{due to diagonal matrix} &=sum_{i=1}^nfrac{1}{sqrt{g_{ii}}}frac{partial f}{partial x^i}mathbf{e}_i end{align} For the specific case of cylindrical coordinates, we thus get begin{align} text{grad}(f)&=frac{1}{sqrt{g_{rr}}}frac{partial f}{partial r}mathbf{e}_r+ frac{1}{sqrt{g_{phiphi}}}frac{partial f}{partial phi}mathbf{e}_{phi}+ frac{1}{sqrt{g_{zz}}}frac{partial f}{partial z}mathbf{e}_z &=frac{partial f}{partial r}mathbf{e}_r + frac{1}{r}frac{partial f}{partial phi}mathbf{e}_{phi}+frac{partial f}{partial z}mathbf{e}_z, end{align} which is precisely the formula you quote. So, just to reiterate, the $frac{1}{r}$ comes from the metric tensor itself, not the $Gamma$ (the $Gamma$'s only appear if you're covariantly differentiating tensor fields of rank $geq 1$ i.e $r+sgeq 1$).

Take a look at this math answer of mine for a similar calculation for polar coordinates in the plane (it's almost exactly the same calculation), and see the various links there.

Comments

YOu write

.... which I've expanded into begin{align} nabla f&=(nabla_rf)hat{r}+(nabla_{phi}f)hat{phi}+(nabla_hf)hat{h} end{align}

Well, this is just wrong; on the LHS you have a covector field (i.e a tensor field of type $(0,1)$) while on the RHS you have a vector field (a tensor field of type $(1,0)$) so of course they cannot be equal. This is also not the definition of $nabla f$ as given in your book (which is the same as $(*)$ which I wrote above). Actually, even the $hat{e}^{mu}$ notation is terrible, because the $hat{}$ somehow suggests you're talking about a unit covector field, which is just wrong. The correct equation is $(*)$ which I wrote above, and surely that equation is super easy to remember.