Consider the 1st and 2nd derivatives (differential operations) of a function $z=f(x)$ with respect to horizontal coordinate $x$ in 2D Cartesian coordinates $(x,z)$,
$$\frac{df}{dx} \quad \text{and} \quad \frac{d^2f}{dx^2}.$$
If $f(x)$ is axisymmetric about the vertical $z$ axis and we want to transform the derivatives into cylindrical coordinates $(r,z)$, can they be written straightforward as follows:
$$\frac{1}{r}\frac{d(rf)}{dr} \quad \text{and} \quad \frac{1}{r}\frac{d}{dr}(r\frac{df}{dr}), $$ where $z=f(r)$ and $r$ is the radial coordinate.
Update:
For the 1st derivative, I believe the answer of @David K is correct. However, I don't understand the 2nd derivative. Actually, I got the transformation of the 2nd derivative by comparing the Laplace operators in Cartesian coordinates ($z=f(x)$) and in axisymmetric cylindrical coordinates ($z=f(r)$):
In Cartesian coordinates:
$$\nabla^2 f=\frac{d^2f}{dx^2}$$
In cylinderical coordiantes:
$$\nabla^2 f=\frac{1}{r}\frac{d}{d r}(r\frac{d f}{d r})$$
You have no angular component in your "cylindrical" coordinates, so all you have actually is Cartesian coordinates in which one coordinate is named $r$ instead of $x.$ The transformation of the derivative is indeed extremely straightforward:
\begin{align} \frac{\mathrm df}{\mathrm dx} &\to \frac{\mathrm df}{\mathrm dr}, \\ \frac{\mathrm d^2f}{\mathrm dx^2} &\to \frac{\mathrm d^2f}{\mathrm dr^2}. \end{align}
Why would it be anything else?
You could extend your function $f$ from a function on the $x$ axis to a function $\bar f$ over a plane $\pi_0$ through the origin, perpendicular to the $z$ axis, by setting up cylindrical coordinates around the $z$ axis, thereby assigning polar coordinates $r,\theta$ to the plane $\pi_0$, and defining the function
$$ \bar f(r, \theta) = f(r). $$
You can then take partial derivatives with respect to $r$ and $\theta$:
\begin{align} \frac{\partial \bar f}{\partial r} &= \frac{\mathrm df}{\mathrm dr}, \\ \frac{\partial \bar f}{\partial \theta} &= 0. \\ \end{align}
The gradient of $\bar f$ is then
$$ \nabla \bar f = \hat{\mathbf r} \frac{\partial \bar f}{\partial r} + \hat{\mathbf \theta} \frac 1r \frac{\partial \bar f}{\partial \theta}. $$
(See How to obtain the gradient in polar coordinates and its answers.) Note that there is a factor of $\frac1r$ outside one of the partial derivatives, but it's on the partial derivative with respect to $\theta,$ not $r,$ and there is no factor $r$ inside any partial derivative.