How to change Cylindrical $\nabla$ to cartesian $\nabla$? If I change the cylindric basis to cartesian cartesian basis, I had to do a lot of algebra. Is there any easiest way to convert?
Cylindrical $\nabla = \vec{e_r} \frac{\partial}{\partial r} + \frac{1}{r} \vec{e_\phi} \frac{\partial}{\partial r} + \vec{e_z} \frac{\partial}{\partial z}$
 to the cartesian one
 A: Note that the position vector is invariant under the coordinate transformation and therefore we can write
$$\hat xx+\hat yy=\hat \rho\rho \tag 1$$
Taking partial derivatives of $(1)$ with respect to $\rho$ and $\phi$, we obtain respectively
$$\begin{align}
\hat \rho&=\hat x \frac{\partial x}{\partial \rho}+\hat y \frac{\partial y}{\partial \rho} \\\\
&=\hat x \frac{\partial \rho}{\partial x}+\hat y \frac{\partial \rho}{\partial y}\tag 2\\\\
\hat \phi&=\frac{1}{\rho}\left(\hat x \frac{\partial x}{\partial \phi}+\hat y \frac{\partial y}{\partial \phi}\right)\\\\
&= \rho\left(\hat x \frac{\partial \phi}{\partial \rho}+\hat y \frac{\partial \phi}{\partial y}\right) \tag 3
\end{align}$$
Using $(2)$ in $(3)$ in the expression for the gradient operator in cylindrical coordinates yields
$$\begin{align}
\nabla &=\hat \rho\frac{\partial }{\partial \rho}+\hat \phi \frac{1}{\rho}\frac{\partial }{\partial \phi}+\hat z\frac{\partial }{\partial z}\\\\
&=\left(\hat x \frac{\partial \rho}{\partial x}+\hat y \frac{\partial \rho}{\partial y}\right)\frac{\partial }{\partial \rho}+\left(\rho\left(\hat x \frac{\partial \phi}{\partial \rho}+\hat y \frac{\partial \phi}{\partial y}\right) \right)\frac{1}{\rho}\frac{\partial }{\partial \phi}+\hat z\frac{\partial }{\partial z}\\\\
&=\hat x \left(\frac{\partial \rho}{\partial x}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}\right)+\hat y \left(\frac{\partial \rho}{\partial y}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial y}\frac{\partial }{\partial \phi}\right)+\hat z \left(\frac{\partial }{\partial z}\right)\\\\
&=\hat x \frac{\partial }{\partial x}+\hat y \frac{\partial }{\partial y}+\hat z \frac{\partial }{\partial z}
\end{align}$$


Note that in arriving at $(2)$ and $(3)$, we made use of the equalities 
$$\begin{align}
\frac{\partial x}{\partial \rho}&=\frac{\partial \rho}{\partial x}\\\\
\frac{\partial y}{\partial \rho}&=\frac{\partial \rho}{\partial y}\\\\
\frac{\partial x}{\partial \phi}&=\rho^2 \frac{\partial \phi}{\partial x}\\\\
\frac{\partial y}{\partial \phi}&=\rho^2\frac{\partial \phi}{\partial y}
\end{align}$$

