Formula for the gradient of $F(\rho,\phi,z)$ Suppose $F(\rho,\phi,z)$ is continuously differentiable, I am interested in showing that the maximum directional derivative of $F$ at any point is given by $$\sqrt{\left(\frac{dF}{d\rho}\right)^2+\left(\frac{1}{\rho}\right)^2\left(\frac{dF}{d\phi}\right)^2+\left(\frac{dF}{dz}\right)^2}$$ 
I understand that the maximum directional derivative is the same as the magnitude but can't really but I am just not very much sure of why the rho is squared in the formula  
 A: I'm assuming $F:\mathbb{R}^3\to\mathbb{R}\;$ i.e. $F$ is a scalar field.
Suppose we move from $(\rho,\phi,z)$ to a point $(\rho+d\rho ,\phi+d\phi,z+d z)$ close by and the vector which terminates at the new point starting at $(\rho,\phi,z)$ is $\vec{dr}$. 
The vector $\hat\rho$ is defined as unit vector along which $\phi$ and $z$ don't change and $\rho$ increases. $\hat\phi,\hat z$ are defined in a similar fashion. $\hat\rho,\hat\phi,\hat z$ are not all constant vectors and vary from point to point but never fail to form an orthogonal system. 
If we move infinitesimally we can imagine $\hat\rho,\hat\phi,\hat z$ doesn't change during our movement. As $\hat\rho,\hat\phi,\hat z$ are orthogonal to each other and hence linearly independent we can express $\vec{dr}$ as a linear combination of these $3$.
$$\vec{dr}=A\hat\rho+B\hat\phi+C\hat z$$
We can learn a lot by looking at the special cases when two of $d\rho,d\phi,dz$ is $0$. In the following pictures the red point is our initial point and the blue point is the final one. 
If we move along $\hat\phi$ to get to $(\rho,\phi+d\phi,z)$ we end up with the following figure. 

In this case $\vec{dr}=\hat\phi\rho d\phi$.
If we move along $\hat\rho$ to get to $(\rho+d\rho,\phi,z)$ we get the following figure:

In this case $\vec{dr}=\hat\rho d\rho$.
If we move along $\hat z$ or along $z$-axis to get to $(\rho,\phi,z+dz)$ we get the following figure.
 
In this case $\vec{dr}=\hat zdz$.
In the general case we can move from our initial to final point in the following way
$$(\rho,\phi,z)\to(\rho,\phi+d\phi,z)\qquad\text{displacement}=\hat\phi\rho d\phi\\
(\rho,\phi+d\phi,z)\to(\rho+d\rho,\phi+d\phi,z)\qquad\text{displacement}=\hat\rho d\rho\\
(\rho+d\rho,\phi+d\phi,z)\to(\rho+d\rho,\phi+d\phi,z+dz)\qquad\text{displacement}=\hat zdz$$
So total displacement is given by $$\vec{dr}=(d\rho)\hat\rho+(\rho d\phi)\hat\phi+(dz)\hat z$$
Just like a single-variable real-valued function satisfies $\Delta f\approx {df\over dx}\Delta x$ for small $\Delta x$, we define gradient to satisfy a similar condition for multi-variable real-valued scalar field, namely $\Delta F\approx\nabla F\cdot \Delta\vec{r}$ where $\Delta\vec r$ represents the vector along which the change in $F$ or $\Delta F$ was achieved which guarantees the maximum directional derivative at some point will be found in the direction of the gradient at that point. So we can get an expression for $\nabla F$ in cylindrical polar in the following way. 
Change in  $F$ along $\vec{dr}$ is $$F(\rho+d\rho ,\phi+d\phi,z+d z)-F(\rho ,\phi,z)={\partial F\over\partial\rho}d\rho+{\partial F\over\partial\phi}d\phi+{\partial F\over\partial z}dz\\
=\left({\partial F\over\partial\rho}\hat\rho+{1\over\rho}{\partial F\over\partial\phi}\hat\phi+{\partial F\over\partial z}\hat z\right)\cdot\vec{dr}=\nabla F\cdot\vec{dr}\\
\therefore|\nabla F|=\sqrt{\left({\partial F\over\partial\rho}\right)^2+{1\over\rho^2}\left({\partial F\over\partial\phi}\right)^2+\left({\partial F\over\partial z}\right)^2}$$
