Cylindrical coordinates and Differential Displacement: unit vectors and differentiation

Question

How to calculate the Differential Displacement (Path Increment)

This is what it starts with:
\begin{align} \text{From the Cylindrical to the Rectangular coordinate system:}& \\ x&=\rho\cos\phi \\ y&=\rho\sin\phi \\ z&=z \\ & \\ \text{From the Rectangular to the Cylindrical coordinate system:}& \\ \rho&=\sqrt{x^2+y^2} \\ \phi&=\arctan(x,y) \\ z&=z \\ & \\ \text{Unit vectors:}& \\ \hat{\rho}&=\frac{\vec{\rho}}{\rho}=\frac{x\,\hat{x}+y\,\hat{y}}{\rho}=\hat{x}\cos\phi+\hat{y}\sin\phi \\ \hat{\phi}&=\hat{z}\times\hat{\rho}=-\hat{x}\sin\phi+\hat{y}\cos\phi \\ \hat{z}&=\hat{z} \\ & \\ \text{Partial derivatives:}& \\ \frac{\partial\hat{\rho}}{\partial\rho}&=0 \\ \frac{\partial\hat{\rho}}{\partial\phi}&=-\hat{x}\sin\phi+\hat{y}\cos\phi \\ \frac{\partial\hat{\rho}}{\partial{z}}&=0 \\ & \\ \frac{\partial\hat{\phi}}{\partial\rho}&=0 \\ \frac{\partial\hat{\phi}}{\partial\phi}&=-\hat{x}\cos\phi-\hat{y}\sin\phi=-\hat{\rho} \\ \frac{\partial\hat{\phi}}{\partial{z}}&=0 \\ & \\ \frac{\partial\hat{z}}{\partial\rho}&=0 \\ \frac{\partial\hat{z}}{\partial\phi}&=0 \\ \frac{\partial\hat{z}}{\partial{z}}&=0 \\ & \\ \text{Differential Displacement (Path increment):}& \\ \vec{r}&=\rho\hat{\rho}+z\hat{z} \\ d\vec{r}&=d(\rho\hat{\rho}+z\hat{z})=\hat{\rho}\,d\rho+\rho\, d\hat{\rho}+\hat{z}\,dz+z\,d\hat{z} \\ & \\ \end{align} This is my question:
This paper says:
\begin{align} \hat{\rho}\,d\rho+\rho\,d\hat{\rho}+\hat{z}\,dz+z\,d\hat{z}&=\hat{\rho}\,d\rho+\rho(\frac{\partial\hat{\rho}}{\partial\rho}d\rho+\frac{\partial\hat{\rho}}{\partial\phi}d\phi+\frac{\partial\hat{\rho}}{\partial z}dz)+\hat{z}\,dz+z(\frac{\partial\hat{z}}{\partial\rho}d\rho+\frac{\partial\hat{z}}{\partial\phi}d\phi+\frac{\partial\hat{z}}{\partial z}dz) \\ \end{align} What I don't get is why
$$d\hat{\rho}=\frac{\partial\hat{\rho}}{\partial\rho}d\rho+\frac{\partial\hat{\rho}}{\partial\phi}d\phi+\frac{\partial\hat{\rho}}{\partial z}dz$$ And in the same vein, why
$$d\hat{z}=\frac{\partial\hat{z}}{\partial\rho}d\rho+\frac{\partial\hat{z}}{\partial\phi}d\phi+\frac{\partial\hat{z}}{\partial z}dz\quad\text{?}$$

The road to a solution:
The derivative of a position vector $\vec{r}$, with respect to coordinate value $l$ (where $l\in\{x,y,z,\rho,\phi,r,\theta\}$) is expressed as:
\begin{align} \frac{\partial\vec{r}}{\partial l}&=\frac{\partial}{\partial l}(x\hat{a}_x+y\hat{a}_y+z\hat{a}_z) \\ &=\frac{\partial(x\hat{a}_x)}{\partial l}+\frac{\partial(y\hat{a}_y)}{\partial l}+\frac{\partial(z\hat{a}_z)}{\partial l} \\ &=\left(\frac{\partial x}{\partial l}\right)\hat{a}_x+\left(\frac{\partial y}{\partial l}\right)\hat{a}_y+\left(\frac{\partial z}{\partial l}\right)\hat{a}_z \\ \end{align} If a point moves such that its coordinate $l$ changes from $l$ to $l+\Delta$, then the position vector that describes that point changes from $\vec{r}$ to $\vec{r}+\vec{\Delta l}$.
This vector $\vec{\Delta l}$ is a directed distance between the vector $\vec{r}$ at coordinate $l$ and the new vector $\vec{r}+\vec{\Delta l}$ at coordinate $l+\Delta l$!
This directed distance $\vec{\Delta l}$ is related to the position vector derivative as:
\begin{align} \vec{\Delta l}&=\Delta l \frac{\partial\vec{r}}{\partial l} \\ &=\Delta l\left(\frac{\partial x}{\partial l}\right)\hat{a}_x+\Delta l\left(\frac{\partial y}{\partial l}\right)\hat{a}_y+\Delta l\left(\frac{\partial z}{\partial l}\right)\hat{a}_z \\ \end{align} As an example, consider the case when $l=\rho$.
Since $x=\rho\cos\phi$ and $y=\rho\sin\phi$ we find that:
\begin{align} \frac{\partial\vec{r}}{\partial\rho}&=\frac{\partial x}{\partial\rho}\hat{a}_x+\frac{\partial y}{\partial\rho}\hat{a}_y+\frac{\partial z}{\partial\rho}\hat{a}_z \\ &=\frac{\partial(\rho\cos\phi)}{\partial\rho}\hat{a}_x+\frac{\partial(\rho\sin\phi)}{\partial\rho}\hat{a}_y+\frac{\partial z}{\partial\rho}\hat{a}_z \\ &=(\cos\phi)\,\hat{a}_x+(\sin\phi)\,\hat{a}_y \\ &=\hat{a}_\rho \\ &=\hat{\rho} \end{align} A change in position from coordinates $\rho,\phi,z$ to $\rho+\Delta \rho,\phi,z$ results in a change in the position vector from $\vec{r}$ to $\vec{r}+\vec{\Delta l}$.
The vector $\vec{\Delta l}$ is a directed distance extending from point $\rho,\phi,z$ to point $\rho+\Delta \rho,\phi,z$, and is equal to:
\begin{align} \vec{\Delta l}&=\Delta \rho\frac{\partial\vec{r}}{\partial\rho} \\ &=\Delta \rho\,(\cos\phi)\,\hat{a}_x+\Delta \rho\,(\sin\phi)\,\hat{a}_y \\ &=\Delta \rho\,\hat{a}_\rho \\ &=\Delta \rho\,\hat{\rho} \end{align} If $\Delta l$ is really small (i.e., as it approaches zero) we can define something called a differential displacement vector $\vec{dl}$:
\begin{align} \vec{dl}&\doteq\lim\limits_{\Delta l \to 0}\vec{\Delta l} \\ &=\lim\limits_{\Delta l \to 0}\Delta l\left(\frac{\partial\vec{r}}{\partial l}\right) \\ &=\lim\limits_{\Delta l \to 0}\left(\frac{\partial\vec{r}}{\partial l}\right)\Delta l \\ &=\left(\frac{\partial\vec{r}}{\partial l}\right)dl \end{align} Using this, the differential displacement vectors for the cylindrical coordinate system are therefore:
\begin{align} \vec{d\rho}&=\frac{\partial\vec{r}}{\partial\rho}d\rho \\ &=\hat{a}_\rho \,d\rho \\ &=\hat{\rho}\,d\rho \\ & \\ \vec{d\phi}&=\frac{\partial\vec{r}}{\partial\phi}d\phi \\ &=\left(\frac{\partial x}{\partial\phi}\hat{a}_x+\frac{\partial y}{\partial\phi}\hat{a}_y+\frac{\partial z}{\partial\phi}\hat{a}_z\right)d\phi \\ &=\left(\frac{\partial\rho\cos\phi}{\partial\phi}\hat{a}_x+\frac{\partial\rho\sin\phi}{\partial\phi}\hat{a}_y+\frac{\partial z}{\partial\phi}\hat{a}_z\right)d\phi \\ &=\left(-\rho\,(\sin\phi)\,\hat{a}_x +\rho\,(\cos\phi)\,\hat{a}_y\right)d\phi \\ &=\left(-(\sin\phi)\,\hat{a}_x+(\cos\phi)\,\hat{a}_y\right)\rho \,d\phi \\ &=\hat{a}_\phi \,\rho \,d\phi \\ &=\hat{\phi}\,\rho\,d\phi \\ & \\ \vec{dz}&=\frac{\partial\vec{r}}{\partial z}dz \\ &=\hat{a}_z \,dz \\ &=\hat{z}\,dz \\ \end{align} This I got from this site.

Solution?:
Thanks for the tip, narasimham.

The total derivative of a function is different from its corresponding partial derivative ($\partial$). Calculation of the total derivative of $f$ with respect to $t$ does not assume that the other arguments are constant while $t$ varies; instead, it assumes that the other arguments too depend on $t$.

Remember that:
$$\hat{\phi}=-\hat{x}\sin\phi+\hat{y}\cos\phi$$

Then, this means that:
\begin{align} d\hat{\rho}&=\frac{\partial\hat{\rho}}{\partial\rho}d\rho+\frac{\partial\hat{\rho}}{\partial\phi}d\phi+\frac{\partial\hat{\rho}}{\partial z}dz \\ &=0\cdot d\rho+(-\hat{x}\sin\phi+\hat{y}\cos\phi)\,d\phi+0\cdot dz \\ &=\hat{\phi}\,d\phi \\ \end{align} And:
\begin{align} d\hat{z}&=\frac{\partial\hat{z}}{\partial\rho}d\rho+\frac{\partial\hat{z}}{\partial\phi}d\phi+\frac{\partial\hat{z}}{\partial z}dz \\ &=0\cdot d\rho+0\cdot d\phi+ 0\cdot dz \\ &=0 \\ \end{align}

So:
\begin{align} \hat{\rho}\,d\rho+\rho \,d\hat{\rho}+\hat{z}\,dz+z\,d\hat{z}&=\hat{\rho}\,d\rho+\rho(\frac{\partial\hat{\rho}}{\partial\rho}d\rho+\frac{\partial\hat{\rho}}{\partial\phi}d\phi+\frac{\partial\hat{\rho}}{\partial z}dz)+\hat{z}\,dz+z(\frac{\partial\hat{z}}{\partial\rho}d\rho+\frac{\partial\hat{z}}{\partial\phi}d\phi+\frac{\partial\hat{z}}{\partial z}dz) \\ &=\hat{\rho}\,d\rho+\rho(\hat{\phi}\,d\phi)+\hat{z}\,dz+\hat{z}\cdot0 \\ &=\hat{\rho}\,d\rho+\hat{\phi}\,\rho\,d\phi+\hat{z}\,dz \\ \end{align} Which is eqal to the summation of the differential displacement vectors calculated above.