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$\mathrm{with}\left(\mathrm{Physics}\left[\mathrm{Vectors}\right]\right)$

$\left[{\mathrm{\&x}}{\,}{\mathrm{`+`}}{\,}{\mathrm{`.`}}{\,}{\mathrm{ChangeBasis}}{\,}{\mathrm{ChangeCoordinates}}{\,}{\mathrm{Component}}{\,}{\mathrm{Curl}}{\,}{\mathrm{DirectionalDiff}}{\,}{\mathrm{Divergence}}{\,}{\mathrm{Gradient}}{\,}{\mathrm{Identify}}{\,}{\mathrm{Laplacian}}{\,}{\nabla}{\,}{\mathrm{Norm}}{\,}{\mathrm{Setup}}{\,}{\mathrm{diff}}\right]$
 (1) 
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$\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$

$\left[{\mathrm{mathematicalnotation}}{=}{\mathrm{true}}\right]$
 (2) 
The radial cylindrical unit vector
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$R\u2254\mathrm{\_\ρ}$

${R}{\u2254}\stackrel{{\wedge}}{{\mathrm{\rho}}}$
 (3) 
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$\mathrm{diff}\left(R\,x\right)$

${}\frac{{y}{}\stackrel{{\wedge}}{{\mathrm{\phi}}}}{{{x}}^{{2}}{+}{{y}}^{{2}}}$
 (4) 
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$\mathrm{diff}\left(R\,y\right)$

$\frac{{x}{}\stackrel{{\wedge}}{{\mathrm{\phi}}}}{{{x}}^{{2}}{+}{{y}}^{{2}}}$
 (5) 
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$\mathrm{diff}\left(R\,z\right)$

Note the difference when you change the order in which derivatives are computed in a 2nd order derivative
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$\mathrm{diff}\left(x\,x\,\mathrm{\rho}\right)$

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$\mathrm{simplify}\left(\mathrm{diff}\left(x\,\mathrm{\rho}\,x\right)\right)$

${\frac{{{y}}^{{2}}}{{\left({{x}}^{{2}}{+}{{y}}^{{2}}\right)}^{\raisebox{1ex}{${3}$}\!\left/ \!\raisebox{1ex}{${2}$}\right.}}}}$
 (8) 
Curvilinear coordinates and related unit vectors can have functional dependency, and so can the differentiation variable. Consider for instance the radial unit vector $\stackrel{\wedge}{\mathrm{\rho}}$ in cylindrical coordinates as a function of the polar angle $\mathrm{\phi}$ which in turn is a function of $t$
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$\mathrm{\_\ρ}\left(\mathrm{\phi}\left(t\right)\right)$

$\stackrel{{\wedge}}{{\mathrm{\rho}}}{}\left({\mathrm{\phi}}{}\left({t}\right)\right)$
 (9) 
The derivative with respect to $\mathrm{\phi}\left(t\right)$ takes into account the geometrical dependency of $\stackrel{\wedge}{\mathrm{\rho}}$ with respect to $\mathrm{\phi}$, while keeping, in the result, the dependency with respect to $t$ of the derivand
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$\mathrm{diff}\left(\,\mathrm{\phi}\left(t\right)\right)$

$\stackrel{{\wedge}}{{\mathrm{\phi}}}{}\left({t}\right)$
 (10) 
The derivative of $\stackrel{\wedge}{\mathrm{\rho}}\left(\mathrm{\phi}\left(t\right)\right)$ with respect to $t$ uses the chain rule taking the result above into account, hence
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$\mathrm{diff}\left(\,t\right)$

$\stackrel{{\wedge}}{{\mathrm{\phi}}}{}\left({t}\right){}\stackrel{{\mathbf{.}}}{{\phi}}{}\left({t}\right)$
 (11) 
You can turn OFF geometric differentiation using Setup
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$\mathrm{Setup}\left(\mathrm{geom}=\mathrm{false}\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}}\mathrm{geom}\mathrm{\text{'}\; against\; keyword\; \text{'}}\mathrm{geometricdifferentiation}\text{'}$
 
$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$
 
$\left[{\mathrm{geometricdifferentiation}}{=}{\mathrm{false}}\right]$
 (12) 
So now, for example, the differentiation above returns
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$\mathrm{diff}\left(\,t\right)$

${\mathrm{D}}{}\left(\stackrel{{\wedge}}{{\mathrm{\rho}}}\right){}\left({\mathrm{\phi}}{}\left({t}\right)\right){}\stackrel{{\mathbf{.}}}{{\phi}}{}\left({t}\right)$
 (13) 
and differentiating $\stackrel{\wedge}{\mathrm{\rho}}$ with respect to x returns 0 instead of eq.(4)
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$\mathrm{diff}\left(R\,x\right)$

You can still use geometric differentiation by passing the optional argument geometricdifferentiation
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$\mathrm{diff}\left(R,x,\mathrm{geometricdifferentiation}\right)$

${}\frac{{y}{}\stackrel{{\wedge}}{{\mathrm{\phi}}}}{{{x}}^{{2}}{+}{{y}}^{{2}}}$
 (15) 