DirectionalDiff(A, B_) computes the directional derivative of $A$ in the direction of $\mathrm{B\_}$, that is, the scalar product of a unitary vector in the direction of $\mathrm{B\_}$ times Nabla - the differential operator - applied to the function A. Two cases can happen:

$A$ is not a vector. Hence $\mathrm{DirectionalDiff}\left(A\,\mathrm{B\_}\right)=\left(\frac{\mathrm{B\_}}{\mathrm{Norm}\left(\mathrm{B\_}\right)}\right)·\nabla \left(A\right)$

$\mathrm{A\_}$ is a vector. Hence $\mathrm{DirectionalDiff}\left(\mathrm{A\_}\,\mathrm{B\_}\right)=\frac{\left(\mathrm{B\_}·\nabla \right)·\mathrm{A\_}}{\mathrm{Norm}\left(\mathrm{B\_}\right)}$

The %DirectionalDiff is the inert form of DirectionalDiff, that is: it represents the same mathematical operation while holding the operation unperformed. To activate the operation use value.

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

$\left[\mathrm{\&x}\,\mathrm{`+`}\,\mathrm{`.`}\,\mathrm{Assume}\,\mathrm{ChangeBasis}\,\mathrm{ChangeCoordinates}\,\mathrm{CompactDisplay}\,\mathrm{Component}\,\mathrm{Curl}\,\mathrm{DirectionalDiff}\,\mathrm{Divergence}\,\mathrm{Gradient}\,\mathrm{Identify}\,\mathrm{Laplacian}\,\nabla \,\mathrm{Norm}\,\mathrm{ParametrizeCurve}\,\mathrm{ParametrizeSurface}\,\mathrm{ParametrizeVolume}\,\mathrm{Setup}\,\mathrm{Simplify}\,\mathrm{`^`}\,\mathrm{diff}\,\mathrm{int}\right]$

$\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$

$\left[\mathrm{mathematicalnotation}=\mathrm{true}\right]$

$\mathrm{DirectionalDiff}\left(a\left(x\,y\,z\right)\,\mathrm{\_i}\right)=\left(\mathrm{\_i}·\nabla \right)\left(a\left(x\,y\,z\right)\right)$

$\frac{\partial }{\partial x}a\left(x\,y\,z\right)=\frac{\partial }{\partial x}a\left(x\,y\,z\right)$

$\mathrm{DirectionalDiff}\left(a\left(r\,\mathrm{\theta }\,\mathrm{\phi }\right)\,\mathrm{\_\θ}\right)=\left(\mathrm{\_\θ}·\nabla \right)\left(a\left(r\,\mathrm{\theta }\,\mathrm{\phi }\right)\right)$

$\frac{\frac{\partial }{\partial \mathrm{\theta }}a\left(r\,\mathrm{\theta }\,\mathrm{\phi }\right)}{r}=\frac{\frac{\partial }{\partial \mathrm{\theta }}a\left(r\,\mathrm{\theta }\,\mathrm{\phi }\right)}{r}$

$R≔a\left(x\,y\,z\right)\mathrm{\_i}+b\left(x\,y\,z\right)\mathrm{\_j}+c\left(x\,y\,z\right)\mathrm{\_k}$

$R≔a\left(x\,y\,z\right)\stackrel{\wedge }{i}+b\left(x\,y\,z\right)\stackrel{\wedge }{j}+c\left(x\,y\,z\right)\stackrel{\wedge }{k}$

$\mathrm{DirectionalDiff}\left(R\,\mathrm{\_i}\right)=\left(\mathrm{\_i}·\nabla \right)\left(R\right)$

$\left(\frac{\partial }{\partial x}a\left(x\,y\,z\right)\right)\stackrel{\wedge }{i}+\left(\frac{\partial }{\partial x}b\left(x\,y\,z\right)\right)\stackrel{\wedge }{j}+\left(\frac{\partial }{\partial x}c\left(x\,y\,z\right)\right)\stackrel{\wedge }{k}=\left(\frac{\partial }{\partial x}a\left(x\,y\,z\right)\right)\stackrel{\wedge }{i}+\left(\frac{\partial }{\partial x}b\left(x\,y\,z\right)\right)\stackrel{\wedge }{j}+\left(\frac{\partial }{\partial x}c\left(x\,y\,z\right)\right)\stackrel{\wedge }{k}$

$\mathrm{DirectionalDiff}\left(a\left(x\,y\,z\right)\,\mathrm{\_r}\right)$

Error, (in Physics:-Vectors:-DirectionalDiff) vectors must be projected over one and the same base

$\mathrm{DirectionalDiff}\left(a\left(x\,y\,z\right)\,\mathrm{ChangeBasis}\left(\mathrm{\_r}\,1\right)\right)$

$\frac{\left(\frac{\partial }{\partial x}a\left(x\,y\,z\right)\right)\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\phi }\right)+\left(\frac{\partial }{\partial y}a\left(x\,y\,z\right)\right)\sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\phi }\right)+\left(\frac{\partial }{\partial z}a\left(x\,y\,z\right)\right)\cos \left(\mathrm{\theta }\right)}{\sqrt{{\cos \left(\mathrm{\phi }\right)}^2{\sin \left(\mathrm{\theta }\right)}^2+{\sin \left(\mathrm{\phi }\right)}^2{\sin \left(\mathrm{\theta }\right)}^2+{\cos \left(\mathrm{\theta }\right)}^2}}$