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The directional derivative of a scalar field $f(x,y,z)$ at a point $P$ along a vector $ \mathbf v$ is the dot product of the gradient (vector differential operator):

$$\begin{align} \nabla &= \begin{bmatrix}\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\end{bmatrix}^\top\\[2ex] &=\frac{\partial}{\partial x}\hat i+ \frac{\partial}{\partial y}\hat j + \frac{\partial}{\partial z}\hat k \end{align}$$

applied to the function $f$

$$\nabla f = \begin{bmatrix}\frac{\partial}{\partial x}f,\frac{\partial}{\partial y}f,\frac{\partial}{\partial z}f\end{bmatrix}^\top$$

and the vector $\mathbf v$:

$$D_{\mathbf v}f=\langle \nabla f, \mathbf v\rangle$$


The divergence is the dot product of the gradient operator $\nabla$ with a vector field $\mathbf F=\left(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)\right):$

$$\mathrm{div} \mathbf F=\nabla\mathbf F =\langle\nabla, \mathbf F\rangle$$


In English, the directional derivative compares a given vector to the direction of maximum steepness along a surface and at a given point, providing an idea of the change of the surface along the particular direction indicated by the vector.

The divergence quantifies the relative change in the vector field at a point.

Here are some parallelisms and differences:

\begin{array}{|c|c|c|c|} \hline &\small\text{Dir. Derivative} & \small\text{Divergence} \\ \hline \text{input}& \small\text{point & vector} &\small\text{point } \\ \hline \text{output} & \small\text{scalar} & \small\text{scalar}\\ \hline \text{defined on...} & \small\text{scalar field} &\small\text{vector field} \\ \hline \text{operation} & \small\text{dot prod} &\small\text{dot prod} \\ \hline \end{array}

Is there a way of getting an intuition in English of their similarities in concept or intention?

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    $\begingroup$ You have $\operatorname{div}{\mathbf{F}} = \sum_i \langle \mathbf{e}_i ,D_{\mathbf{e}_i} \mathbf{F} \rangle $ for any orthonormal basis $\{\mathbf{e}_i\}_i$: it's like a trace of the total derivative. $\endgroup$
    – Chappers
    Commented May 4, 2018 at 16:23

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When you want to find intuition for the divergence, I suggested considering physical situations. If one uses a vector field to describe the flow of a fluid. Then the divergence describes the sources and sinks of this fluid. On can get the flux through an closed surface (i.e. quantize the source or sink) by integrating the divergence.

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  • $\begingroup$ Yes, the intuition for divergence is clear. The question is about the similarity of the two separate concepts of directional derivative and divergence. $\endgroup$ Commented May 4, 2018 at 16:18

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