Differential of a vector which is a function of a vector and a scalar variable

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Consider a vector $\vec{\xi}=\vec{X}(\vec{x},t).$ Because $\vec x$ has 3 components x, y,z then we can say that: $\vec \xi$ has 3 components u, v, w and u = u(x, y, z, t) ; v = v(x, y, z, t) ; w = w(x, y, z, t).

Fix t at some value, in a paper I am reading, the author wrote:

$$\vec{\xi}+d\vec{\xi} = \vec{X}(\vec{x}+d\vec{x},t)=\vec{X}(\vec{x},t) + d\vec{x}\cdot\nabla \vec{X}.$$

Here $\nabla$ differentiate with respect to $\vec x$. Hence, $$d\vec{\xi} = d\vec{x}\cdot\nabla X = \sum_{i=1}^3 dx_i \frac{\partial X}{\partial x_i}=dx_i \frac{\partial X}{\partial x_i},$$

where the last form uses the convention that there is an implicit summation on repeated indices, a convention which greatly reduces clutter!


I don't understand the part: $$d\vec{\xi} = d\vec{x}\cdot\nabla X = \sum_{i=1}^3 dx_i \frac{\partial X}{\partial x_i}=dx_i \frac{\partial X}{\partial x_i},$$


Here is what I understand if I have to derive $d\vec{\xi}$:

$d\vec{\xi}$ = du.$\vec i$+ dv.$\vec j$+ dw.$\vec k$ , with:

$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy + \frac{\partial u}{\partial z}dz = \nabla u.d\vec x$ (dt = 0 because t fixed at some value)

$dv =...= \nabla v.d\vec x$

$dw =...= \nabla w.d\vec x$

So we have: $d\vec{\xi}$ = ($\nabla u.d\vec x$).$\vec i$+ ($\nabla v.d\vec x$).$\vec j$+ ($\nabla w.d\vec x$).$\vec k$ (Let me know if I was wrong !!)

I know some about the directional derivative but it still does not help me to see the common of my thought with the author's thought

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@Chapper:

My equation is:

$d\vec{\xi}$ = ($\nabla u.d\vec x$).$\vec i$+ ($\nabla v.d\vec x$).$\vec j$+ ($\nabla w.d\vec x$).$\vec k$

$$= (\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy + \frac{\partial u}{\partial z}dz).\vec i+(\frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy + \frac{\partial v}{\partial z}dz).\vec j+(\frac{\partial w}{\partial x}dx + \frac{\partial w}{\partial y}dy + \frac{\partial w}{\partial z}dz).\vec k $$

$$ = dx.(\frac{\partial u}{\partial x}.\vec i+\frac{\partial v}{\partial x}.\vec j+\frac{\partial w}{\partial x}.\vec k)+dy.(\frac{\partial u}{\partial y}.\vec i+\frac{\partial v}{\partial y}.\vec j+\frac{\partial w}{\partial y}.\vec k)+dz.(\frac{\partial u}{\partial z}.\vec i+\frac{\partial v}{\partial z}.\vec j+\frac{\partial w}{\partial z}.\vec k)$$

$$= dx.\frac{\partial \vec X}{\partial x}+dy.\frac{\partial \vec X}{\partial y}+dz.\frac{\partial \vec X}{\partial z}.$$

$\frac{\partial \vec X}{\partial x}$ is the directional derivative of vector $\vec X$ in the $\vec {Ox}$ direction. $\frac{\partial \vec X}{\partial x}$ is vector with 3 components, each component is the directional derivative of the corresponding scalar component of vector $\vec X$ in the $\vec {Ox}$ direction. $\frac{\partial \vec X}{\partial y}$ and $\frac{\partial \vec X}{\partial z}$ are similar to $\frac{\partial \vec X}{\partial x}$