# Directional derivative of a vector

Suppose $\vec X = (u, v, w)$ with u = u (x, y, z) ; v = v (x, y, z) ; w = w (x, y, z). And a fixed vector $\vec Y =(a, b, c)$ ; a, b, c are constant.

I am just informed that: $(\vec{Y} \cdot \nabla) \vec{X}$ is the directional derivative of $\vec X$ in the direction $\vec Y$. But I don't understand this information.

What I think is:

$$(\vec{Y} \cdot \nabla) \vec{X} = \left (a\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}+c\frac{\partial}{\partial z}\right )\vec X = \left (a\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}+c\frac{\partial}{\partial z}\right )(u\vec i+v\vec j+w\vec k)$$

$$=\left (a\frac{\partial u}{\partial x}+b\frac{\partial u}{\partial y}+c\frac{\partial u}{\partial z}\right )\vec i+\left (a\frac{\partial v}{\partial x}+b\frac{\partial v}{\partial y}+c\frac{\partial v}{\partial z}\right )\vec j+\left (a\frac{\partial w}{\partial x}+b\frac{\partial w}{\partial y}+c\frac{\partial w}{\partial z}\right )\vec k \space \space (*)$$

And if I have to derive the directional derivative of $\vec X$ in the direction $\vec Y$ by myself, thing will be (actually this is what I learned from school): $$\frac{\partial \vec X}{\partial \vec Y} = \frac{\partial u}{\partial \vec Y}\space \vec i+\frac{\partial v}{\partial \vec Y}\space \vec j+\frac{\partial w}{\partial \vec Y}\space \vec k$$

$$=\left (cos \alpha\frac{\partial u}{\partial x}+cos \beta\frac{\partial u}{\partial y}+cos \gamma\frac{\partial u}{\partial z}\right )\vec i+\left (cos \alpha\frac{\partial v}{\partial x}+cos \beta\frac{\partial v}{\partial y}+cos \gamma\frac{\partial v}{\partial z}\right )\vec j+\left (cos \alpha\frac{\partial w}{\partial x}+cos \beta\frac{\partial w}{\partial y}+cos \gamma\frac{\partial w}{\partial z}\right )\vec k \space \space (**)$$

with cos $\alpha$, cos $\beta$, cos $\gamma$ are the coordinates of a unit vector in the direction $\vec Y$. Clearly, $cos\alpha = a/\sqrt{a^2+b^2+c^2}$, $cos\beta = b/\sqrt{a^2+b^2+c^2}$, $cos\gamma = c/\sqrt{a^2+b^2+c^2}$.

Compare (*) and (**), you will understand my problem, which way of thinking is wrong ?

Consider a scalar function $f:\>{\mathbb R}^3\to{\mathbb R}$ for simplicity. Given a point ${\bf p}$ in the domain of $f$ there are two notions of directional derivatives of $f$ at ${\bf p}$ in use:
(i) Given a unit vector ${\bf u}=(u_1,u_2,u_3)\in S^2$, whereby $u_i=\cos\bigl(\angle({\bf u},{\bf e}_i)\bigr)$ one can define the directional derivative of $f$ at ${\bf p}$ in direction ${\bf u}$ as $$D_{\bf u}f({\bf p}):=\lim_{t\to0+}{f(p+t{\bf u})-f(p)\over t}=\nabla f({\bf p})\cdot{\bf u}\ .$$ Here the onesided limit may exist even in cases where $f$ is not differentiable at ${\bf p}$, e.g. for $f({\bf x}):=|{\bf x}|$ at the origin. This directional derivative corresponds to the formula you remember from high school (your second formula). But I'd never write a vector in the denominator of anything, even a mere "symbol".
(ii) Assume that $f$ is indeed differentiable at ${\bf p}$. Then, given any tangent vector ${\bf X}$ at ${\bf p}$, one defines the directional derivative of $f$ at ${\bf p}$ in direction ${\bf X}$ as $$D_{\bf X}f({\bf p}):=\lim_{t\to0}{f({\bf p}+t{\bf X})-f({\bf p})\over t}=\nabla f({\bf p})\cdot{\bf X}\ .\tag{1}$$ The last equality sign results from the chain rule. Here you can of course replace the RHS by ${\bf X}\cdot \nabla f({\bf p})$, which is your first formula. This writing of the directional derivative conveys the interpretation that ${\bf X}$ somehow "acts" on the function $f$, whereas $(1)$ conveys the interpretation that we want to know the increment of $f$ as a function of the variable increment ${\bf X}$.
In the second ("more modern") version (ii) of the directional derivative the vector ${\bf X}$ does not only carry directional information, as the ${\bf u}$ in (i) did, but is a fullfledged vector having direction and length.
• so the 2 formulas in my question are both true ? In my question, the directional derivative in second version is equal the directional derivative in first version dividing by the length of $\vec Y$, is that ok ?? so how to know which version people choose when they show me a value of a directional derivative? – Dat Sep 10 '17 at 10:39
• The people "informing" you about $Y\cdot \nabla X$ had my explanation (ii) in mind. – Christian Blatter Sep 10 '17 at 12:18