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If I have a function like f(x,y,z) = xyz + 1, it´s easy to calculate the directional derivative by ∇f•v.

But if we have a vector f(x,y,z)= (2xy, xz², 3y²z), ¿how we can calculate his directional derivative in the direction v(1,4,4) in the point p(1,1,1)?

Thankyou very much!

Laxus D
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    These don't really have a "directional derivative" in that sense, we usually use the term for scalar functions. You can consider the directional derivative of each component, in which case you end up with $3$ directional derivative vectors, that you can combine (and transpose) to get the Jacobian (or total derivative). – Osama Ghani May 28 '20 at 00:06

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I'll clear a few things for you first. If $f: \Bbb R^m \to \Bbb R^n$ and $v \in \Bbb R^m$, then the directional derivative of $f$ at a point $a \in \Bbb R^m$ along the vector $v$ is defined as the following limit (if it exists of course)

\begin{equation} f'(a;v) = \lim_{t \to 0}\frac{f(a+tv)-f(a)}{t} \end{equation}

If the function $f$ was differentiable to begin with, then using the above definition it is easy to see that

\begin{equation} f'(a;v) = \begin{bmatrix} \nabla f_1 \cdot v\\ \nabla f_2 \cdot v \\ \dots \\ \nabla f_n \cdot v \end{bmatrix} \end{equation}

where $f = (f_1,\dots f_n)$ and $f_i: \Bbb R^m \to \Bbb R$.

General Grievous
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