The name directional suggests they are vector functions. However, since a directional derivative is the dot product of the gradient and a vector it has to be a scalar. But, in my textbook, I see the special case of the directional derivatives $F_x(x,y,z)$ and $F_y(x,y,z)$ being treated as vectors. I want a clarification for this.
[Math] Are directional derivatives a scalar or vector
derivativespartial derivativevector-spacesvectors
Best Answer
Typically, we think of the directional derivative of a scalar-valued function $f:\mathbb R^n\to\mathbb R$ in a (unit) direction $v\in\mathbb R^n$ at a point $x\in\mathbb R^n$; this is just defined as $$\nabla f|_x \cdot v$$. At a particular point $x$, this is just a scalar; we can also view $\nabla f \cdot v$ more generally as another function $\mathbb R^n \to \mathbb R$, assigning to each point $x$ $f$’s derivative in direction $v$ at that point.
In the case that $f$ is a vector-valued function $f:\mathbb R^n\to\mathbb R^m$, we could define a sort of “vector directional derivative” (nonstandard vocabulary) $\mathbb R^n \to \mathbb R^m$ of $f$ at $x$ in direction $v$ by taking the directional derivative of each component function $f_i : \mathbb R^n \to \mathbb R$ and assembling these into a vector in $\mathbb R^m$. Equivalently, we could take the $m \times n$ Jacobian matrix $J(f)$ of $f$’s partial derivatives, whose $(i,j)$-th entry is $\frac{\partial f_i}{\partial x_j}$; then this “vector directional derivative” is simply equal to $J(f) \; v$, as a matrix product.