Are directional derivatives a scalar or vector?
Typically, we think of the directional derivative of a scalarvalued 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 vectorvalued 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.
Fadel Hossam Manour
Updated on August 11, 2020Comments

Fadel Hossam Manour over 2 years
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.

Elchanan Solomon over 2 yearsIt'll give you a scalar if the function is scalarvalued, and a vector is the function is vectorvalued.

Anindya Prithvi over 2 yearsAs far as i can think of you're interested in $\frac{\partial U}{\partial x}= F_x$ ?

Fadel Hossam Manour over 2 yearsElchanan Solomon how can the directional derivative be a vector when it is the dot product of of the grad and a vector? The dot product of two vectors is always a scalar

J.G. over 2 years@FadelHossamManour The grad acts on a scalar. The result is a vector. Its components are neither scalars nor vectors; they lack indices, as scalars do, but unlike true scalars they're not invariant under orthogonal transformations of Cartesian coordinates.
