# Are directional derivatives a scalar or vector?

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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.

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Updated on August 11, 2020

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.
As far as i can think of you're interested in $-\frac{\partial U}{\partial x}= F_x$ ?
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