Jtdylktuy

Please forgive how simple this is, but I can't seem to find any explanations for how to differentiate single-variable equations of the following form:

$f(boldsymbol{x}) = 5boldsymbol{x}$, where $boldsymbol{x}$ is a $n$-dimensional column-vector of scalar values; i.e., $boldsymbol{x} = langle a_1, a_2, a_3, ... a_n rangle$ for $a_i in mathbb{R}$).

Also, if vector-valued functions are those that map a vector to a scalar (i.e. $mathbb{R}^n to mathbb{R}$), what are functions like this called?

The differential of a mapping $f: mathbb{R}^n rightarrow mathbb{R}^n$ at a point $p$, if it exists, is a linear transformation $df_p: mathbb{R}^n rightarrow mathbb{R}^n$ which best approximates the change in $f$ near $p$. In particular, the differential $df_p$ is implicitly defined by the Frechet quotient:
$$ lim_{h rightarrow 0} frac{f(p+h)-f(p)-df_p(h)}{| h |} = 0$$
For small $h$, $f(p+h) simeq f(p) + df_p(h)$. The relation of the differential and the partial derivatives more commonly taught in introductory calculus is given by the definition $frac{partial f}{partial x_i}(p) = df_p(e_i)$ where $(e_i)_j = delta_{ij}$ or equivalently $e_i cdot e_j = delta_{ij}$. Here I use $e_1,e_2,dots , e_n$ to denote the standard basis for $mathbb{R}^n$. Incidentally, this definition of partial derivatives equally well applies to a basis for some abstract finite dimensional normed linear space. That said, $| h | = sqrt{ h cdot h}$ is the length of $h$. Notice, we cannot divide by $h$ since generally division by a vector is not defined. Getting back to the main story,
$$ J_f(p) = [df_p] = [df_p(e_1)|df_p(e_2)| cdots | df_p(e_n)] = left[ frac{partial f}{partial x_1}(p)bigg{|}frac{partial f}{partial x_2}(p)bigg{|}cdots bigg{|}frac{partial f}{partial x_n}(p) right] $$
is the Jacobian matrix of $f$ at $p$. The relation between $df_p$ and $J_f(p)$ is given by matrix multiplication:
$$ df_p(h) = J_f(p)h $$
We can view the Jacobian as a stack of gradient vectors, one for each component function of $f = (f_1,f_2, dots , f_n)$; $nabla f_j = [partial_1 f_j, dots , partial_n f_j]^T$ and
$$ J_f = left[ begin{array}{c} (nabla f_1)^T \ (nabla f_2)^T \ vdots \ (nabla f_n)^T end{array}right] $$
Thus,
$$ df_p(h) = left[ begin{array}{c} (nabla f_1)^T \ (nabla f_1)^T \ vdots \ (nabla f_n)^T end{array}right]left[ begin{array}{c} h_1 \ h_2 \ vdots \ h_n end{array}right] = left[ begin{array}{c} (nabla f_1) cdot h \ (nabla f_2)cdot h \ vdots \ (nabla f_n) cdot hend{array}right]. $$
In fact, the derivative (differential) of $f$ involves many gradients at once working in concert as above. You see, the larger confusion here is the tendency for students to assume the derivative of a function on $mathbb{R}^n$ should be another function on $mathbb{R}^n$. It's not. The first derivative is naturally identified with the pointwise assignment of a linear map at each such point as the Frechet quotient exists. Then, it turns out the higher derivatives of a function on $mathbb{R}^n$ can be identified with the pointwise assignment of a completely symmetric $k$-linear mapping. These things are explained rather nicely in Volume 2 of Zorich's Mathematical Analysis. However, this material is standard in any higher course in multivariate analysis.

Getting back to your actual function $f(x) = 5x$, this function is linear so the best linear approximation to the function is essentially the function itself. We can calculate $J_f(p) = 5I_n$ where $I_n$ is the $n times n$ identity matrix. Or, if you prefer, $df_p(h) = 5I_nh = 5h$ for each $p in mathbb{R}^n$.

Actually, a multivariate derivation is called gradient. For a function $f(x_1,x_2,cdots ,x_n):Bbb R^nto Bbb R$ we define a gradient vector as following:

A gradient vector contains $n$ components namely $g_i$ , $i=1,2,cdots ,n$. We therefore define $$g_i={partial f(x_1,cdots , x_i,cdots , x_n)over partial x_i}$$as if other $x_j$s are constant. Then the gradient vector would be$$nabla f=[g_1 g_2 cdots g_n]$$for a function $f:Bbb R^nto Bbb R^n$ we define a gradient matrix instead whose entries (namely $g_{ij}$) are:$$g_{ij}={partial f_i(x_1,cdots , x_n)over partial x_j}$$where$$f=[f_1 f_2 cdots f_n]$$In this question, the gradient matrix will become$$nabla f=5I_n$$where $I_n$ is the identity matrix of order $n$ (why?).

P.S. for higher dimension input and/or output functions, the gradient should be defined using tensors.