littlemathquark said:
Yes, and as in your example where ##f=(f_1)## has only one component, this matrix becomes a vector, and in this sense
$$
\nabla_a f = \left.Df\right|_a
$$
But let's be careful. Consider
$$
\left. \dfrac{d}{dx}\right|_{x=a}f(x)
$$
This is a simple derivative, but what are the variables? We technically have three possibilities:
\begin{align*}
a&\longmapsto f'(a)=\left. \dfrac{d}{dx}\right|_{x=a}f(x)\quad \text{location, result: a vector of partial derivatives}\\[6pt]
\mathbb{R}^n&\longrightarrow \mathbb{R}^n\\[6pt]
f'(a)&=\left.Df(x)\right|_a=\left.D\right|_af(x)=D_a f(x)=\nabla_a(f(x))\\[6pt]
&\text{It is often written as}\\[6pt]
f'(x)&=Df(x)=(Df)(x)=\nabla(f(x))\\[6pt]
&\text{neglecting the location}\\[18pt]
v&\longmapsto f'(a)\cdot v=\left(\left. \dfrac{d}{dx}\right|_{x=a}f(x)\right)\cdot v\quad \text{direction, result: a number, product of }f'(a)\text{ and }v\\[6pt]
\mathbb{R}^n&\longrightarrow \mathbb{R}\\[6pt]
f'(x)\cdot v&=Df(x)\cdot v=\nabla(f(x))\cdot v\\[18pt]
f&\longmapsto f'=\dfrac{df}{dx}\quad \text{differentiation, result: a function of a function space}\\[6pt]
\mathcal{C}^\infty(\mathbb{R}^n) &\longrightarrow \mathcal{C}^{\infty }(\mathbb{R}^n)\\[6pt]
f'&=Df =\nabla f
\end{align*}
A notation like ##f'(x)## does not distinguish between those meanings but they are important if you want to know the domain and the range of them.